2018-G12-Math-E

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.PunjabExample: Find dy and dy of the function deined as f(x) = x2 , when x = 2 and dx = 0.01Solution: As f(x) = x2, so f ‘ (x) = 2x (a dx = dx) dy = f(x + dx) - f(x) = (x + dx)2 - x2 = 2x dx + (dx)2 = 2x dx + (dx)2 Thus f(2 + 0.01) - f(2) = 2(2) (0.01) + (0.01)2 = 0.04 + 0.0001 = 0.0401, that is dy = 0.0401 when x = 2 and dx = dx = 0.01 Also dy = f ‘ (x) dx = 2(2) x (0.01) = 0.04 (a f ‘ (x) = 2x, x = 2 and dx = 0.01) Thus dy - dy = 0.0401 - 0.04 = 0.0001.3.1.3 Finding dy by using differentials dx We explain the process in the following example.Example: Using diferentials ind dy when y - In x = Inc dx xSolution: Finding diferentials of both sides of the given equation, we get d  y - lnx= d [lnc]= 0 x using d(f ± g) = df ± dg, we have d  y  - d ( ln x⇒) =0 ddx  y-. 1  1=. dx 0 x x x Using d(fg) = fdg + gdf, we get yd  1  + 1 dy - 1 dx =0 x x x y ×  - 1 dx  + 1 dy - 1 dx = 0 ⇒ 1 dy = 1 dx + y dx x2 x x x x x2 version: 1.1 4

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab or 1 dy =+ 1x y =dx  x+ y =dx 1  x + y  dx x x2 x2 x x ⇒ dy = x + y  dx x Thus dy = x + y dy = f ' ( x)dx dx x3.1.4 Simple application of differentials Use of diferentials for approximation is explained in the following examples.Example 1: Use diferentials to approximate the value of 17 .Solution: Let f(x) = x Then f (x + dx) = x + d x As the nearest perfect square root to 17 is 16, so we take x = 16 and dx = dx = 1 Then y = f(16) = 16 = 4 Using f (x + dx) c f (x) + dy c f(x) + f ‘ (x) dx. we have f (16 +1) ≈ f (16) + 1 × (1)  f ' ( x ) =1 x  2 2 16 ≈ 4+ 1 = 4+ 1 = 4.125 2×4 8 Hence 17 ≈ 4.125Example 2: Use diferentials to approximate the value of 3 8.6Solution: Let f (x) = 3 x theny + d y = f ( x + d x) = 3 x + d x = 3=x=+ dx (d x dx) and f ' ( x) 1 2 3x3 version: 1.1 5

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.PunjabAs the nearest perfect cube root to 8.6 is 8, so we take x = 8and dx = 0.6, then f (=8) =3 8 2 =and f '=(8) 3=(18) 23 1 1, 3× 4 12so dy =f ' ( x)dx =1 × (0.6) =0.05 12Using f ( x + d x) = f ( x) + dy, we have f (8 + 0.6) = f (8) + 0.05 +=2 0.=05 2.05But using calculator, we ind that 3 8.6 is approximately equal to 2.0488.Example 3: Using diferentials, ind the approximate value of sin 460Solution: Let y = sinx, then (dx = dx) y + dy = sin (x + dx) = sin (x + dx) We take x = 450 = p and dx = 10 =0.01745 4 Hence dy = cos x dx  d ( sin x) = cos x  dx c (cos 45°)(0.01745) =1 (0.01745) 2 c 0.7071 (0.01745) c 0.01234 Using f (x + dx) c f(x) + dy we have sin (460) c sin 45° + dy c 0.7071 + 0.01234 = 0.71944 c 0.7194Using calculator, we ind sin 460 is approximately equal to 0.71934.Example 4: The side of a cube is measured to be 20 cm with a maximum error of 12 cmin its measurement. Find the maximum error in the calculated volume of the cube. version: 1.1 6

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.PunjabSolution: Let x be the side and V be the volume of the cube, then V = x3 and dV = (3x2) dx Taking x = 20 (cm) and dx = 0.12 (cm), we get dV = [3(20)2] (0.12) = 1200 x (0.12) = 144 (cubic cm) The error 144 cubic cm in volume calculation of a cube is either positive or negative. EXERCISE 3.11. Find dy and dy in the following cases: (i) y = x2 - 1 when x changes from 3 to 3.02 (ii) y = x2 + 2x when x changes from 2 to 1.8 (iii) y = x when x changes from 4 to 4.412. Using diferentials ind dy and dx in the following equations dx dy (i) xy + x = 4 (ii) x2 + 2y2 = 16 (iii) x4 + y2 = xy2 (iv) xy - lnx = c3. Use diferentials to approximate the values of (i) 4 17 (ii) (31)1/5 (iii) cos 290 (iv) sin 6104. Find the approximate increase in the volume of a cube if the length of its each edge changes from 5 to 5.02.5. Find the approximate increase in the area of a circular disc if its diameter is ?3.2 INTEGRATION AS ANTI - DERIVATIVE (INVERSE OF DERIVATIVE) In chapter 2, we have been inding the derived function (diferentialcoeicient) of a given function. Now we consider the reverse (or inverse) processi.e. we ind a function when its derivative is known. In other words we can say that iff’(x) = f(x), then f(x) is called an anti-derivative or an integral of f(x). For example, ananti-derivative of f(x) = 3x2 is f(x) = x3 because f’(x) = d (x3) = 3x2 = f(x). dx version: 1.1 7

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab The inverse process of diferentiation i.e. the process of inding such a function whosederivative is given is called anti-diferentiation or integration. While inding the derivatives of the expressions such as x2 + x, x2 + x + 5, x2 + x - 3etc., we see that the derivative of each of them is 2x + 1, that is, d (x2 + x) = d (x2 + x + 5) = d (x2 + x - 3) = 2x + 1 dx dx dx Now if f(x) = 2x + 1 (i) Then f(x) = x2 + x is not only anti-derivative of (i). But all anti-derivatives of f(x) = 2x + 1 are included inx2 + x + c where c is the arbitrary constant which can be found if further information is given. As c is not deinite, so f(x) + c is called the indeinite integral of f(x) , that is, ∫ f (x) dx = Φ (x) + c (ii) In (ii), f(x) is called integrand and c is named as the constant of integration.The symbol ∫ .... dx indicates that integrand is to be integrated w.r.t. x.Note that d and ∫ .... dx are inverse operations of each other. dx3.2.1 Some Standard Formulae for Anti-Derivatives We give below a list of standard formulae for anti-derivatives which can be obtainedfrom the corresponding formulae for derivatives: General Form Simple FormIn formulae 1-7 and 10-14, a ≠ 01. ∫(ax=+ b)n dx (ax + )b n+1 + c ,( n ≠ -1) ∫ x=ndx x n+1 + c ( n ≠ -1) a(n + 1) n +12. ∫ sin ( ax + b) dx =- 1 cos ( ax + b) + c ∫ sin x-dx =c+os x c a ∫ cos xdx =sin x + c ∫ sec2 xdx =ta+n x c3. ∫ cos(ax + b=)dx 1 sin ( ax + b ) + c a4. ∫ sec2 (ax + b=)dx 1 tan(ax + b) + c a version: 1.1 8

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab5. ∫ cosec2 (ax + b)dx =- 1 cot (ax + b) + c ∫ cosec2 x-dx =c+ot x c ∫ sec xtan xdx =+ sec x c a ∫ cosecx cot-x dx =c+osec x c6. ∫ sec(ax + b)tan(ax + b=)dx 1 sec(ax + b) + c ∫ exd=x ex + c a ∫ axd=x 1 .ax + c.(a〉0,a ≠ 1)7. ∫ cosec ( ax + b ) cot ( ax + b ) dx =- 1 cosec ( ax + b) + c ln a a ∫ 1 dx = ln x + c,x ≠ 08. ∫ eλx+µdx =λ1 × eλx+µ + c (λ ≠ 0) x9. ∫ a=λx+µdx λ 1 .aλx+µ + c.( a〉 0,a ≠ 1,λ ≠ 0) ∫ tan xdx= ln sec( x) + c ln a =l+n cos x c10. ∫ 1 b dx = ∫ (ax + b)-1 dx ∫ cotx dx = ln sinx + c ax + ∫ sec xdx= ln sec x + tan x + c = 1 ln ax + b + c ,( ax + b ≠ 0) ∫ cosec x=dx ln cosec x - cot x + c a11. ∫ tan(ax=+ b)dx 1 ln sec ( ax + b) + c a -=1 ln cos(+ax b+) c - a12. ∫ cot (ax =+ b)dx 1 ln sin ( ax + b) + c a13. ∫ sec(ax=+ b)dx 1 ln sec ( ax + b) + tan ( ax + b) + c a14. ∫ cos ec(=ax + b)dx 1 ln co sec(ax + b) - cot (ax + b) + c a These formulae can be veriied by showing that the derivative of the right hand side ofeach with respect to x is equal to the corresponding integrand.Examples: x5+1 x6 + c ( )  =16 x6  ∫1. x5 dx = 5+1 + c = 6 d 1 d=x x6 1=.6 x 6-1 x5 dx 6 6 =1 dx =x - 32 dx - 3 +1  d  -2  = d ( ) 1  x3 dx x dx 2∫ ∫2. x2 -2 x -3 +1 2 version: 1.1 9

