Calculus Practice Test
12 problems — mixed review for the retake
Show all work; a correct answer with no steps earns little partial credit, and steps are where you’ve
been losing points. Suggested time: 75 minutes, no notes. Worked solutions begin on page 3 —
attempt every problem before you flip to them.
Part I — Problems
Z
Problem 1. Evaluate
x cos(2x) dx.
integration by parts
Z
Problem 2. Evaluate
sin3 (x) cos2 (x) dx.
trig integral, odd power
Z
Problem 3. Evaluate
cos2 (x) dx.
trig integral, even power
Z
Problem 4. Evaluate
(ln x)2
dx.
x
u-substitution
Z Problem 5. Evaluate
√
3
4
√
+ 2x − 5 x dx.
3
x
power rule with roots
Problem 6. Find
dy
for y = (3x4 + 2x)5 .
dx
chain rule
Problem 7. Find f ′ (x) for f (x) = log3 (2x2 + 5x).
logarithm-base differentiation
Problem 8. Find the 3rd-degree Maclaurin polynomial P3 (x) for f (x) = e2x .
Maclaurin polynomial
Problem 9. Find the 2nd-degree Taylor polynomial P2 (x) for f (x) =
√
x centered at a = 4.
Taylor polynomial, center ̸= 0
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Problem 10. Find the radius and interval of convergence for
∞
X
(x − 3)n
n=1
n 5n
.
interval of convergence
Problem 11. Determine whether
∞
X
3
2 n
converges. If it does, find its sum.
5
n=0
geometric series
Problem 12. Given the parametric equations x = 2t+1, y = t2 −3: eliminate the parameter to find
the rectangular equation, describe the curve, and state its orientation (the direction of the arrows) as
t increases.
parametric equations + orientation
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Part II — Answer Key & Worked Solutions
Solution 1 (by parts).
Let u = x, dv = cos(2x) dx, so du = dx, v = 12 sin(2x).
Z
Z
x
x cos(2x) dx = 2 sin(2x) − 21 sin(2x) dx = x2 sin(2x) − 12 − 21 cos(2x) + C
x
1
2 sin(2x) + 4 cos(2x) + C
Solution 2 (odd-power trig).
Sine is odd; split sin3 x = (1 − cos2 x) sin x. Let u = cos x, du = − sin x dx.
Z
Z
3
5
2 2
(1 − u )u (−du) = − (u2 − u4 ) du = − u3 + u5 + C
5
3
1
1
5 cos x − 3 cos x + C
Solution 3 (even-power trig).
No factor to peel off, so use the power-reduction identity cos2 x =
Z
1 + cos(2x)
.
2
1 + cos(2x)
dx = 12 x + 21 · 12 sin(2x) + C
2
x
1
2 + 4 sin(2x) + C
Recognition tip: an odd power lets you substitute; an even power forces a power-reduction identity
instead.
Solution 4 (u-substitution).
Let u = ln x, so du = x1 dx — and the x1 dx is already in the integral.
Z
3
u2 du = u3 + C
3
1
3 (ln x) + C
Solution 5 (power rule with roots).
√
3
Rewrite roots as fractional exponents: √
= 3x−1/3 and 5 x = 5x1/2 .
3
x
x2/3
= 29 x2/3
2/3
ˆ
R
3x−1/3 dx = 3 ·
ˆ
R
2
2x4 dx = x5
5
ˆ
R
−5x1/2 dx = −5 ·
x3/2
3/2
= − 10
3 x
3/2
3
3/2
9 2/3
+ 25 x5 − 10
+C
2x
3 x
Solution 6 (chain rule).
Power down, inner kept, power lowered, times inner derivative. Inner derivative of 3x4 + 2x is 12x3 + 2.
dy
= 5(3x4 + 2x)4 (12x3 + 2)
dx
dy
= 5(12x3 + 2)(3x4 + 2x)4 = 10(6x3 + 1)(3x4 + 2x)4
dx
Solution 7 (log-base differentiation).
Rule:
d
g′
loga (g) =
. Here a = 3, g = 2x2 + 5x, g ′ = 4x + 5.
dx
g ln a
f ′ (x) =
4x + 5
(2x2 + 5x) ln 3
Solution 8 (Maclaurin).
f (x) = e2x : f ′ = 2e2x , f ′′ = 4e2x , f ′′′ = 8e2x . At x = 0: f (0) = 1, f ′ (0) = 2, f ′′ (0) = 4, f ′′′ (0) = 8.
P3 (x) = 1 + 2x +
8
4 2
x + x3
2!
3!
P3 (x) = 1 + 2x + 2x2 + 43 x3
Solution 9 (Taylor at a = 4).
f ′′ (4)
(x − 4)2 .
2!
1
f (x) = x1/2 ⇒ f (4) = 2. f ′ (x) = 21 x−1/2 ⇒ f ′ (4) = 41 . f ′′ (x) = − 14 x−3/2 ⇒ f ′′ (4) = − 14 · 18 = − 32
.
Centered at a = 4, the polynomial uses powers of (x − 4): P2 (x) = f (4) + f ′ (4)(x − 4) +
P2 (x) = 2 + 41 (x − 4) +
−1/32
(x − 4)2
2
1
P2 (x) = 2 + 14 (x − 4) − 64
(x − 4)2
Only difference from a Maclaurin: the center is 4, not 0, so terms are powers of (x − 4) and you
evaluate derivatives at 4.
Solution 10 (interval of convergence).
(x − 3)n+1
n 5n
|x − 3|
n
|x − 3|
=
·
·
→
.
(n + 1)5n+1 (x − 3)n
5
n+1
5
|x − 3|
Converge when
< 1 ⇒ |x − 3| < 5 ⇒ −2 < x < 8, so R = 5.
5
Endpoints:
Ratio test:
P 5n
P1
=
— harmonic series (p = 1), diverges. Exclude (round bracket).
n
n5
n
P (−5)n P (−1)n
ˆ x = −2:
=
— alternating harmonic, converges. Include (square bracket).
n 5n
n
ˆ x = 8:
4
R = 5,
Interval [−2, 8)
Note how the two endpoints behave differently — that’s why you always test each one separately.
Solution 11 (geometric series).
P n
Form
ar with first term a = 3 (at n = 0) and ratio r = 25 . Since |r| = 25 < 1, it converges, with
a
.
sum
1−r
3
3
2 = 3 =5
1− 5
5
Converges; sum = 5
a
Recognition tip: n in the exponent means geometric (sum = 1−r
); n raised to a power in the denominator means p-series.
Solution 12 (parametric + orientation).
Solve the simpler equation for t: from x = 2t + 1, t =
y=
x−1
2
2
−3=
x−1
. Substitute into y:
2
(x − 1)2
−3
4
This is an upward-opening parabola with vertex (1, −3).
Orientation: as t increases, x = 2t + 1 always increases, so the point moves left to right. Quick check:
t = −1 → (−1, −2), t = 0 → (1, −3) (vertex), t = 1 → (3, −2) — down the left branch to the vertex,
then up the right branch.
y = 14 (x − 1)2 − 3,
parabola, vertex (1, −3), traced left → right
When the parameter isn’t trig, eliminate it by solving the easy equation for t and substituting — no
identity needed.
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