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WorksheetsBC Unit 6 Review
Total questions: 51
Worksheet time: 4hrs 15mins
∫01x2+3x+25x+8dx
ln8
ln227
ln18
ln288
∫1exx2+1dx
1e2−1
2e2+1
2e2+2
e2e2−1
∫−14x(x2−3)5dx is equivalent to which of the following?
21∫−213u5du
2∫−213u5du
∫−213u5du
∫−14u5du
21∫−14u5du
The graph of a differentiable function f is shown. If h(x)=∫0xf(t)dt , which of these following is true?
h(6)<h′(6)<h"(6)
h(6)<h"(6)<h′(6)
h′(6)<h(6)<h"(6)
h"(6)<h(6)<h′(6)
h"(6)<h′(6)<h(6)
Let f be a differentiable function such that
∫f(x)sinxdx=−f(x)cosx+∫4x3cosxdx . Which of the following could be f(x)?cosx
sinx
4x3
−x4
x4
The graph of f', the derivative of f, consists of two line segments and a semicircle, as shown. If f(2)=1, then f(-5)=
2π−2
2π−3
2π−5
6−2π
4−2π
If ∫1xf(t)dt=4x2+2120x−4 , then ∫1∞f(t)dt=
6
1
-3
-4
divergent
If f′(x)>0 for all real numbers x and ∫47f(t)dt=0 , which of the following could be a table of values for the function f?
∫xex2dx=
21ex2+C
ex2+C
xex2+C
21e2x+C
e2x+C
The graph of the piecewise linear function f is shown. What is the value of ∫−19(3f(x)+2)dx ?
7.5
9.5
27.5
47
48.5
∫5x(x−x2)dx
215x−15x2+C
5x+45x4+C
2x25−45x4+C
225x25−45x4+C
35x27−65x6+C
∫(2x−3)(x+2)7xdx
23ln∣2x−3∣+2ln∣x+2∣+C
3ln∣2x−3∣+2ln∣x+2∣+C
3ln∣2x−3∣−2ln∣x+2∣+C
−(2x−3)26−(x+2)22+C
−(2x−3)23−(x+2)22+C
∫(3x+1)5dx
18(3x+1)6+C
6(3x+1)6+C
2(3x+1)6+C
2(23x2+x)6+C
(23x2+x)5+C
∫x2−6x+81dx
ln∣(x−2)(x−4)∣+C
21ln∣(x−2)(x−4)∣+C
21ln∣∣∣∣x−2x−4∣∣∣∣+C
21ln∣∣∣∣x−4x−2∣∣∣∣+C
∫01x1+3x2dx
7/9
1/6
2
7
∫1∞(x3−3)3x2dx
241
−241
81
divergent
∫12x2x−4dx
-1/2
ln2-2
ln2
2
ln2+2
∫−21x∣x∣dx
-3
-1
2
3
nonexistent
∫(−3xcos4x)dx
−163cos4x−43xsin4x+C
−43xcos4x−163sin4x+C
−3cos4x−3xsin4x+C
−163xcos4x+43sin4x+C
∫x3lnxdx
4x4lnx−4x4+C
4x4lnx+4x4+C
4x4lnx−16x4+C
4x4lnx+16x4+C
∫x2−2x1dx
ln∣x(x−2)∣+C
ln∣∣∣∣xx−2∣∣∣∣+C
21ln∣∣∣∣xx−2∣∣∣∣+C
21ln∣∣∣∣x−2x∣∣∣∣+C
∫8∞x341dx
-3/2
-1/2
1/2
3/2
divergent
The graph of f(x) is given and consists of three line segments. Find ∫14xf′(x)dx .