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab -1 =-2.  - 1 -1 -3 =x 2 2 +c 1 = 1 - +c =- x - 2 x2 =x 2 x3 1 2    3. ∫ ( 2 x 1 3)4 d∫x =(+2 x )3 -4 dx d - 6( 1 3)3 + dx 2x + = (22x(-+43+)=-14)+1 + c (2x + )3 -3 -+ c ( )=1+ d (2x )3 -3 6 dx -6 - =6+(2x1+ 3)3 c =- 1 ( -3) ( 2 x + )3 -3-1 ( 2 ) =( 2 x 1 3)4 6 +4. ∫ cos 2x=dx sin 2x =+ c 1 sin 2x + c  d  1 sin 2x  = 1 d ( sin 2x)  22 dx 2 2 dx = 1 (cos 2=x) × 2 cos 2x 25. ∫ sin3 xdx =-cos3x + c =- 1 cos3x + c  d  - 1 c-os 3x  =13 ddx (cos 3x)  33 dx 36. ∫ cos ec2x-dx =c+ot x c  d (- cot x) = -(- co sec2 x) = co sec2 x  dx7. ∫ sec5xtan5xdx =+sec5 x c  d=dx  sec5=5 x   15×(sec5 xtan5 x) 5 sec5xtan5x 58. ∫ eax+bdx =eax+b + c ( )d  eaax+=b  eax+b  a dx 1 eax+b ×=a a 3λ x ( )== ddx  λ3lλnx3  1 3λx 9. ∫ 3λx dx = λ ln3 + c λ ln 3 3λx (ln3)λ10. ∫ 1 b dx∫ =(+ax b)-1 dx  d  1 ln ( =ax + b) 1 . a=x1+ b .a 1 b   ax + dx a a ax + = 1 ln ( ax + b) + c ,( ax + b > 0) a11. ∫ 1 dx = ln x + x2 + a2 + c ( ( ))d 1 1 + 1 2x  dx ln x+ =x2 + a2 x+ x2 + a2 x2 + a2 × ( )x2 + a2 2 version: 1.1 10

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab 1× x2 + a2 + x =x21+ a2  x2 + a2 x2 + a2 x+3.2.2 Theorems on Anti-DerivativesI. The integral of the product of a constant and a function is equal to the product of theconstant and the integral of the function.In symbols, ∫ af ( x)dx= a ∫ f ( x)dx where a is a constant.II. The integral of the sum (or diference) of two functions is equal to the sum (or diference)of their integrals.In symbols, ∫  f1 ( x) ± f2 ( x) dx = ∫ f1 ( x)dx ± ∫ f2 ( x)dx3.2.3 Anti-Derivatives of [f(x)]n f ’(x) and [f(x)]-1 f ’(x)  f ( x ) n+1 Prove that: (i) ∫  f ( x)=n f ' ( x)dx + n +1 c, (n ≠ -1) (ii) (f(x) > 0) ∫  f ( x)-1 f '=( x)dx ln f ( x) + c,Proof:  f ( )x n+1 + c1(i) Since d ([f(x)n+1) = (n + 1) [f(x)]n f ‘ (x) dx ∴ by deinition, ∫ (n +1)  f ( x)n f '=( x)dx (n +1) ∫  f ( x)n f ' ( x)dx =+ f ( x)n+1 c1 (by theorem I) ∫or [ f (x)]n f ' =(x)dx [ f (x)]n+1 + c =where c ≠nc+-1 1(n 1) n +1 version: 1.1 11

13.. IQntueagdrraatitoicn Equations eLearn.Punjab(ii) Since d [In f(x)] = 1 . f ‘ (x) eLearn.Punjab dx f(x) version: 1.1 By deinition, we have∫ 1 . f ' ( x)dx =ln+f ( x) c (>f ( x) 0) f (x)∫or [ f (x)]-1 f '=(x)dx I n f (x) + c.Thus we can prove that∫(i) =xndx xn+1 + c, (n ≠ -1) n +1 (a ≠ 0, n ≠ -1)(ii) ∫(ax + b)n=dx (ax + b)n+1 + c, (a ≠ 0) a(n +1)(iii) ∫=1x dx ln x + c(iv) ∫ 1 =dx 1 ln ax + b + c, ax + b aExamples: Evaluate(i) ∫( x +1)( x - 3)dx (ii) ∫ x x2 -1 dx(iii) ∫ x x 2 dx, ( x > -2) (iv) ( )∫1 dx, (x > 0) + x +1 x(v) ∫ dx , (x > 0) (vi) ∫ sinx + cos3x dx x +1- x cos2x sin x(vii) ∫ 3 - cos 2x dx, (cos 2x ≠ -1) 1 + cos 2xSolution: (By theorems I and II)(i) ∫( x +1)( x - 3) dx= ∫( x2 - 2x - 3) dx =∫ x2dx - 2∫ x dx - 3∫1 dx = x3 - 2. x2 - 3.x + c  ∫ xndx = xn+1 + c1 and 32 n +1 12

31.. IQntueagdraratitoicn Equations eLearn.Punjab = 1 x3 - x2 - 3x + c ∫1 d∫x =x0=dx x+0+1 c2  eLearn.Punjab 3 1 = 1 version: 1.1( )∫ ∫(ii) x x2 -1 dx = x2 -1 2 x dx =∫  f ( x) 12 f ' ( x) dx i (If×( f )(x=) x2 -1. = 1 ∫  f ( x ) 1 f '(x) then f ' ( x)= 2x⇒ x= 1 f ' ( x)) 2 2 2  f ( x) 3 3 2 3 x2 +1 2 + c. ( )= 1 +=c 1 2 3 2(iii) ∫ x dx = ∫ x + 2 - 2 dx, ( x > -2) + x + 2 x 2 = ∫ 1 - x 2 2  dx = ∫ dx - 2 ∫ ( x + 2)-1 .1 dx = x - 2 ln ( x + 2) + c + ∫ 1 dx =x1+1 . 1x d∫x (x 0) > x +1( )(iv) x = ∫  f ( x)-1 .2 f ' ( x) dx =f ' ( x) 1 =if f (+x) x1 2x ( )2∫  f ( x)-1 f ' ( x) dx or 1 2 f ' ( x) x = 2 ln f ( x) + c = 2 ln x +1 + c(v) ∫ dx , (x > 0) x +1- x Rationalizing the denominator, we have dx x +1+ x x +1- ( )( )∫ x =∫ dx x +1- x x +1+ x 13

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab x +1+ x (+x 1)+12 1  x +1- x ∫ ∫= dx= x2 dx ∫ ∫=+( x )1 d+x 1 12 x 2 dx = ( x + 1) 3 + 3 + c= 2(x + 1) 3 + 2 3 + 2 2 x2 x2 3 3 33 c 22∫(vi) sin x + cos3 x dx cos2x sin xSolution: ∫ ∫sincoxs=2+x scionsx3 x dx +  sin x x cos3 x x  dx cos2 x sin cos2 x sin = ∫  1 x + cos x  dx cos2 sin x = ∫ sec2 x dx + ∫ cot x dx =tan+x ln sin+x c(vii) ∫ 3 - cos 2x dx, (cos 2 x ≠ -1) 1 + cos 2xSolution: ∫ 13=+- ccooss22xx =∫ 4 -1 (+1 +co-sc2osx2x) dx ∫  1 + 4 2x 1 dx cos = ∫ 2 4 2 x dx - ∫1=dx ∫ 2 sec2x dx - ∫1 dx cos = 2tan x - x + c version: 1.1 14

31.. IQntueagdraratitoicn Equations eLearn.Punjab EXERCISE 3.2 eLearn.Punjab version: 1.11. Evaluate the following indeinite integrals(i) ∫(3x2 - 2x + 1) dx (ii) ∫ x + 1  dx, (x > 0) x( )(iii) ∫ x x + 1 dx, )1 ( x > 0) (iv) ∫(2x + dx 32( )(v) ∫ x + 1 2 dx, ( x > 0) (vi) ∫  x- 1 2 dx, (x > 0) x(vii) ∫ 3x + 2 dx, (x > 0) (viii) ∫ y(y + 1) dy , (y > 0) x y( )(ix) q - 12 ( )(x) - 2 q ∫ dq , (q > 0) ∫ 1 x (x > 0) x dx,∫(xi) e2x + ex dx ex2. Evaluate(i) ∫ dx x + b  x + a > 0  ∫(ii) 1 - x2 dx x+a + x + b > 0 1 + x2(iii) ∫ dx , ( x > 0,a > 0) (iv) ∫(a - 2 x ) 3 dx x +a + 2 x ( )∫ 1 + ex 3 (vi) ∫ sin(a + b) x dx(v) ex dx(vii) ∫ 1 - cos 2x dx, (1 - cos 2x > 0) (viii) ∫(ln x)× 1 dx , (x > 0) x(ix) ∫ sin2x dx (x) ∫1+ 1 dx,  - p < x < p  cos x 2 2(xi) ∫ ax2 ax +b c dx (xii) ∫ cos3x sin2x dx + 2bx +(xiii) ∫ cos 2x - 1 dx, (1 + cos 2x ≠ 0) (xiv) ∫tan2 x dx 1+ cos 2x 15

