1 callculus

d s kwd k

1.[PAT1/58] R : ( ( (1) = ) (1 2.[PAT1/57] () 1.) (0,3)

: ) = 3 , (1) = 8 1) =? = )

3.[PAT1/58] R 6 1+ ( ) 0 =4 = ( ) 4. [PAT1/52] ( ) = lim ( ) ( ) =? ( ) ()

: = lim () =2 6 (2) =? (+ )

3. (antideriv (antide ’( ) = ( ) : ( )= ( )= :

vative) erivative) 1 ( )=3

3. ( ) = 5. ( ) = 2 + 1 8. ( ) = ( 1)(4 10. ( ) =

)

’( ) = ( ) () = , + = : 2. = + : 25 3. ( ) = : 25

= ( )+ 1 ()

4. ( ± )( ) = : (22 + 12 = () = () = () = () : =5 +3

() ± () 15) y ) ) 2

, () ( () ( ! ( )= ( ) + ** , >> t

() () () () =0 ( ) 0 >> ( ) = 0

:t t 3m 1. 2.

-3t m/s2 1 m/s

1. 1.1) ( + 3 + 5 ) 1.5) 1.7) ( ) 1.14) ( + 8 ) 2. ( ) = (2) =

=2 ( )

3. = ( ) (, ) 3.2) = 2 + 4 (0 4. ( ) () 4.3) ( ) = + 5 + 4,0 2, (0) = 3 5. g=9.8m/s2 5.1) 5.2)

) 0,5) () =0 0 15, (0) = 98 m/s

5.3) 5.4) 6. t (20 ) m/s2

249.9 m

(The Fundam f [, ] () =

mental Theorem of Calculus) Ff () ()

2.11 1. ( + 3) 4. 5. + 8. ( + 1)



f [, == 1. ( ) 0 x 2. ( ) 0 = x X = = X ()

]A f X [, ] A X = () [, ] A () () () [, ] X * () () ( )=0 =

= 2 =1 f [0,5] F’(x)=f(x) f f F(0) = 10 F

( )= 9 [0,5] F f [0,2] 14 [2,5] 20 F(2) F(5)

1. ( ) = 0 ( 2. () = ** ***

X X ) + ()

1= 3. = =1

= = 3 =0 =6+ =1

5. = =3 6f () (0) = 0

= 25 = 1 ’( ) = ( ) {1,2,3,4,5}

[, ] [() () () () - 1. ( ) = ( ) - 2. () = [() () ** ***

( ), ( ) ( )] () = )] + [ ( ) ( )]

= =0 =3 +1 =2 =0

2 =2 24 = =3

1.[PAT1/54] 2 + 3 = 0 (0,3) ( )= +2 ( ) 2. [PAT1/52] = A A=B X B X=

= () =3 ) () (2) =? (2) = 0 =1 =

3.[PAT1/58] ( ,

;<1 ) + ; 1 <1 3 +2; 1 ( ) =?