5.5
14.5
23
25
∫15x−31dx
0
ln2
2ln2
diverges
∫3xe−2xdx
−61xe−2x−121e−2x+C
31xe−2x−31e−2x+C
61xe−2x−121e−2x+C
−32xe−2x−34e−2x+C
∫(2x+1)(x−2)4x−3dx
2ln∣2x+1∣+ln∣x−2∣+C
ln∣2x+1∣+ln∣x−2∣+C
2ln∣2x+1∣−ln∣x−2∣+C
ln∣2x+1∣−ln∣x−2∣+C
Given k is a constant, find ∫x2+kxdx in terms of x and k.
karctan(kx)+C
arctan(kx)+C
21ln∣∣x2+k∣∣+C
ln∣∣x2+k∣∣+C
∫3∞x2+91dx
9π
6π
21ln18
ln18
diverges
∫1+x2x2dx
x−arctanx+C
x+ln∣∣x2+1∣∣+C
x+arctanx+C
31x3+x+C
∫x2+4x+81dx
2x+41ln∣∣x2+4x+8∣∣+C
21arctan(2x+2)+C
arctan(2x+2)+C
ln∣∣x2+4x+8∣∣+C
∫9−4x21dx
21arcsin(32x)+C
23arcsin(32x)+C
−419−4x2+C
−169−4x2+C
∫9−4x2xdx
21arcsin(32x)+C
23arcsin(32x)+C
−419−4x2+C
−169−4x2+C
∫x+1x2−3x+2dx
21x2−4x+6ln∣x+1∣+C
21x2−3x+2ln∣x+1∣+C
21x2−2x+4ln∣x+1∣+C
21x2−2x+C
∫1∞x1dx
divergent
π
e
ln∞
Let h be a continuous function. Using the substitution u=2−4x , the integral ∫−3−1h(2−4x)dx is equal to which of the following?
41∫614h(u)du
−41∫614h(u)du
−41∫−3−1h(u)du
∫−3−1h(u)du
52(x−4)25+4(x−4)23+C
52(x−4)25+8(x−4)23+C
54(x−4)25+32(x−4)23+C
32(x−4)23+12(x−4)21+C
Evaluate: ∫2∞x2 dx
21
ln2
1
3
divergent
∫4∞(9−x2)31−2xdx
divergent
(23)32
932+732
23(932+732)
∫0∞x2e−x3dx
−31
1
31
divergent
∫14x+1x2−2x−4dx
−74
920
−23+ln52
29+5ln52
∫x2+3x−101dx
71lnx+5x−2+C
71lnx−5x+2+C
71lnx+2x−5+C
71lnx−5x+2+C
∫3x e2x dx
−2xe2x+4e2x+C
23xe2x−43e2x+C
xe−2x+2(1−x2)+C
−2xe2x+4lne2x+C
∫x2+2x−15dx
81ln∣x−3∣−81ln∣x+5∣+c
41ln∣x−3∣−81ln∣x+5∣+c
81ln∣x+5∣−81ln∣x−3∣+c
41ln∣x+5∣−41ln∣x−3∣+c
∫01(x−1)sin(πx)dx=
−1
0
−π1
π
−π
Evaluate ∫−∞∞(x2+3)2xdx
diverges
0
-2
4
Which of these integrals are improper integrals?
∫−∞1 x4 dx
∫−75 x−5x+2dx
∫−710 x−5x+2dx
∫−7−1 x−5x+2dx
∫−1∞ x4 dx
∫25 5−x1dx
31
23
diverges
3
∫tsint dt
−tcost+sint+C
tcost+sint+C
tcost−sint+C
−tsint−cost+C
∫xe4xdx
4xe4x+16e4x+C
4xe4x−16e4x+C
4xe4x+16e4x+C
4xe4x−16e4x+C
∫x4 lnx dx
4x+4ex+C
32x23ln4x−94x23+C
32(4x2−1)⋅e4x2+C
5x5lnx−25x5+C
What is the formula for Integration by parts?
∫u dv = uv−∫v du
∫v dv =∫u du −uv
∫u du = vu −∫v dv
∫u dv = uv+∫v du