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab3.3 INTEGRATION BY METHOD OF SUBSTITUTIONSometimes it is possible to convert an integral into a standard form or to an easyintegral by a suitable change of a variable. Now we evaluate ∫ f ( x) dx by the method ofsubstitution. Let x be a function of a variable t, that is, if x = f(t), then dx = f’(t) dtPutting x = f(t) and dx = f’(t) dt, we have ∫ f (x)dx =∫ f (f(t)f '(t) dt.Now we explain the procedure with the help of some examples.Example 1: Evaluate ∫2 a dt b (at + b > 0) at +Solution: Let at + b = u. Then a dt = duThus ∫ 2 =aadt t+ b =∫ 2duu 2∫1 -1 u 2 du = 1  -1 +1  + c = 1  1  + c = 1 at + b + cExample 2: 2 2 u2 +1 u2 u2 + c = -1 1 2 2 Evaluate ∫ x dx. 4 + x2Solution: Put 4 + x2 = t ⇒ =2x dx dt or =x dx 1 dt, therefore 2 x=d=x 1  1 =dt 12 -1 + 1 . t1/ 2 ∫ ∫ ∫4 2 2 1/ 2 c + x2 t t 2 dt = t + c = 4 + x2 + c version: 1.1 16

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.PunjabExample 3: Evaluate ∫ x x - a dx, ( x > a)Solution: Let x - a = t ⇒ x = a + t ⇒ dx = dtThus ∫ x x - a dx = ∫(a + t ) t dt  1 3  1 3 2 2 t 2 dt t 2 dt ∫ ∫ ∫=at + t dt = a + 35 = t2 + t2 + c= 2a t 3 + 2 t 5 +c a3 5 2 2 35 22 = 2t 3  a + 1 t  + c= 2(x - a ) 3  a + 1 ( x - a )  + c 2 3 5 2 3 5 = 2(x - a ) 3  5a + 3( x - a ) + = c 2 (-x a ) 3 ( 5+a 3-x 3a+) c 2 15 2 15 = 2 (x - a ) 3 ( 2a + 3x) + c 2 15Example 4: Evaluate ∫ cot x dx , (x > 0). xSolution: Put x = z,( )then d x =⇒dz 1 =dx dz 2xor 1 dx = 2dz xthu=s=∫ cot x x dx ∫ cot x . 1 dx ∫ cot z.(2dz) x ===2∫ cot z dz 2∫ cos z dz 2∫(sin z)-1 cos z dz 0 as = 2ln sin z + c, si(nzz> x > 0) = +2ln sin x c version: 1.1 17

13.. IQntueagdrraatitoicn Equations eLearn.PunjabExample 5: Evaluate (i) ∫cosec x dx (ii) ∫sec x dx eLearn.Punjab version: 1.1Solution: ∫ cos ec x dx = ∫ cos ec x(cos ec x - cot x) dx cot x cos ex x - ( )Put cosec x cot x =t, th-en cosec x co+t x cosec2 x =dx dt or cosec x (cosec x - cot x) dx =dt so ∫ cosec x(cosec x- cot x)=dx ∫ 1=dt ln t + c (cosec x - cot t x)Thus cosec x dx = ln cosec x - cot x + c [t = cosec x - cot x](ii) ∫ sec x dx = ∫ sec x ( sec x + tan x) dx + tan x) ( sec x ( )Put sec x =+ tan x t, then sec xtan x + se=c2 x dx dt or sec x(sec x + tan x) dx =dt so ∫ sec x ( sec x tan x )=dx ∫ =dt ln t + c t ( sec x tan x) (t = sec x + tan x) Thus ∫ sec x dx =ln sec x + tan x +cExample 6: Evaluate ∫cos3 x sin x dx,(sin x > 0).Solution: Pu=t sin x t=, then dt  2 1 x .cos x dx sin==or 2t dt cos x dx  sin x t  Put=t=ing sin x t and cos x dx 2t dt in the integral, we have, ( ) ( )∫ ∫cos2 x sin x cos x dx =1 - t4 . t × 2t dt, co-s2 x =1=- sin2 x 1 t4 ( )∫ ∫ ∫= 2 t2 - t6 dt = 2 t2dt - 2 t6 dt = t3 - t7 +c 2. 2 37 = 2 ( sin )3 - 2 ( sin x ) 7 =+ c 2 3 - 2 7 +c 2 x2 sin2 x sin2 x 37 37 18

31.. IQntueagdraratitoicn Equations eLearn.PunjabExample 7: Evaluate ∫ 1 + sin x dx ,  - p < x < p  eLearn.Punjab 2 2 version: 1.1Solution: ∫ 1 + sin x dx =∫+ 1 sin x . 1- sin x dx =∫ 11 - sin2 x dx 1- sin x - sin x = ∫ cos x dx 1 - sin x Put sin x = t, then cos x dx = dt, therefore∫ 1+ sin x dx =∫ 1-1sin x .cos x dx-=∫ 1d-t t ∫=(1 t - 1 dt 2 ) = (1 )-t - 1 +1 +c = -2 1 -t +c 2 -1  2 + 1(- 1) -=2- 1 si+n x cExample 8: Find ∫ dx , (x> 0)Solution: x (ln 2x)3Example 9:Solution: Put In 2x = t, then 1 .2 dx = dt or 1 dx = dt 2x x 1 .1 =dx 1 t -2 x t3 -2 (ln 2x)3 ∫ ∫ ∫Thus =.dt t -3=dt + c =- 1 + c =- 1 )2 +c 2t 2 2(ln 2x ∫Find a x2 x dx, (a > 0, a ≠ 1) Put x2 = t, then x dx = 1 dt 2 ∫ ∫Thus ax2 x =dx at × 1 dt 2 19

13.. IQntueagdrraatitoicn Equations eLearn.Punjab ∫= 1 at d=t 1 at + =c ax2 + c eLearn.Punjab 2 2 ln a 2ln a version: 1.1Example 10: Evaluate(i) ∫ 1 dx, (- a < x < a) (ii) ∫ x 1 a2 dx, (x > a or x < - a) a2 - x2 x2 -where a is positive.Solution: (i) Let x = a Sin q, that is, x = a Sinq for - p < q < p , then dx =a cosq dq 22 dx = a cosq dq a2 - x2 a2 - a2 sin2 q ∫ ∫Thus == ∫ a cos q dq ∫ a cosq dq 1- sin2 a cosq a q = ∫1 dq= q + c =Sin-1  +x  c =ax Sin q  a(ii) Put x = a Sec q i.e., x = a sec q for 0 < q < p or p < q < p . Then dx = a sec q tan q dq 22 dx = a secq tanq d q x x2 - a2 a secq a2sec2q - a2 ∫ ∫Thus ( ( ) a2 sec2q 1 -∫ a secq tanq d q a secq .a tanq )∫= 1 1 dq= 1 .q + c = a2 tan2= a tanq aa =1 Sec-1+x c.  Sec=q x  aa a 20

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab3.4 SOME USEFUL SUBSTITUTIONSWe list below suitable substitutions for certain expressions to be integrated.Expression Involving Suitable Substitution(i) a2 - x2 x = a sin q(ii) x2 - a2 x = a sec q(iii) a2 + x2 x = a tan q(iv) x + a (or x - a ) x + a = t (or x - a =t)(v) 2ax - x2 x - a =a sin q(vi) 2ax + x2 x + a =a sec qExample 1. ∫Evaluate 1 (a > 0) dx a2 + x2Solution: =Let x a tan q for - p < q < p . Then Thus 22 dx = a sec2 q dq 1 a sec2 q dq + a2 tan2 q a 1 + tan2 q 1 =dx∫ ∫ ∫a2 + x2 a2 × a sec2 q =dq== ∫ a sec2 q dq ∫ secq dq a secq secq (secq + tanq )=dq = ∫ ln (sec +q tan q+) c1 secq + tanq=ln  a2 + x2 +x  +c1 sec2q 1=t+an2q 1=+ax22 =a2 a+2 x2 i.e., a a==ln  =a2+ + ax2=++x+a c+1secq c1secqa+ a2 + xq2 as secq is a( ( ) )==ln x + +a2 + x+2 - l-n aln +a c+1 pco1 spositi-ve-fo<r p<<pq< qp p2  2 version: 1.1 21

13.. IQntueagdrraatitoicn Equations eLearn.Punjab ( ( ) )=ln x + =a2 + x+2 + c+wh + c=whe-re c =c1 - ln a eLearn.Punjab version: 1.1Note: x + a2 + x2 is always positive for real values of a.Example 2. Evaluate ∫ dx , (x> 0) 2x + x2Solution: ∫ dx =∫ dx 2x + x2 ( x +1)2 -1 Let x + 1 = sec q. Then 0 < q < p  2 dx = sec q tan q dq Thus ∫ dx = ∫ sec q tan dq = ∫ sec q tan dq = ∫ secq dq sec2 q -1 tan q ( x +1)2 -1 ( )+ln (secq + tanq ) + c = ln x + 1 + 2x x2 + c1 EXERCISE 3.3Evaluate the following integrals:1. ∫ - 2x dx 2. ∫ x2 dx +13 ∫3. 4 x2 dx 4 - x2 + 4x + x24. ∫ 1 dx ∫5. ex dx ln 3 x x ex +6. ∫(x x+b )1 dx 7. ∫ sec2 x dx + 2bx + tan x c2( )∫8. dx a2= +c (a) Show that x2 - ln x + x2 - a2 ∫(b) show that a2 - x=2 dx a Sin 1 x + x a2 - x2 + c aa 22

