THE MIT PRECISION BLUEPRINT REAL EXAM ADVANCED INTEGRATION PROMPTS WITH VERIFIED ANSWERS AND FULL ANALYTICAL RATIONALES

THE MIT PRECISION BLUEPRINT REAL
EXAM ADVANCED INTEGRATION
PROMPTS WITH VERIFIED ANSWERS
AND FULL ANALYTICAL RATIONALES


Calculus II Final Exam
1. Evaluate the definite integral: \(\int_{0}^{1} x e^{-x^2} \,
dx\).
A. \(\frac{1}{2}\left(1 - e^{-1}\right)\)
B. \(\frac{1}{2}e^{-1}\)
C. \(1 - e^{-1}\)
D. \(\frac{1}{2}\left(e - 1\right)\)
Correct Answer: A
Rationale: Let \(u = -x^2\), then \(du = -2x \, dx\), which implies \(x \,
dx = -\frac{1}{2} \, du\). Changing the limits of integration: when \(x =
0\), \(u = 0\); when \(x = 1\), \(u = -1\). The integral becomes \(-
\frac{1}{2} \int_{0}^{-1} e^u \, du = \frac{1}{2} \int_{-1}^{0} e^u \,
du = \frac{1}{2} \left[e^u\right]_{-1}^{0} = \frac{1}{2}\left(1 - e^{-
1}\right)\).


2. Find the area of the region bounded by the curves \(y =
x^2\) and \(y = \sqrt{x}\).
A. \(\frac{1}{2}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{4}\)
D. \(\frac{2}{3}\)
Correct Answer: B
Rationale: The curves intersect where \(x^2 = \sqrt{x}\), which gives
\(x^4 = x \implies x(x^3 - 1) = 0\), so \(x = 0\) and \(x = 1\). In the
interval \([0, 1]\), \(\sqrt{x} \geq x^2\). The area is \(\int_{0}^{1}
\left(\sqrt{x} - x^2\right) \, dx = \left[\frac{2}{3}x^{3/2} -
\frac{1}{3}x^3\right]_{0}^{1} = \frac{2}{3} - \frac{1}{3} =
\frac{1}{3}\).

,3. Evaluate the indefinite integral using integration by parts:
\(\int x \cos(x) \, dx\).
A. \(x \sin(x) + \cos(x) + C\)
B. \(x \sin(x) - \cos(x) + C\)
C. \(-x \sin(x) + \cos(x) + C\)
D. \(x \cos(x) + \sin(x) + C\)
Correct Answer: A
Rationale: Let \(u = x\) and \(dv = \cos(x) \, dx\). Then \(du = dx\) and
\(v = \sin(x)\). Using the integration by parts formula \(\int u \, dv =
uv - \int v \, du\), we get \(x \sin(x) - \int \sin(x) \, dx = x \sin(x) - (-
\cos(x)) + C = x \sin(x) + \cos(x) + C\).


4. Determine the volume of the solid generated by revolving
the region bounded by \(y = x^3\), \(y = 0\), and \(x = 2\)
about the x-axis.
A. \(\frac{128\pi }{7}\)
B. \(\frac{64\pi }{7}\)
C. \(\frac{32\pi }{5}\)
D. \(\frac{256\pi }{7}\)
Correct Answer: A
Rationale: Using the disk method, the volume \(V = \pi \int_{a}^{b}
[f(x)]^2 \, dx\). Here, \(V = \pi \int_{0}^{2} (x^3)^2 \, dx = \pi
\int_{0}^{2} x^6 \, dx = \pi \left[\frac{x^7}{7}\right]_{0}^{2} = \pi
\left(\frac{128}{7} - 0\right) = \frac{128\pi}{7}\). [1]


5. Evaluate the improper integral: \(\int_{1}^{\infty}
\frac{1}{x^3} \, dx\).
A. \(1\)
B. \(\frac{1}{2}\)
C. \(2\)
D. Diverges
*Correct Answer: B
Rationale: The improper integral is defined as \(\lim_{t \to \infty}
\int_{1}^{t} x^{-3} \, dx = \lim_{t \to \infty} \left[-
\frac{1}{2x^2}\right]_{1}^{t} = \lim_{t \to \infty} \left(-

,\frac{1}{2t^2} + \frac{1}{2}\right) = 0 + \frac{1}{2} = \frac{1}{2}\).
Since the limit exists and is finite, the integral converges to
\(\frac{1}{2}\).


