( ) # "
A.P. Practice Calculus Exam - Definite Integrals, MVT
Don’t skip steps that you cannot skip. Calculators allowed for arithmetic only.
17
1.
#(
"1
)
2x 7 + 1 dx
"17
# "2x
2.
4
dx
"2
!
!
"
"1
2
3. # ("x " 6x + 3) dx
4.
2
6
# 3sin x dx
0
!
!
9
5.
3
#
4 &
) %$ x " x (' dx
1
!
6.
x 2 "1 dx
#
0
!
3
7.
" x( x
0
2
4
3
+ 1) dx
!
8.
#
4 " x dx
3
!
3" x)
(
#
3
3#
1
9.
9
x
10.
dx
2
$ ( x " cos x ) dx
0
!
!
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Stu Schwartz
11. Given that a( t ) = 2t with v o = "1, find the following from t = 0 to t = 4. If there is a breakup of an integral,
show that before any calculations take place.
Displacement
!
!
Distance
12. Find the exact average value of f ( x ) = x 2 + x over [1, 4] (exact)
!
13. Find the value of c guaranteed by the Mean Value Theorem for Integrals for f ( x ) = 4 + x over [1, 4]
(3 dec pl)
!
14. Use the Second Fundamental Theorem of Calculus to find F "( x )
x
a) F ( x ) =
#
"4
5t 5 "17 dt
b) F ( x ) =
8
5
# cot (3t " 4) dt
x
!
!
!
x
15. Show that the Second Fundamental Theorem of Calculus holds for F ( x ) =
#
3t " 3 dt
4
!
www.MasterMathMentor.com
Stu Schwartz
A.P. Practice Calculus Exam - Definite Integrals, MVT
Don’t skip steps that you cannot skip. Calculators allowed for arithmetic only.
17
"1
# (2x
7
"17
+ 1) dx
# "2x
$"2x 5 '"1
2. &
)
% 5 ("2
17 + 17 = 34
2
2 64
62
"
="
5 5
5
=
"1
# ("x
dx
"2
$2x 8
'17
+ x)
1. &
% 8
("17
!
4
"
" 6x + 3) dx
# 3sin x dx
!
2
$ x3
'"1
2
3. &" " 3x + 3x)
% 3
(2
6
0
"
4. [$3cos x ] 0 6
% 3(
% 3 (
$3' * + 3(1) = $3'
$1*
& 2 )
& 2
)
* "8
1
" 3 " 3 " , "12 + 6/ = 3
+ 3
.
3
3
9
#
4 &
) %$ x " x (' dx
1
!
!
0
1
*2
5. , x 3 2 " 8x1 2 /
+3
.1
#2 & 4
2
(27) " 8( 3) " % " 8( =
$3 ' 3
3
3
!
0
7.
1
2
2
6.
1
u 3 du
u = x 2 + 1,du = 2x dx
$x
' $x
'3
"& " x) + & " x)
%3
(0 % 3
(1
#
3
4 " x dx
u = 4 " x,du = "1 dx
3
!
x = 0,u = 1, x = 3,u = 4
1
0
8. " # u1 2 du
x = 3,u = 1, x = 4,u = 0
1
(3 " x )
#
x
9
2 32 1 2
2
u ] = (1" 0) =
[
0
3
3
3
3
dx
u = 3 " x,du = "
1
2 x
3#
dx
!
2
$ ( x " cos x ) dx
0
2
9.
1
4
3
+ 1) dx
1
"2 # u 3 du
3#
%x2
(
10. ' " sin x*
&2
)0
x = 9,u = 0, x = 1,u = 2
0
$ u 4 '2
"2& ) = "8
% 4 (0
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!
0
2 20 22
+
=
3 3
3
4
1 # u4 &
1 255
= 31.875
% ( = 32 ) =
2 $ 4 '1
8
8
!
" # ( x "1) dx + # ( x 2 "1) dx
3
4
"
3
2
9
" x( x
x 2 "1 dx
#
2
9# 2
+1
8
!
Stu Schwartz
11. Given that a( t ) = 2t with v o = "1, find the following from t = 0 to t = 4. If there is a breakup of an integral,
show that before any calculations take place.
Displacement
!
!
Distance
4
Dist =
v ( t ) = t 2 "1
"1 dt
1
4
# (t 2 "1) dt
Dist = " # ( t 2 "1) dt +
52
3
Dist =
0
Disp =
2
0
4
Disp =
#t
0
# (t
2
0
"1) dt
2
56
+ 18 =
3
3
12. Find the exact average value of f ( x ) = x 2 + x over [1, 4] (exact)
!
4
3
$
x
x2'
*1 164
2
x + x ) dx & + )
+ 8 # , + / 57
(
"
2 (1
%3
+ 3 2 . 2 19
!
3
f avg = 1
=
=
=
=
4 #1
3
3
3
2
!
4
13. Find the value of c guaranteed by the Mean Value Theorem for Integrals for f ( x ) = 4 + x over [1, 4]
!
(3 dec pl)
4
4
$
2x 3 2 '
*
" 4 + x dx &%4 x + 3 )( 16 + 163 #!,+ 4 + 23/. 50 50
1
f avg = 1
=
=
= 3 =
4 #1
3
3
3
9
50
14
156
4+ x =
0 x=
0x=
= 2.420
9
9
81
(
)
14. Use the Second Fundamental Theorem of Calculus to find F "( x )
! x
"4
F ( x ) = # 5t 5 "17 dt
F ( x ) = # cot 5 ( 3t " 4 ) dt
a)
b)
8
x
5
!
F $( x ) = "cot 5 ( 3x " 4 )
F $( x ) = 5x "17
x
15. Show that the Second Fundamental Theorem of Calculus holds for F ( x ) =
!
!
'
d $x
2nd FTC says
& # 3t " 3 dt ) = 3x " 3
dx % 4
(
x
3 !
2
# 3t " 3 dt = 9 (3x " 3) 2 " 6
4
#
3t " 3 dt
4
3
$2
'
d& ( 3x " 3) 2 " 6)
%9
(
= 3x " 3
dx
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!
Stu Schwartz
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