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.PunjabEvaluate the following integrals: ∫ dx 1 1+x x2 Tan-1 1-x 3 ( )∫ ∫10. 11.( )9. 1 + x dx dx 1 + x2 212. ∫1 sin q dq ∫13. ax 14. ∫ dx + cos2 q 7 - 6x - x2 a2 - x415. ∫ cos x dx 16. ∫ cos x  ln sin x  dx x ln sin sin x sin x17. ∫4 + x dx x2 ∫18. x4 + x + dx 2x + 2x2 519. ∫ cos  x - x  ×  1 - 1 dx 20. ∫ x+2 dx 2 x x+321. ∫ sin x 2 x dx 22. ∫ 1 sin x dx 3 cos x + cos + 223.5 INTEGRATION BY PARTS We know that for any two functions f and g. d  f=( x) g ( x) +f ′( x) g ( x) f (x)g′(x) dxor =f ( x) g′( x) -d  f ( x) g ( x) f ′(x) g(x) dxIntegrating both the sides with respect to x, we get,=∫ f ( x) g′( x) dx -∫  d ( f (x) g(x)) f ′( x) g ( x) dx dx= -∫  d  f ( x) g ( x)  dx ∫ f ′( x) g ( x) dx dx version: 1.1 23

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab = f ( x) g ( x) + c - ∫ f ′( x) g ( x) dx (By Deinition)i.e., ∫ f ( x) g′( x) =f ( x) g-( x) ∫ g ( x) f ′( x+) dx c (i)or ∫ f ′( x=) g′( x) dx - f ( x) g ( x) ∫ g ( x) f ′( x) dx (i) A constant of integration is written, when ∫ g ( x) f ′( x) dx is evaluated. The equation (i)or (i)’ is known as the formula for integration by parts.If we put u = f(x) and dv = g ’ (x) dxthen du = f ‘ (x) dx and v = g(x).The equation (i) and (i)’ can be written as ∫u dv =-uv ∫ v d+u c (ii) ∫u d=v uv - ∫ v du (ii)'Example 1. Find ∫ x cos x dx.Solution: If f(x) = x and g ‘ (x) = cos x, then f ’(x) = 1 and g(x) = sin x Thus ∫ x co=s x dx x sin x - ∫(sin x) (1) dx = x sin x - (- cos x) + c = +x sin x + cos x cExample 2. Find ∫ x e x dxSolution: Let u = x and dv = ex dx, then du = 1 .dx and v = ex Applying the formula for integration by parts, we have ∫ ∫x ex dx =x ex ex x 1 dx-= x ex ex + c - version: 1.1 24

31.. IQntueagdraratitoicn Equations eLearn.PunjabExample 3. Evaluate ∫ x tan2 x dx eLearn.Punjab version: 1.1Solution: ∫ x =tan2 x dx ∫ x(sec2 x - 1) dx ( ) 1 + tan2x = sec2x = -∫ x sec2 x dx ∫ x dx (I)Integrating the ist integral by parts on the right side of (I), we get∫ ∫x tan2 x d=x  x2 + c1  [x tan x - tan x . 1 dx] - 2= x tan x-dx + ∫ 1 x=-. ( sin+x) dx +x22 - c  - x tan x + ln cos x c2 x2 c1 cos 2 =x tan x + ln cos x - x2 + c, where c =c2 - c1 2Example 4. Evaluate ∫ x5 ln x dxSolution: ∫ x5 ln x dx = ∫(ln x) x5 dxExample 5. ∫ ∫=(ln x) x6 - x6 . 1 dx =x6 ln x - 1 x5dx 6 6x 6 6 = x6 ln x - 1  x6 + c1 6 6  6 =x6 ln x x6 + c w- here c = c1 - 6 36 6 ( )Evaluate ∫ ln x + x2 + 1 dxSolution: ( )Let f ( x) = ln x + x2 +1 and g′( x) = 1. Then=f ′( x) × 1+ (+1 1 x2 )1 1 -1 2x  x + x2 + 1 2 2.=+ 1 + 1 . 1 x 1  x+ x2 x2 + 25

13.. IQntueagdrraatitoicn Equations eLearn.Punjab=× 1  x2 + 1 + x  = 1 and g(x) = x eLearn.Punjab x2 + 1 x2 + version: 1.1 x + x2 + 1 1Using the formula ∫ f ( x)=g′( x) dx f ( x) g ( x) - ∫ g ( x) f ′( x) dx, we get(∫ln x + )x2 + 1 . 1 dx = [ln(x + ∫x2 + 1)] .x - x . 1 dx (∫ln x + x2 + 1 ) ( )∫x2 + 1 x - 1 x2 + 1 - 1 ( 2 x ) dx 2 2 = x ln (x + ( )x2 +1) - 1  x2 1  2  1++1 2 c1   2 ( )=x ln x + x2 + 1 + c1, where c =1- x2 +1 - 2 c1Example 6. ∫Evaluate x2. a eax dxSolution: If we put f(x) = x2 and g ’ (x) = a eax, then f ’ (x) = 2x and g(x) = eaxUsing the formula ∫ f ( x)=g′( x) dx f ( x) g ( x) - ∫ g ( x) f ′( x) dx, we get ∫ x2 . axa=x dx ∫x2 eax - eax.(2x) dx ∫x2eax - 2 x eax dx =∫ ∫But x eax dx = x  eax  -  eax  × 1 . dx a a ∫=1 xeax - 1 eax dx =1 x eax - 1 . eax  + c1 a a a aa∫Thus x2a eax dx =x2eax - 2  1 .x eax - 1 eax + c1 a a2 =x2eax - 2 .xeax + 2 eax + c1 where c = 2c1 a a2 26

31.. IQntueagdraratitoicn Equations eLearn.PunjabExample 7. ∫Find eaxcos bx dx. eLearn.Punjab version: 1.1Solution: Let f(x) = eax and g ’ (x) = cos bxthe=n=f ′( x) a . eax and g ( x) sin bx b∫Thus eax cos bx dx× =eax  sin- bx  ∫  si×n bx  (aeax ) dx b b = - 1 eax sin bx ∫a eax sin bx dx (I) b b∫Integrating eax sin bx dx , by parts, we get  cos bx   cos bx ∫ ∫eax sin bx dx = eax × - b - - b × (aeax ) dx + c1 =1 eax cos b-x ∫a eax cos+bx dx c1 (II) + b b∫Putting the value of eax sin bx dx in (I), we get 1 eax a - 1 eax cos bx +a c1  b b b b∫ ∫eax co=s bx dx - eax c+os bx dx sin bx ∫= 1 eax sin bx + a - a2 eax cos bx dx - a b2 b2 b b eax cos bx .c1∫or1 a2  eax cos bx dx = 1 eax sin bx + a eax cos bx - a b2 b b2 b .c1∫i.e. eax=cos bx dx b2 b2 +b12 eax sin bx ba2-eax cos bx× b2 a . c1 a2 + a2 + b2 b[=a2e+axb2 b sin bx + a cos bx-] + c, ( )where c =b a2a+b b2 c1If we put a = r cos q and b = r sin q;then a2 + b2 = r2 ⇒ r = a2 + b2 b= r csoinsqq= tanq ⇒ q= tan-1 b a r a 27

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjaband a cos bx + b sin bx = r cos q cos bx + r sin q sin bx = r [cos bx cos q + sin bx sin q] = r cos (bx - q) =+ a2 b2 co-s  bx tan -1 b  , =q tan -1 b  a aThe answer can be written as:∫ eax c=os bx dx 1 eax cos bx - tan -1 b  + c a2 - b2 aExample 8. ∫Evaluate a2 + x2 dx( ) ( )∫Solution: 1 1 a2 + x2 . 1 dx = ∫a2 + x2 x - x. a2 + x2 2 . 2x dx 2 ∫= x a2 + x2 - x2 dx a2 + x2 ∫= x a2 + x2 - a2 + x2 - a2 dx a2 + x2 ∫ ∫= x a2 + x2 - a2 dx a2 + x2 dx + a2 + x2∫ ∫2 a2 + x2 dx= x a2 + x2 + a2. 1 dx a2 + x2 ( )= x a2 + x2 + a2 ln x + a2 + x2 + c1 (See Example 1 Article 3.4) a2 + x2 dx = x a2 + x2 + a2 ln x + )a2 + x2 + c, where c = a2c1 2∫ (2 2 x2 - a2 can be evaluated.Similarly integrals ∫ a2 - x2 dx and ∫Example 9. Evaluate ∫ sin4 x dx.Solution: ( )∫ ∫ ∫=sin4 x dx =sin2 x . sin2 x dx - sin2 x 1 cos2 x dx version: 1.1 28

