MAT 251 CALCULUS II ULTIMATE EXAM
PREP QUESTIONS AND ANSWERS WITH
STEP-BY-STEP RATIONALES
This study package contains 200 comprehensive multiple-choice
questions specifically engineered for Calculus II assessments,
featuring integrated correct answers and detailed step-by-step
rationales. Each question aligns with advanced integration
techniques, sequences, series, and polar coordinate transformations
to ensure thorough concept mastery. Ideal for student looking to
maximize their test scores, this document serves as an excellent
resource for final exam preparation and independent self-study.
Part 1: Integration Techniques (Questions 1–15)
Find the indefinite integral \(\int x e^{2x} \, dx\).
A) \(\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + C\)
B) \(\frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x} + C\)
C) \(2xe^{2x} - 4e^{2x} + C\)
D) \(x^2e^{2x} + C\)
Answer: A
Rationale: This problem requires integration by parts using the
formula \(\int u \, dv = uv - \int v \, du\). Let u = x and \(dv = e^{2x}dx\).
This gives du = dx and \(v = \frac{1}{2}e^{2x}\). Substituting these into
the formula yields \(\frac{1}{2}xe^{2x} - \int \frac{1}{2}e^{2x} \, dx\),
which simplifies directly to \(\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} +
C\).
,Evaluate the definite integral \(\int_{0}^{\pi/2} \sin^3(x) \cos(x) \, dx\).
A) \(\frac{1}{2}\)
B) \(\frac{1}{4}\)
C) 1
D) \(\frac{1}{3}\)
Answer: B
Rationale: Use u-substitution. Let \(u = \sin(x)\), which means \(du =
\cos(x)dx\). Changing the integration bounds: when x = 0, u = 0; when
x = π/2, u = 1. The integral transforms into \(\int_{0}^{1} u^3 \, du =
[\frac{1}{4}u^4]_{0}^{1} = \frac{1}{4}\).
Which substitution is most appropriate to evaluate \(\int \frac{1}{x^2
\sqrt{16-x^2}} \, dx\)?
A) \(x = 4\tan(\theta)\)
B) \(x = 4\sec(\theta)\)
C) \(x = 4\sin(\theta)\)
D) \(x = 16\sin(\theta)\)
Answer: C
Rationale: The integrand contains the algebraic form \(\sqrt{a^{2}-
x^{2}}\) where a = 4. The standard trigonometric substitution for this
form is \(x = a\sin(\theta)\), which allows the identity \(1 - \sin^2(\theta)
= \cos^2(\theta)\) to eliminate the square root radical.
Find the form of the partial fraction decomposition for \(\frac{2x+1}{(x-
1)^{2}(x^{2}+4)}\).
A) \(\frac{A}{x-1} + \frac{B}{x^2+4}\)
B) \(\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+4}\)
C) \(\frac{A}{x-1} + \frac{Bx+C}{(x-1)^2} + \frac{Dx+E}{x^2+4}\)
D) \(\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x^2+4}\)
Answer: B
Rationale: A repeated linear factor (x-1)² requires separate terms for
every power up to its multiplicity: \(\frac{A}{x-1} + \frac{B}{(x-1)^2}\).
The irreducible quadratic factor (x²+4) requires a linear numerator of
the form Cx+D.
,Evaluate the improper integral \(\int_{1}^{\infty} \frac{1}{x^3} \, dx\).
A) The integral diverges
B) 1
C) \(\frac{1}{2}\)
D) \(\frac{1}{3}\)
Answer: C
Rationale: Rewrite the improper integral as a limit: \(\lim_{t \to \infty}
\int_{1}^{t} x^{-3} \, dx = \lim_{t \to \infty} [-\frac{1}{2x^2}]_{1}^{t}\).
Evaluating the limit gives \(\lim_{t \to \infty} (-\frac{1}{2t^2} +
\frac{1}{2}) = 0 + \frac{1}{2} = \frac{1}{2}\). This means the integral
converges to \(\frac{1}{2}\).