6. Which of the following substitutions is most appropriate to
evaluate \(\int \frac{1}{x^2 \sqrt{4 - x^2}} \, dx\)?
A. \(x = 2 \tan(\theta)\)
B. \(x = 2 \sec(\theta)\)
C. \(x = 2 \sin(\theta)\)
D. \(x = 4 \sin(\theta)\)
Correct Answer: C
Rationale: The integrand contains an expression of the form
\(\sqrt{a^{2}-x^{2}}\) where \(a = 2\). The standard trigonometric
substitution for this form is \(x = a \sin(\theta)\), hence \(x = 2
\sin(\theta)\) is the correct choice because it simplifies the radical via
the identity \(4 - 4\sin^2(\theta) = 4\cos^2(\theta)\).


7. Find the partial fraction decomposition form for the
integrand: \(\int \frac{2x + 1}{(x-1)^2(x^2 + 1)} \, dx\).
A. \(\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx + D}{x^2 + 1}\)
B. \(\frac{A}{x-1} + \frac{B}{x-1} + \frac{C}{x^2 + 1}\)
C. \(\frac{A}{(x-1)^2} + \frac{Bx + C}{x^2 + 1}\)
D. \(\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x + 1} + \frac{D}{x -
1}\)
Correct Answer: A
Rationale: The denominator contains a repeated linear factor \((x-
1)^2\) and an irreducible quadratic factor \((x^2+1)\). The repeated
linear factor requires terms for each power up to its multiplicity:
\(\frac{A}{x-1} + \frac{B}{(x-1)^2}\). The irreducible quadratic factor
requires a linear numerator: \(\frac{Cx+D}{x^{2}+1}\). Combining
these gives option A.


8. Find the length of the arc of the curve \(y =
\frac{2}{3}x^{3/2}\) from \(x = 0\) to \(x = 3\).
A. \(\frac{14}{3}\)
B. \(\frac{16}{3}\)

, C. \(\frac{26}{3}\)
D. \(4\)
Correct Answer: A
Rationale: The arc length formula is \(L = \int_{a}^{b} \sqrt{1 +
\left(\frac{dy}{dx}\right)^2} \, dx\). Here, \(\frac{dy}{dx} =
\frac{2}{3} \cdot \frac{3}{2}x^{1/2} = x^{1/2}\). Thus, \(L =
\int_{0}^{3} \sqrt{1 + x} \, dx\). Let \(u = 1+x\), then \(du = dx\). The
limits change from \([0,3]\) to \([1,4]\). \(L = \int_{1}^{4} u^{1/2} \,
du = \left[\frac{2}{3}u^{3/2}\right]_{1}^{4} = \frac{2}{3}(4^{3/2} -
1^{3/2}) = \frac{2}{3}(8 - 1) = \frac{14}{3}\).


9. Determine whether the sequence \(a_n = \frac{3n^2 +
5}{2n^2 - n}\) converges, and if so, find its limit.
A. Converges to \(0\)
B. Converges to \(\frac{3}{2}\)
C. Converges to \(3\)
D. Diverges
Correct Answer: B
Rationale: To find the limit of the sequence as \(n \to \infty\), divide the
numerator and the denominator by the highest power of \(n\) in the
denominator, which is \(n^{2}\): \(\lim_{n \to \infty} \frac{3 +
5/n^2}{2 - 1/n} = \frac{3 + 0}{2 - 0} = \frac{3}{2}\). Since the limit is a
finite number, the sequence converges to \(\frac{3}{2}\).


10. Evaluate the geometric series: \(\sum_{n=1}^{\infty} 3
\left(\frac{2}{5}\right)^{n-1}\).
A. \(5\)
B. \(\frac{5}{2}\)
C. \(2\)
D. \(\frac{15}{2}\)
Correct Answer: A
Rationale: A geometric series \(\sum_{n=1}^{\infty} a r^{n-1}\)
converges if \(\vert{}r\vert{} < 1\), and its sum is given by \(S =
\frac{a}{1 - r}\). For this series, the first term \(a = 3\) (when \(n=1\))
and the common ratio \(r = \frac{2}{5}\). Since
\(\left\vert{}\frac{2}{5}\right\vert{} < 1\), the sum is \(S = \frac{3}{1 -
2/5} = \frac{3}{3/5} = 3 \cdot \frac{5}{3} = 5\).