31.. IQntueagdraratitoicn Equations eLearn.Punjab ∫ ∫= - sin2 x dx sin2 x cos2 x dx eLearn.Punjab - -∫ 1 - cos 2x dx ∫ sin2 x cos2 x dx (I) 2 version: 1.1 Integrating ∫ sin2 x cos2 x dx by parts, we have∫ ∫sin2 x cos2 x dx = cos x sin2 x cos x dx  sin3 x  sin3 x × (- sin x) dx [ If f (x) = cos x and 3 g'(x) = sin2 x cos x.= - cos x ∫3∫=1 cos x sin3 x 1 sin4 x dx ..... (II) + then f '(x) = sin x 33 and g(x) = sin2 sin3 x  3 Putting the value of ∫ sin2 x cos2 x dx in (I), we obtain, ∫sin4 x dx =-∫ 12 cos 2 x- dx  1 cos x s+in3 x 1 ∫ sin 4 x dx 2 3 3 =12 ∫-1 dx 1 ∫ cos 2-x dx 1 cos x s-in3 x 1 ∫ sin 4 x dx 2 3 3∫or 1 +1  sin4 x dx= 1 × - 1  sin22x+ c-1 1 cos x sin3 x 3 2 2 3 ∫sin4 x=dx 3  1 × -1 sin 2x - 1 cos x +sin3 x c  4 2 4 3 =- 3 x 3 si-n 2x 1 cos x +sin3 x c wh=ere c 3 c1 8 16 4 4Example 10. Evaluate ∫ e x (1 + sin x ) dx. 1 + cos xSolution: e x (1 + sin x ) ex 1 + 2 sin x cos x   1 + cosx = 1 + 2cos2 x 1 cos x 2 2 2 ∫ 1+ dx ∫- dx 2cos2 x 2 29

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjabi.e. ∫ ex1=(1++cosisnxx) dx +∫ e x  1 sec2 x tan x  dx 2 2 2 ∫ ∫= + 1 ex sec2 x dx ex tan x dx (I) 22 2But ∫ tan x  . ex dx = tan 2x - . ex ∫ ex  sec2 x  .+1 dx c, (Integrating by parts) 2 2 2 ∫i.e. ex tan x dx =ex-tan x ∫1 ex sec+2 x dx c (II) 22 22 ∫Putting the value of ex tan x dx in (I), we get 2x (1 +∫ ∫esin x )=dx 1 ex sec2 +x dx e x ta-n x ∫1 ex sec2+x=dx c ex t+an x c cos x 22 2 21+ 22Example 11. Show that ∫ eax a f ( x) + f ' ( x) dx = eax f (c) + c.Solution: ∫ eax a f ( x) + f ′( x) =dx ∫ eax. a f ( x) dx + ∫ eax. f ′( x) dx ...(i) In the second integral, let j ( x) = eax and g′( x) = f ′( x), then j=′( x) ( )eax × a and g ( x) = f ( x) ( )so ∫ eax . f ′( x) dx =eax × f ( x) - ∫ f ( x) × aeax dx + c =eax f-( x) ∫ a eax f ( x+) dx c thus ∫ eax a f ( x) + f ′( x) d=x ∫ aeax f ( x) dx + ∫ eax f ′( x) dx + c = ∫ a eax f ( x) dx + eax f ( x) - ∫ a eax f ( x) dx + c = eax f ( x) + c. version: 1.1 30

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab EXERCISE 3.41. Evaluate the following integrals by parts add a word representing all the functions are deined.(i) ∫ x sin x dx (ii) ∫ ln x dx (iii) ∫ x ln x dx (v) ∫ x4 ln x dx(iv) ∫ x2 ln x dx (viii) ∫ x3 ln x dx (vi) ∫ x2 Tan-1x dx (xi) ∫ x3 cos x dx∫(vii) Tan-1x dx (xiv) ∫ x2 sin x dx (ix) (xvi)∫(x) x Tan-1x dx (xviii) ∫ x3 Tan-1x dx (xii) (xx)∫(xiii) Sin-1x dx ∫ x Sin-1x dx(xv) ∫ ex sin x cos x dx ∫ x sin x cos x dx(xvii) ∫ x cos2 x dx ∫ x sin2 x dx(xix) ∫ (ln 2 dx ∫ (ln (tan x) sec2 x dx x)∫(xxi) x Sin-1x dx 1 - x22. Evaluate the following integral.(i) ∫ tan4 x dx (ii) ∫sec4 x dx (iii) ∫ ex sin 2x cos x dx(iv) ∫ tan3 x sec x dx (v) ∫ x3 e5x dx ∫(vi) e-x sin 2x dx∫(vii) e2x cos3x dx (viii) ∫ cosec3 x dx∫3. Show that eax=sin bx dx 1 - eax sin  bx + Tan -1 b  c. a2 + a b 24. Evaluate the following indeinite integrals.∫(i) a2 - x2 dx ∫(ii) x2 - a2 dx(iii) ∫ 4 - 5x2 dx (iv) ∫ 3 - 4x2 dx(v) ∫ x2 + 4 dx ∫(vi) x2 eax dx version: 1.1 31

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab5. Evaluate the following integrals.(i) ∫ e x  1 + ln x  dx (ii) ∫ ex (cos x + sin x)dx x∫(iii) eax  Sec-1x + 1 -  dx ∫(iv) e3x  3 sin x - cos x  dx a x2 1  sin2 x x(v) ∫ e2x [- sin x + 2cos x] dx (vi) ∫ (1 xe x )2 dx + x(vii) ∫ e-x (cos x - sin x) dx em Tan-1 x 1 + x2 dx ∫ ( )(viii) ex (1 x)(ix) ∫1 2x dx (x) ∫ (2 + + )2 dx - sin x x(xi) ∫  1 - sin x  ex dx 1 - cos x3.5 INTEGRATION INVOLVING PARTIAL FRACTIONS If P(x), Q(x) are polynomial functions and the denominator Q(x)( ≠ 0), in the rationalfunction P(x) ,can be factorized into linear and quadratic (irreducible) factors, then the rational Q(x)function is written as a sum of simpler rational functions, each of which can be integrated bymethods already known to us. Here we will give examples of the following three cases when the denominator Q(x)containsCase I. Non-repeated linear factors.Case II. Repeated and non-repeated linear factors.Case III. Linear and non-repeated irreducible quadratic factors or non repeatedirreducible quadratic factors. version: 1.1 32

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.PunjabEXAMPLES OF CASE IExample 1: Evaluate ∫ 2 x 2 -x +6 6 dx, (x > 2)Solution: - 7x + The denomicator 2x2 - 7x + 6 = (x - 2) (2x - 3),Let ( x --2x)(+2x=6 - 3) A + B x-2 2x - 3or -x + 6 = A(2x - 3) + B(x - 2) which is true for all x Putting x = 2, we get -2 + 6 = A(4 - 3) + B x 0 ⇒ A = 4and Putting =x 3 , we get - 3 + 6= A(0) + B  3 - 2  22 2or 9 =B - 1  ⇒B =- 9 2 2Thus ∫(x -x + 6 3) dx = ∫ x 4 2 + -9 3  dx - 2x - - 2)(2x - )-1 =4 ∫( x - 2)-11 . dx - 9 ∫(2x - . 2dx 2 3 = 4 ln (x - 2) - 9 ln (2x - 3) + c, (x > 2) 2Example 2: ∫Evaluate 2x3 - 9x2 + 12x dx, (x > 2) 2x2 - 7x + 6Solution: After performing the division by the denominator, we get∫ ∫2x3 - 9x2 + 12xdx=  x - 1 + -x +6 6  dx 2x2 - 7x + 2x2 - 7x + 6 = ∫ x dx - ∫1 dx + ∫(x 4 2) dx + ∫ -9 3 dx, (See the Example 1) - 2x - = x2 - x + 4 ln ( x - 2) - 9 (2x - 3) + c, ( x > 2) 22 version: 1.1 33

13.. IQntueagdrraatitoicn Equations eLearn.PunjabExample 3: Evaluate (i) ∫ 2a dx, (x > a) eLearn.Punjab (ii) (x < a) version: 1.1 x2 - a2 ∫ a2 2a dx, - x2Solution: (i) The denominator x2 - a2 = (x - a)(x + a),Let ( x 2a + a) = x A + B -a x+a - a)(x = 1 - x 1 , (Applying the method of partial fractions) x -a + aThus ∫(x - 2a + a=) dx ∫ x 1 - x 1 a  dx -= ∫( x a)--1.1 dx + ( x )a -1 . 1 dx -a + a)(x = ln x - a - ln x + a + =c ln x - a + c, (x > a) x + a(ii) It is left as an exercise.EXAMPLES OF CASE IIExample 4: Evaluate ∫(x 7 1 1) dx ( x 1)Solution: - 1) (x + We write∫(x 7x - 1 1) dx = x A 1 + (x B + C - x +1 - 1)2 ( x + - 1)2 = 2 1 + 3 - 2  Applying the method  x - x +1 of Partial Fractions ( x - 1)2Thus ∫(x 7x - 1 1) dx = ∫  x 2 + 3 - 2 1 dx - + - 1)2 ( x + 1 ( x - 1)2 x = 2∫( x - )1 -1.1 dx + 3∫( x - )1 -2 .1 dx - 2∫( x + )1 -1.1 dx ( )1 -2+1 = 2 ln ( x - 1) + 3 x- - 2 ln ( x + 1) +c ( x >1) -2 +1 34