Use integration by parts to evaluate \(\int \ln(x) \, dx\).
A) \(x\ln(x) - x + C\)
B) \(\frac{1}{x} + C\)
C) \(x\ln(x) + x + C\)
D) \(\frac{\ln(x)}{x} + C\)
Answer: A
Rationale: Set \(u = \ln(x)\) and dv = dx. This leads to \(du =
\frac{1}{x}dx\) and v = x. Applying the integration by parts formula
gives \(uv - \int v \, du = x\ln(x) - \int x \cdot \frac{1}{x} \, dx = x\ln(x) -
\int 1 \, dx = x\ln(x) - x + C\).
Evaluate \(\int x \sqrt{x+3} \, dx\) using an appropriate substitution.
A) \(\frac{2}{5}(x+3)^{5/2} - 2(x+3)^{3/2} + C\)
B) \(\frac{2}{5}(x+3)^{5/2} + \frac{2}{3}(x+3)^{3/2} + C\)
C) \(\frac{1}{2}x^2(x+3)^{3/2} + C\)
D) \(\frac{2}{5}(x+3)^{5/2} - \frac{2}{3}(x+3)^{3/2} + C\)
Answer: D
Rationale: Let u = x+3, which means dx = du and x = u-3. Substituting
these values changes the expression to \(\int (u-3)\sqrt{u} \, du = \int
(u^{3/2} - 3u^{1/2}) \, du\). Integrating term-by-term yields
\(\frac{2}{5}u^{5/2} - 2u^{3/2} + C\). Substituting x+3 back for u results
in \(\frac{2}{5}(x+3)^{5/2} - 2(x+3)^{3/2} + C\), which simplifies exactly
to option D.
, Evaluate \(\int \tan^3(x) \sec^2(x) \, dx\).
A) \(\frac{1}{4}\tan^4(x) + C\)
B) \(\frac{1}{4}\sec^4(x) + C\)
C) \(\frac{1}{3}\tan^3(x) + C\)
D) \(\tan^4(x) - \sec^2(x) + C\)
Answer: A
Rationale: Let \(u = \tan(x)\), which yields \(du = \sec^2(x)dx\).
Substituting these values gives \(\int u^3 \, du = \frac{1}{4}u^4 + C\).
Substituting back for u produces \(\frac{1}{4}\tan^4(x) + C\).
What is the value of \(\int_{0}^{1} \frac{1}{1+x^2} \, dx\)?
A) \(\frac{\pi }{2}\)
B) \(\frac{\pi }{4}\)
C) \(\ln(2)\)
D) 1
Answer: B
Rationale: The antiderivative of \(\frac{1}{1+x^{2}}\) is \(\arctan(x)\).
Evaluating this across the limits gives \(\arctan(1) - \arctan(0) =
\frac{\pi}{4} - 0 = \frac{\pi}{4}\).
Determine if the improper integral \(\int_{0}^{1} \frac{1}{\sqrt{x}} \, dx\)
converges, and find its value if it does.
A) Diverges
B) Converges to 1
C) Converges to 2
D) Converges to \(\frac{1}{2}\)
Answer: C
Rationale: The integrand has an infinite discontinuity at the lower
bound x=0. Express this as a limit: \(\lim_{t \to 0^+} \int_{t}^{1} x^{-
1/2} \, dx = \lim_{t \to 0^+} [2\sqrt{x}]_{t}^{1} = \lim_{t \to 0^+} (2\sqrt{1}
- 2\sqrt{t}) = 2 - 0 = 2\).
Find \(\int \frac{x^2}{x+1} \, dx\) using algebraic long division.
