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WorksheetsBC Unit 6 Test Review
Total questions: 78
Worksheet time: 4hrs 32mins
x2
4x-7
(1/3)x3-5x
∫ 2x cos(x2) dx
sin(x2) + C
2 sin ( 2x ) + C
(1/2) sin(x2) + C
4 cos(2x ) + C
What should be assigned to u in the integral?
2x2
sin (2x2)
5x
no u needed
What is the choice of u for ∫ ex(1+ex)(1/2) dx?
ex
1+ex
(1+ex)(1/2)
(ex)(1/2)
What is the Integration by Parts Formula?
What would you choose for your u here if you used integration by parts?
t
3t
e2t
et
don't use IBP, let u = 2t
Evaluate the indefinite integral using integration by parts.
∫3x e2x dx
−2xe2x+4e2x+C
23xe2x−43e2x+C
xe−2x+2(1−x2)+C
−2xe2x+4lne2x+C
Evaluate the indefinite integral using integration by parts.
∫t2lnt dt
42t2ln2t−t2+C
3t3 ln3−9t3+C
2t+2et+C
4e2t−2t−1+C
Evaluate the indefinite integral using integration by parts. u and v' are provided.
∫x4 lnx dx ; u=lnx, v ′=x4
4x+4ex+C
32x23ln4x−94x23+C
32(4x2−1)⋅e4x2+C
5x5lnx−25x5+C
Evaluate the indefinite integral using integration by parts. u and v ' are provided.
∫tsint dt ; u=t, v ′=sint
tcos−1t−(1−t2)21+C
−tcost+sint+C
tsin−1t+(1−t2)21+C
tsint+cost+C
What would you choose for your u here if you used integration by parts?
x
sin(x)
cos(x)
ex
What method would you use here?
antiderivative rules.
u-substitution
integration by parts
Stare at the question... forever.
Use substitution to evaluate the integral ∫4sec24x tan4x dx
2(tan4x)23+c
415(tan4x)34+c
34(tan4x)23+c
32(tan4x)23+c
Break this into partial fractions.
∫x2−13x+2dx
ln∣∣x2−1∣∣+c
21ln∣x+1∣+25ln∣x−1∣+c
23ln∣x+1∣+21ln∣x−1∣+c
25ln∣x+1∣+21ln∣∣∣(x−1)2∣∣∣+c
∫x2+2x−151dx
81ln∣∣∣∣x+5x−3∣∣∣∣+C
41ln∣∣∣∣x+5x−3∣∣∣∣+C
81ln∣∣∣∣x−5x+3∣∣∣∣+C
41ln∣∣∣∣x−5x+3∣∣∣∣+C
We're interested in calculating the area under the curve for x between -6 and 6. Order the areas from LEAST to GREATEST.
left-hand, right-hand, midpoint
right-hand, left-hand, midpoint
right-hand, midpoint, left-hand
left-hand, midpoint, right-hand
What kind of Riemann sum is described by the diagram?
Left-hand
Right-hand
Midpoint
What is the correct description of the subdivisions in the Riemann sum?
five uniform subdivisions
five nonuniform subdivisions
six uniform subdivisions
six nonuniform subdivisions
Approximate the area under y=h(x) from x=-2 to x=4 using a Right-hand sum and three equal subdivision
20 units2
26.5 units2
28 units2
What method would you use here?
inverse trig
u-substitution
integration by parts
partial fractions
∫5x4(x5−7)3dx can be solved by
integration by parts
u-substitution
partial fractions
natural log pattern
Which technique should be used to integrate this:
natural log pattern
u-substitution
integration by parts
partial fractions
Find the partial fraction decomposition...
∫ 4+9x2 dx can be solved by
natural log pattern
inverse trig integration
partial fractions
u-substitution
∫ x2−x+1x2dx must be solved by first
completing the square
finding the partial fractions
using u-substitution
performing long division
∫ x2+13x−1dx can be solved by first
using inverse trig
separating the integral into two fractions
completing the square in the denominator
using the natural log pattern
What does this picture represent?
Left Riemann Sum
Right Riemann Sum
Middle Riemann Sum
Trapezoidal Sum
What does picture represent?
Left Riemann Sum
Right Riemann Sum
Middle Riemann Sum
Trapezoidal Sum
Use a midpoint Riemann Sum to approximate the area between 0 to 3 with 3 subintervals.
14
7
26
11
Find the Left-hand Riemann Sum, with three sub-intervals indicated by the table.
28
16
34
23
Use 3 trapezoids to determine the approximate area of the shaded area.
12
9
10
5
Based on the table, use a trapezoidal sum of 4 sub-intervals to estimate the area under the curve.
32.5
40.5
78
160
Determine ∫08x2dx using an estimate of 4 equivalent based trapezoids
176
352
420
488
If ∫25 f(x)dx=5 and ∫45 f(x)dx=π, find ∫55 f(x)dx.
−π
0
−0
π
If∫25 f(x)dx=5 and ∫45 f(x)dx=π, find ∫54 f(x)dx.
0
−1
−π
π
∫25 f(x)dx=5 and ∫45 f(x)dx=π, find ∫24 f(x)dx. If
π−5
2
5−π
−(5−π)
∫020C(n)dn =
1000
250
750
1125
∫410f(x)dx =
2π + 3
4π + 3
2π - 3
π - 3
∫25(x3+3)dx as limit of a sum is equivalent to
n→∞limi=1∑n[(2+n3i)3+3]⋅n1
n→∞limi=1∑n[(n3i)3+3]⋅n3
n→∞limi=1∑n[(2+n3i)3+3]⋅n3i
n→∞limi=1∑n[(2+n3i)3+3]⋅n3
∫0πcosxdx as limit of a sum is equivalent to
n→∞limi=1∑n[cos(nπi)]⋅ni
n→∞limi=1∑n[cos(ni)]⋅ni
n→∞limi=1∑n[cos(nπi)]⋅nπ
n→∞limi=1∑n[cos(ni)]⋅nπ
n→∞limi=1∑n[(n5i)2+n5i+1]⋅n5 in integral notation would be
∫05(x2+x+1)dx
∫56(x2+x+1)dx
∫01((5x)2+5x+1)dx
∫010(2x2+2x+1)dx
Find the antiderivative of
f'(x) = x2 when f(3) = 11.
(1/3)x3+ 2
x3 + 29
(1/3)x3 + 9
x3 + 11
10
20
23
35