31.. IQntueagdraratitoicn Equations eLearn.Punjab = 2 ln ( x - 1) - ln ( x + 1) + 3 ( x --1+1)-1  c eLearn.Punjab = version: 1.1 2 ln  x - 1  - 3 +c x + 1 x -1Example 5: ( )∫ ex x2 + 1 Evaluate ( x + 1)2 dx ( )∫ ex x2 + 1Solution: dx = ∫ ex 1 - 2 + (x 2 1)2  dx, (By Partial Fractions) ( x + 1)2 + ( x + 1) ( )⇒ ∫ ex x2 + 1 dx = ∫ ex dx - 2 ∫ x ex 1 dx + 2 ∫(x ex dx (I) + ( x + 1)2 + 1)2 We integrate by parts the last integral on the right side of (I). (x + )1 -1  ( x + )1 -1  -1 -1 ∫ ∫ex ( x + )1 -2 dx = ex . - . ex dx∫or (x ex 1)2 -dx =+x e+x 1 ∫ x ex 1 dx (II) + +Using (II), (I) becomes ( )ex x2 + 1 ex + 2 - ex + ex dx  x+ x+ +∫ ∫ ∫ ∫( x + 1)2 dx = ex dx - 2 dx 1 x 1 1 ( ) ∫ ∫= ex - 2ex + ex ex + c - 2 x + 1 dx x +1 2 x + 1 dx = ex - 2ex + c = xex + ex - 2ex + c= ex ( x - 1) + c. x+ x +1 1 x +1Example 6: Evaluate ∫ x3 1 1 dxSolution: - The denominator x3 - 1 = (x - 1 )(x2 + x + 1), 35

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab 1 A + Bx + C - x2 + x + 1 ( x - 1) x2 + x + 1( )Let = x 1 1  - 1  x - 2 ( )Applying the method of partial fractions 3 3 + 3 = x -1 + 1 , x2 + x= -1 . x 1 1 1 . x2 x+2 , 3 - 3 +x+ 1( )Thus(x- 1) 1 + x + 1 dx = ∫  1 . x 1 1 - 1 . 2x + 4 1 dx x2 3 - 6 x2 + x +∫=  1 . x 1 1 1.dx - 1. 2x + 1 - 1. x2 3 + 1 dx 3 - 6 x2 + x + 1 6 +x( )∫ ∫ ∫= 1 ( x - )1 -1 dx - 1 1 36 x2 + x + 1 -1 (2x + 1) dx - 1 2  2 dx 2  + 1 + 3 2 2 x( )= - 1. Tan-1  x + 1  1 ln x - 1 - 1 ln x2 + x + 1 1 +2 c 36 23  3  2 2( )= 1  2 ++ 1  1 ln x - 1 - 1 ln x2 + x + 1 - 3 Tan -1 x 3 c 36Note: x2+ x + 1 is positive for real values of x.Example 7: Evaluate ∫ 2 x 1 dxSolution: x6 - Put x2 = t, then 2x dx = dt and∫ x6 2x 1 dx = ∫ t3 1 1 dt = ∫ (t - 1 t + 1) - - 1)(t2 +  +  ( )= 1 2t 3 +1 1 ln t - 1 - 1 ln t2 + t + 1 - 3 Tan -1 c 36 (See the example 6) version: 1.1 36

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab( )= 1 Tan-1  2x2 ++ 1 1 ln x2 - 1 - 1 ln x4 + x2 + 1 - 3 3 c36Example 8: ( )Evaluate ∫ x 3 1 dx, x ≠ 0, x ≠ -1 x3 - ( )3 =+A x B-+1 Cx + D x x +x +1Solution: Let x x3 - 1 =-+3 2x + 1 x x 1 + x2 + x + 1 (By the method of partial fractions) - 1( )Let∫ x(x-1) 3 dx = ∫  -3 + 1 + 2x + 1 1  dx x2 +x +1 x x -1 x2 + x + ( )=- 3∫( x)-11 . dx + ∫( x - )1 -11 . dx + ∫ x2 + x + 1 -1 (2x + 1) dx ( )= - 3ln x + ln x - 1 + ln x2 + x + 1 + c ( )= - 3ln x + ln x - 1 x2 + x + 1 + c = - 3ln x + ln x3 - 1 + cExample 9: ( )Evaluate ∫ x2 2x2 + 6x + 3) dxSolution: + 1 (x + 2x We write 2x2 + 6x Ax + B + Cx + Dx2 + 1 x2 + 2x + 3 x2 + 1 x2 + 2x + 3( )( )Let = = 2x + 1 - 2x + 3 (Applying the method of partial fractions) x2 + 1 x2 + 2x + 3 2x2 + 6x  2x + 1 2x + 3 3  dx + 1 x2 + 2x x2 + 1 + 2x +( )( )∫ ∫Thusx2 + 3 dx = - x2 = ∫ x2 2x dx + ∫ x2 1 1 - ∫ x2 2x + 2 dx - ∫ x2 + 1 + dx + 1 + + 2x + 3 2x 3 version: 1.1 37

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab ( )( )= -1 (2x) 1 1 ∫ x2 +1 dx + ∫ x2 + 1 dx - ∫( x + 2x + )3 -1 (2x + 2) dx - ∫(x + 2 dx 1)2 + 2 ( ) ( )= ln x2 + 1 + Tan-1x - ln x2 + 2x + 3 - 1 Tan-1 x ++1 c 22 EXERCISE 3.5Evaluate the following integrals.1. ∫ x 2 3x + 1 6 dx 2. ∫ ( x + 5x +8 1) dx - x - 3) (2x - (a - b)∫3. x2 + 3x - 34 dx 4. ∫(x a)(x x b) dx, (a > b) x2 + 2x - 15 - -5. ∫1 3-x dx ∫6. x2 2x dx - x - 6x2 - a27. ∫ 6x2 1 dx ∫8. 2x3 - 3x2 - x - 7 dx + 5x 2x2 - 3x - 2 - 49. ∫(x 3x2 - 12x + 11 3) dx 10. ∫ x(x 2x - 1 3) dx - 1) (x - 2) (x - - 1)( x - ∫ 5x2 + 9x + 6 ∫ (1 4 + 7x x2 - 1 (2x + 3) x)2 (2 +( )11. dx 12. + 3x) dx13. ∫(x - 2x2 dx 14. ∫(x - 1 dx 1)2 ( x + 1) 1)( x + 1)2∫15. x3 x+4 4 dx 16. ∫ ( x3 - 6x2 + 25 dx - 3x2 + x+ 1)2 ( x - 2)217. ∫ x3 + 22x2 + 14x - 17 dx 18. ∫(x x-2 1) dx ( x - 3)( x + 2)3 + 1)( x2 +19. ∫(x - x + 1) dx 20. ∫(x 9x - 7 1) dx 1)( x2 + 3)( x2 + version: 1.1 38

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab21. ∫(x 1 + 4x dx 22. ∫ 12 dx +8 - 3)(x2 + 4) ∫ 9x + 6 2x2 + 5x + 3 x3 - 8 ( x - 1)2 x2 + 423. dx ( )∫24. dx 2x2 - x - 7 3x + 1 - ( x + 2)2 x2 + x + 1 + 1)( x2( )∫25. dx 26. ∫ (4x2 x + 1) dx27. ∫ (x2 4x + 1 + 5) dx ( )( )∫28. 6a2 dx x2 + a2 x2 + 4a2 + 4)( x2 + 4x 2x2 - 2 3x - 8 x4 + x2 + 1( )∫29. dx 30. ∫ (x2 - x + 2)(x2 + x + 2) dx 3x3 + 4x2 + 9x + 5 x2 + x + 1 x2 + 2x + 3( )( )∫31. dx3.6 THE DEFINITE INTEGRALS We have already discussed in section 3.2 about the indeinite integral that is, if f‘ (x) =f(x), then ∫ f ( x=) dx f ( x) + c, where c is an arbitrary constantIf ∫ f ( x=) dx f(x) + c, then the integral of f from a to b is denoted by b f ( x) dx (read asintergral from a to b of f of x, dx) and is evaluated as: ∫a b f (x) dx = b f ′ ( x ) dx (if f ( x) = f′( x)) ∫ ∫ aa = f(x) + b = f (b) + c - f (a) + c = f (b) - f(a) c ab f (x) dx has a deinite value f(b) - f(a), so it is called the deinite integral of f from a to b.∫ a f ( x)ba or f ( x)ba - f(a) is denoted as (read f(x) from a to b)f(b) version: 1.1 39

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab The interval [a, b] is called the range of integration while a and b are known as thelower and upper limits respectively.As f(b) - f(a) is a deinite value, so the variable of integration x in b f (x) dx can bereplaced by any other letter. a∫i.e. b f ( x=) dx b f (=t ) dt f(b) - f(a) ∫ ∫ aaNote: If the lower limit is a constant and the upper limit is a variable, then the integral is xa function of the upper limit, that is, ∫ f (t ) dt = f (=t ) ax f ( x) - f (a) a x ∫For Example, 3t2 dt = t3=ax x3 - a3 a The relation f’ (x) = f(x) shows that f(x) gives the rate of change of f(x), so the totalchange in f(x) from a to b as f(b) - f(a) shows the connection between anti-derivatives anddeinite integral b f (x) dx . a∫3.6.1 The Area Under The Curve About 300 B.C. and around this, mathematicians succeeded to ind area of planeregion like triangle, rectangle, trapezium and regular polygons etc. but the area of thecomplicated region bounded by the curves and the x-axis from x = a to x = b was a challengefor mathematicians before the invention of integral calculus. Now we give attention to the use of integration for evaluating areas. Suppose that afunction f is continuous on interval a 7 x 7 b and f(x) > 0. To determine the area under thegraph of f and above the x-axis from x = a to x = b, we follow the idea of Archimedes(287-212 B.C.) for approximating the function by horizontal functions and the area under fby the sum of small rectangles. version: 1.1 40