A) \(\frac{1}{2}x^2 - x + \ln\vert{}x+1\vert{} + C\)
B) \(\frac{1}{2}x^2 + x - \ln\vert{}x+1\vert{} + C\)
C) x² - x + C
PREP QUESTIONS AND ANSWERS WITH
STEP-BY-STEP RATIONALES
This study package contains 200 comprehensive multiple-choice
questions specifically engineered for Calculus II assessments,
featuring integrated correct answers and detailed step-by-step
rationales. Each question aligns with advanced integration
techniques, sequences, series, and polar coordinate transformations
to ensure thorough concept mastery. Ideal for student looking to
maximize their test scores, this document serves as an excellent
resource for final exam preparation and independent self-study.
Part 1: Integration Techniques (Questions 1–15)
Find the indefinite integral \(\int x e^{2x} \, dx\).
A) \(\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + C\)
B) \(\frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x} + C\)
C) \(2xe^{2x} - 4e^{2x} + C\)
D) \(x^2e^{2x} + C\)
Answer: A
Rationale: This problem requires integration by parts using the
formula \(\int u \, dv = uv - \int v \, du\). Let u = x and \(dv = e^{2x}dx\).
This gives du = dx and \(v = \frac{1}{2}e^{2x}\). Substituting these into
the formula yields \(\frac{1}{2}xe^{2x} - \int \frac{1}{2}e^{2x} \, dx\),
which simplifies directly to \(\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} +
C\).
,Evaluate the definite integral \(\int_{0}^{\pi/2} \sin^3(x) \cos(x) \, dx\).
A) \(\frac{1}{2}\)
B) \(\frac{1}{4}\)
C) 1
D) \(\frac{1}{3}\)
Answer: B
Rationale: Use u-substitution. Let \(u = \sin(x)\), which means \(du =
\cos(x)dx\). Changing the integration bounds: when x = 0, u = 0; when
x = π/2, u = 1. The integral transforms into \(\int_{0}^{1} u^3 \, du =
[\frac{1}{4}u^4]_{0}^{1} = \frac{1}{4}\).
Which substitution is most appropriate to evaluate \(\int \frac{1}{x^2
\sqrt{16-x^2}} \, dx\)?
A) \(x = 4\tan(\theta)\)
B) \(x = 4\sec(\theta)\)
C) \(x = 4\sin(\theta)\)
D) \(x = 16\sin(\theta)\)
Answer: C
Rationale: The integrand contains the algebraic form \(\sqrt{a^{2}-
x^{2}}\) where a = 4. The standard trigonometric substitution for this
form is \(x = a\sin(\theta)\), which allows the identity \(1 - \sin^2(\theta)
= \cos^2(\theta)\) to eliminate the square root radical.
Find the form of the partial fraction decomposition for \(\frac{2x+1}{(x-
1)^{2}(x^{2}+4)}\).
A) \(\frac{A}{x-1} + \frac{B}{x^2+4}\)
B) \(\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+4}\)
C) \(\frac{A}{x-1} + \frac{Bx+C}{(x-1)^2} + \frac{Dx+E}{x^2+4}\)
D) \(\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x^2+4}\)
Answer: B
Rationale: A repeated linear factor (x-1)² requires separate terms for
every power up to its multiplicity: \(\frac{A}{x-1} + \frac{B}{(x-1)^2}\).
The irreducible quadratic factor (x²+4) requires a linear numerator of
the form Cx+D.
,Evaluate the improper integral \(\int_{1}^{\infty} \frac{1}{x^3} \, dx\).
A) The integral diverges
B) 1
C) \(\frac{1}{2}\)
D) \(\frac{1}{3}\)
Answer: C
Rationale: Rewrite the improper integral as a limit: \(\lim_{t \to \infty}
\int_{1}^{t} x^{-3} \, dx = \lim_{t \to \infty} [-\frac{1}{2x^2}]_{1}^{t}\).
Evaluating the limit gives \(\lim_{t \to \infty} (-\frac{1}{2t^2} +
\frac{1}{2}) = 0 + \frac{1}{2} = \frac{1}{2}\). This means the integral
converges to \(\frac{1}{2}\).