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab To explain the idea mentioned above, we irstdraw the graph of f deined as: f ( x) = 1 x2 2 The graph of f is shown in the igure. We divide the interval [1, 3] into four sub-intervals of equal length= =3 -1 1 . 42 As the subintervals are [x0, x1], [x1, x2], [x2, x3], [x3, x4], so x0 = 1, x1 = 1.5, x2 = 2, x3 = 2.5, x4 = 3 In the igure MA = f(x0), NB = f(x1) and MN = dx, so itis obvious that the area of the rectangle AMNC < the area of the shaded region AMNB < areaof the rectangle DMNB, that is, f(x0).dx < area of the shaded region AMNB < f(x1).dx ** * * Let x 1 , x 2, x 3, x 4 be the mid point of four sub-intervals mentioned above. ** Then the value of f at x 1 , is f (x 1) , so the area of therectangle FMNE = * dx f (x 1) (See the rectangle FMNE shown in the igure) We observe that the area of the rectangle FMNE isapproximately equal to the area of the region AMNB underthe graph of f from x0 to x1. Now we calculate the sum of areas of the rectangles shown in the igure, that is,f ( ∗ ) dx + f ( ∗ ) dx + f ( ∗ ) d x + f ( ∗ ) d x x1 x2 x3 x4 =  f ( ∗ ) + f ( ∗ ) + f ( ∗ ) + f ( ∗ )d x  x1 x2 x3 x4 version: 1.1 41

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab =  1  x0 + x1 2 + 1  x1 + x2 2 + 1  x2 + x3 2 + 1  x3 + x4 2  1 2 2 2 2 2 2 2 2 2 = 1  1 + 1.5 2 +  1.5 + 2 2 +  2 + 2.5 2 +  2.5 + 3 2  4 2 2 2 2 = 1 (1.25)2 + (1.75)2 + ( 2.25)2 + (2.75)2  4 = 1 (1.5625 + 3.0625 + 5.0625 + 7.5625) 4 = 1 (17.25) = 4.3125 4∫But 3 1 x2 dx =  1 . x3 13 = 1 ( 27 - 1) = 26 = 4.3 1 2 2 3 6 6 If we go on increasing the number of intervals, then the sum of areas of small rectanglesapproaches closer to the number 4.3. * If we divide the interval [1, 3] into n intervals and take x i the coordinate of any pointof the ith interval and dx i = x i - x i - 1, i = 1, 2, 3, ..., n, then the sum of areas of n rectangles is∑n f  * d x which tends to the number 4.3 when n gT and each dxig0.i=1 xi ∑Thus lim n f ( ∗ ) d xi = 4.3 and we conclude that n→∞ i=1 d xi →0 xi lim ∑ ∫( )n ∗ dix = 31 x2 dx. 12 n→∞ f xi di x→0 i=1Thus the area above the x-axis and under the curve y = f(x) from a to b is the deinite bintegral ∫ f (x) dx. a version: 1.1 42

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab Consider a function f which is continuous on the interval a 7 x 7 b and f(x) > 0.The graph of f is shown in the igure.We deine the function A(x) as the area above thex-axis and under the curve y = f(x) from a to x. Let dxbe a small positive number and x + dx be any numberin the interval [a, b] such that a < x < x + dx. Let P(xi f(x)) and Q(x + dx, f(x + bx)) be two pointson the graph of f. The ordinates PM and QN are drawnand two rectangles PMNR, SMNQ are completed. According to above deinition, the area abovethe x-axis and under the curve y = f(x) from a to x + dxis A(x + dx), so the change in area is A(x + dx) - A(x) which is shaded in the igure Note that the function f is increasing in the interval [x, x + dx]. From the igure, it is obvious that area of the rectangle PMNR < A(x + dx) - A(x) < area of the rectangle SMNQ, i.e., f(x) dx < A(x + dx) - A(x) < f(x + dx) dx Dividing the inequality by dx, we have f (x) < A( x + d x) - A(x) < f (x + d x) (I) dx lim f ( x) = f (x) and (lim f x + d x) =f ( x) d x→0 d x→0Since the limits of the extremes in (I) are equal, so A( x + d x) - A( x) → f ( x) when d x → 0. dx A( x + d x) - A(x) =Thus lim f (x). dx d x→0or A ‘ (x) = f(x)that is, A(x) is an antiderivative of f, so ∫ f ( x=) dx A(x) + cand x f ( =x ) dx  A(=x)ax A(x) - A(a) ∫a version: 1.1 43

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab Since A(x) is deined as the area under the curve y = f(x) from a to x, so A(a) = 0 Thus A( x) = x f (x) dx (I) ∫a Putting x = b in the equation (I), gives A(b) = b f (x) dx ∫awhich shows that the area A of the region, above the x-axis and under the curve y = f (x) froma to b is given byb f (x) dx, that is, b f (x) dx∫ A=∫aa * If the graph of f is entirely below the x-axis, then the value of each f (x i ) is negative andeach product f ( ∗ ) d xi , is also negative, so in such a case, the deinite integral is negative. xi Thus the area, bounded in this case by the curve y = f(x), the x-axis and the lines=x a=, x b is - b f (x) dx. ∫a For example, sin x is negative for - p < x < 0and is positive for 0 < x < p. Therefore the area bounded by the x-axisand graph of sin function from -p to p is given by -p p sin x dx + sin x dx 00 ∫ ∫ ∫ ∫ ∫ ∫0 p  b f (x) dx =ba f ( x) dx a- sin x dx + sin x dx = -p 0 = [ - cos x ]-p + [ - cos ]x p = - cos(-p ) - cos0 + -(cosp - cos0) 0 0 = - (-1) -1 - (-1) -1 = 2 + 2 = 4 version: 1.1 44

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.PunjabNote: p [ - cos ]x p = - cos p- cos-( p ) =- -1- -(1 ) = 0 -p -p ∫ sin x dx =3.6.2 Fundamental Theorem and Properties of Deinite Integrals The Deinite integral b f (x) dx a∫ gives the area under the curve y = f(x) from x = a to x = b and the x-axis (proof is given in the article 3.6.1) (b) Fundamental Theorem of Calculus If f is continuous on [a, b] and f‘ (x) = f(x), that is, f(x) is any anti-derivative of f on [a, b], then b f ( x=) dx f(b) - f(a) ∫a Note that the diference f(b) - f(a) is independent of the choice of anti-derivative of thefunction f. (c) b f (x) dx = - a f ( x) dx ∫ab ∫=(d) ∫b f ( x) dx c f<( x) <dx + b f ( x) dx, a c b ∫ ∫ a ac= P-roof of (c) and (d): (c) If f‘ (x) = f(x), that is, if f is an anti-derivative of f, then using the Fundamental Theorem of Calculus, we get b f (x) dx = f(b) - f(a) = - f (a) - f (b) = - a f ( x) dx ∫ ∫ ab version: 1.1 45

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab(d) If f’ (x) = f(x), that is, if f(x) is an anti-derivative of f(x), then applying the FundamentalTheorem of Calculus, we have c f ( x) dx =f (c) f (a) and b f-( x) dx f (b) f(c) = - ∫ ∫ acThus c f (x) dx + b f (x) dx = f (c) - f(a) + f(b) - f (c) ∫ ∫ ac = f (b) - f (c) = b f (x) dx a∫Other properties of deinite integrals can easily be proved by applying the FundamentalTheorem of Calculus.Now we evaluate some deinite integrals in the following examples.Example 1: Evaluate (i) ( )3 ∫2 x2 + 1 dxSolution: x + 1 ∫ x3 + 3x2 dx (ii) 1 -1 3 33∫ ( ) ∫ ∫(i) x3 + 3x2 dx = x3 dx + 3x2 dx -1 -1 -1 =  x4 3 +  x3 3-1 = -(3)4 (-+1)4  -(3)3- ( 1)3  4  -1  4 4  =  841 - 1  + 27 - ( -1)= 81 - 1 + ( 27 + 1) 4 4 = 20 + 28 = 48∫ ∫(ii)2 x2 +1 dx = 2 x2 - 1 + 2 dx 1 x +1 1 x +1 2  x2 - 1 2 1 dx= 2  2 1  ∫ ∫= 1 x + 1 + x + 1 x - 1 + x + dx ∫ ∫ ∫2 = x dx - 1 2 2 x 1 1 dx 1 + 1 dx + 2 1 version: 1.1 46

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab =  x2 2 - [ x ]2 + 2 ln ( x +1)12  2 1 1 =  ( 2 )2 - (1)2  - [2 - 1] + 2 ln (2 + 1) - ln (1 + 1)  2 2  =  2 - 1  -1 + 2[ln 3 - ln 2] 2 = 1 + 2ln 3 22Example 2: Evaluate (i) ∫3 x3 + 9x + 1 dx pSolution: x2 +9 0 (ii) ∫4 sec x(sec x + tan x) dx 03 x3 + 9x + 1 dx = 3  x3 + 9x 1 9 ∫ ∫(i)x2 +9 0 x2 + 9 + x2 + dx0 ∫ ( ) ∫=3  x x2 + 9 + 1 9  3  + 1 9  0 x2 + 9 + dx = 0 x + dx x2 x2 3 3 1 + x dx + 0 x2 0∫ ∫= 9 dx ∫= x22 03 13 Tan-1 3x 03  x2 +1(3)2 dx =+1 Tan-1 x c  + 33 ( )=  2 - ( 0)2  + 1  Tan -1 3 - Tan -1 0  3 3 3 3 2 2 =  3 - 0  + 1  Tan -1 1 - Tan-10  2 3 3 = 3 + 1  p - 0  = 3 + p  Tan -1 1= p and Tan-10= 0  2 3 6 2 18 3 6 version: 1.1 47

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjabpp( )(ii) ∫4 sec x(sec x + ∫tan x) dx = 4 sec2 x + sec x tan x dx00 pp 44= +∫ sec2 x dx ∫ sec x tan x dx 00= [tan p p =  tan p- tan 0  sec p sec 0 4 4 x] 4-+ [sec x]+4 0 0 ( )= (1 - 0) + 2 - 1 = 2 pExample 3: ∫Evaluate 41 0 1 - sinx dx pp 4 1 4 1 + sin x ∫ - sin x ∫Solution: 1 dx = (1 - sin x)(1 + sin x) dx 0 0 pp 4 1 + sin x 4 1 + sin x 0 1 - sin2 x 0 cos2 x ∫ ∫= dx = dx pp ∫ ∫ (=04  1 x + sin x  dx 4 )sec x tan x dx cos2 cos2 x = se+c2 x 0 =2 (See the solution of example 2(ii))Example 4: 2 Evaluate ∫ ( x + x ) dx -1Solution: 2 02 (by property (d)) ∫ (x + x ) dx = ∫ (x + x ) dx + ∫ (x + x ) dx -1 -1 0 02  -x =x if x < 0  =x if x > 0 = ∫ [x + (-x)] dx + ∫ (x + (x)] dx -1 0 version: 1.1 48

31.. IQntueagdraratitoicn Equations eLearn.Punjab 02 2 eLearn.Punjab = ∫ 0 dx + ∫ 2x dx = 0 + 2 ∫ version: 1.1 x dx -1 0 0 = 2  x2 02 = 2 4 - 0  = 4 2 2 2Example 5: Evaluate ∫7 3x dx 0 x2 + 9Solution: Let f(x) = x2 + 9. Then f ‘ (x) = 2x, so 3 (2x) -1 =2∫ ∫ ∫3x=dx dx + 3 (x2 9) 2 (2x) dxx2 + 9 x2 + 9 2 ∫= 3 -1 2 [ f (x)] 2 f (x) dx - 1 +1 3 [ f (x)] 2 1 1 = 2 -1 +1 = 3 [ f (x)]2 + c = 3(x2 + 9)2 + c 2 7 3(x2 + 1  7 3 (7 1 1  2 0 2Thus ∫ 3x dx = 9) = + 9) 2 - (0 + 9) 0 x2 + 9 = 3 (16) 1 - (9) 1  = 3(4 - 3) = 3 2 2Example 6: Evaluate 3 x ≠ - 1, 1 ∫2 Sin-1 x dx, 1 1 - x2 2Solution: Let t = Sin-1 x. Then x = sin t for - p ≤ t ≤ p 22and dx = cos t dt = 1 - sin2 t dt  cos t is +ve for -p ≤t ≤ p 2 2  = 1 - x2 dt 49

13.. IQntueagdrraatitoicn Equations eLearn.Punjab or 1 dx = dt (x ≠ -1, 1) eLearn.Punjab 1 - x2 version: 1.1 if x = 1 , then 1 =S in t ⇒ t = Sin-1 1 =p 22 26and if x = 3ð , then 3 = Sin t ⇒ t = Sin-1 3 = 22 23 33 Sin-1x 2 dx = (Sin-1 x) .Thus ∫ ∫2 1 dx 1 1 - x2 1 - x2 1 22 p 3 Sin-1 x t) = ∫ t=dt ⇒( x = Sin t p 6 = p =1  p 2 -  p 2  = 1  p2 - p2  2 3 6 2 9 36 t2 3  2 p 6 = 1  4p 2 - p2  = 3p 2 = p2 2 36 72 24Example 7: p 6 Evaluate ∫ x cos x dx 0Solution: Applying the formula ∫ f (x) f ' (x) dx = f (x) f (x) - ∫f (x) f ' (x) dx, we have ∫ x cos x dx = x sin x - ∫ (sin x) (1) dx = x sin x - [(- cos x) + c1] = x sin x + cos x + c where c = - c1, Thus p 6p ∫ x cos x dx = [x sin x + cos x ]6 0 0 50

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab =  p sin p + cos p  - (0 sin 0 + cos 0) 6 6 6 = p . 1 + 3 - (0+1) = p + 3 - 1 62 2 12 2Example 8: e Evaluate ∫ x In x dx 1Solution: Applying the formula∫ f (x) f ' (x) dx = f (x) f(x) - ∫f(x) f ' (x) dx, we have  ∫ (In x) x dx = (In x) . x2 - ∫ x2 . 1 dx 2 2 x ∫= 1 x2 In x - 1 x dx = 1 x2 In x- 1  x2 + c 2 2 2 22Thus ∫e -  1 x 2 In x x2 e 2 4 1 =x In x dx 1 =  1 e2 In e - e  -  1 (1)2 In 1 - (1)2  2 4 2 4 =  e2 .1 - e2  -  1 .0 - 1  ( In e = 1 and In 1 = 0) 2 4 2 4 = e2 + 1 44 13 1 ∫ f ( x) dx = 5, ∫ f ( x) = 3 and ∫ g( x) dx = 4, thenExample 9: If -2 1 -2evaluate th e following deinite integrals: 3 1(i) ∫ f ( x) dx (ii) ∫ [2 f ( x) + 3 g(x)] dx -3 -2 11(iii) ∫ 3 f ( x) dx - ∫ 2 g( x) dx -2 -2 version: 1.1 51

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab 313 ∫ f (x) dx = ∫ f (x) dx + ∫ f (x) dx = 5 + 3 = 8Solution: (i) -2 -2 1 1 11(ii) ∫ [2 f (x) + 3g(x)] dx = ∫ 2 f (x) dx + ∫ 3 g(x) dx -2 -2 -2 11 = 2 ∫ f (x) dx + 3 ∫ g(x) dx -2 -2 = 2(5) + 3(4) = 10 + 12 = 22 11 11(iii) ∫ 3 f (x) dx - ∫ 2 g(x) dx = 3 ∫ f (x) dx - 2 ∫ g(x) dx-2 -2 -2 -2 = 3 × 5 - 2 × 4 = 15 - 8 = 7 EXERCISE 3.6Evaluate the following deinite integrals. 2 1 ∫3.01 -2 (2x - 1)2 dx1. ∫ (x2 +1) dx ∫2. (x1/3 + 1) dx 1 -1 2 5. ∫ (2t 1) dt 54. ∫ 3 - x dx 6. ∫ x x2 - 1 dx -6 2∫7.2x ∫8. 3  x - 1 2 dx ∫9.1  x + 1  x2 + x + 1 dx 1 x2 + 2 dx 2 x -1 2 3 dx p 2   1 1  0 x2 + 9 3 1∫10. ∫12. x + 1 2 - 1 dx 11. ∫ cos t dt x x2 p 6 p 2 ∫14.2  e x - - x  dx ∫15.4 cos q + sin q dq 0 2 2 0 2cos2 q13. ∫ In x dx e 1 p p p 4 4 6 ∫17. cos2 q cot2 q dq 18. ∫ cos4 t dt16. ∫ cos3q dq p 0 0 6 version: 1.1 52

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab p pp sec q 3 q + cos q 4 419. ∫ cos2 q sin q dq 0 (1 + cos2 q ) tan2q dq 0 sin 0 ∫ ∫20. 21. dq 5 1  1 + 2 2 3 x2 - 2 1 x + 122. ∫ x - 3 dx x3 -1 ∫23. 2 dx ∫24. dx x1/8 3 p p∫25.3 3x2 - 2x +1 dx ∫26. 4 sin x - 1 ∫27. 4 + 1 x dx 2 (x - 1)(x2 + 1) 0 cos2 x sin 01 p p∫28.1 3x ∫2 cos x dx ∫30. 2 sin x dx 0 4 - 3x dx x)(2 + cos x) 29. p sin x (2 + sin x) 0 (1 + cos 63.7 APPLICATION OF DEFINITE INTEGRALS. Here we shall give some examples involving area bounded by the curve and the x-axis.Example 1. Find the area bounded by the curve y = 4 - x2and the x-axis.Solution: We irst ind the points wherethe curve cuts the x-axis. Putting y = 0,we have 4 - x2 = 0 ⇒ x = ± 2. So the curve cuts the x-axis at (-2, 0) and (2, 0) The area above the x-axis and under the curve y = 4 - x2 isshown in the igure as shaded region..Thus the required area =∫ (4 - x ) dx =4x -  version: 1.1 53