Use integration by parts to evaluate \(\int \ln(x) \, dx\).
A) \(x\ln(x) - x + C\)
B) \(\frac{1}{x} + C\)
C) \(x\ln(x) + x + C\)
D) \(\frac{\ln(x)}{x} + C\)
Answer: A
Rationale: Set \(u = \ln(x)\) and dv = dx. This leads to \(du =
\frac{1}{x}dx\) and v = x. Applying the integration by parts formula
gives \(uv - \int v \, du = x\ln(x) - \int x \cdot \frac{1}{x} \, dx = x\ln(x) -
\int 1 \, dx = x\ln(x) - x + C\).
Evaluate \(\int x \sqrt{x+3} \, dx\) using an appropriate substitution.
A) \(\frac{2}{5}(x+3)^{5/2} - 2(x+3)^{3/2} + C\)
B) \(\frac{2}{5}(x+3)^{5/2} + \frac{2}{3}(x+3)^{3/2} + C\)
C) \(\frac{1}{2}x^2(x+3)^{3/2} + C\)
D) \(\frac{2}{5}(x+3)^{5/2} - \frac{2}{3}(x+3)^{3/2} + C\)
Answer: D
Rationale: Let u = x+3, which means dx = du and x = u-3. Substituting
these values changes the expression to \(\int (u-3)\sqrt{u} \, du = \int
(u^{3/2} - 3u^{1/2}) \, du\). Integrating term-by-term yields
\(\frac{2}{5}u^{5/2} - 2u^{3/2} + C\). Substituting x+3 back for u results
in \(\frac{2}{5}(x+3)^{5/2} - 2(x+3)^{3/2} + C\), which simplifies exactly
to option D.
, Evaluate \(\int \tan^3(x) \sec^2(x) \, dx\).
A) \(\frac{1}{4}\tan^4(x) + C\)
B) \(\frac{1}{4}\sec^4(x) + C\)
C) \(\frac{1}{3}\tan^3(x) + C\)
D) \(\tan^4(x) - \sec^2(x) + C\)
Answer: A
Rationale: Let \(u = \tan(x)\), which yields \(du = \sec^2(x)dx\).
Substituting these values gives \(\int u^3 \, du = \frac{1}{4}u^4 + C\).
Substituting back for u produces \(\frac{1}{4}\tan^4(x) + C\).
What is the value of \(\int_{0}^{1} \frac{1}{1+x^2} \, dx\)?
A) \(\frac{\pi }{2}\)
B) \(\frac{\pi }{4}\)
C) \(\ln(2)\)
D) 1
Answer: B
Rationale: The antiderivative of \(\frac{1}{1+x^{2}}\) is \(\arctan(x)\).
Evaluating this across the limits gives \(\arctan(1) - \arctan(0) =
\frac{\pi}{4} - 0 = \frac{\pi}{4}\).
Determine if the improper integral \(\int_{0}^{1} \frac{1}{\sqrt{x}} \, dx\)
converges, and find its value if it does.
A) Diverges
B) Converges to 1
C) Converges to 2
D) Converges to \(\frac{1}{2}\)
Answer: C
Rationale: The integrand has an infinite discontinuity at the lower
bound x=0. Express this as a limit: \(\lim_{t \to 0^+} \int_{t}^{1} x^{-
1/2} \, dx = \lim_{t \to 0^+} [2\sqrt{x}]_{t}^{1} = \lim_{t \to 0^+} (2\sqrt{1}
- 2\sqrt{t}) = 2 - 0 = 2\).
Find \(\int \frac{x^2}{x+1} \, dx\) using algebraic long division.
A) \(\frac{1}{2}x^2 - x + \ln\vert{}x+1\vert{} + C\)
B) \(\frac{1}{2}x^2 + x - \ln\vert{}x+1\vert{} + C\)
C) x² - x + C
