Integrals MCQs in Calculus

By: Prof. Dr. Fazal Rehman Shamil | Last updated: February 12, 2025

\[
\textbf{Question 1:}
\]

\[
\text{What is the integral of } \int 5x^4 \, dx?
\]

\[
\text{(a) } x^5 + C, \quad \text{(b) } \frac{5x^5}{5} + C, \quad \text{(c) } x^5 + 5C, \quad \text{(d) } \frac{x^5}{5} + C
\]

\[
\text{Answer: D}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\int 5x^4 \, dx = 5 \int x^4 \, dx
\]

\[
= 5 \times \frac{x^{5}}{5} + C
\]

\[
= x^5 + C
\]

\[
\text{Thus, the correct answer is } \boxed{x^5 + C}.
\]

\[
\textbf{Question 2:}
\]

\[
\text{Evaluate the integral: } \int_0^1 3x^2 \, dx
\]

\[
\text{(a) } 3, \quad \text{(b) } 1, \quad \text{(c) } 2, \quad \text{(d) } 0
\]

\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\int 3x^2 \, dx = 3 \times \frac{x^3}{3} = x^3
\]

\[
\text{Evaluating from } 0 \text{ to } 1:
\]

\[
\left[ x^3 \right]_0^1 = 1^3 – 0^3 = 1 – 0 = 1
\]

\[
\text{Thus, the correct answer is } \boxed{1}.
\]

\[
\textbf{Question 3:}
\]

\[
\text{Find the integral of } \int \sin(x) \, dx.
\]

\[
\text{(a) } \cos(x) + C, \quad \text{(b) } -\cos(x) + C, \quad \text{(c) } \sin(x) + C, \quad \text{(d) } -\sin(x) + C
\]

\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\int \sin(x) \, dx = -\cos(x) + C
\]

\[
\text{Thus, the correct answer is } \boxed{-\cos(x) + C}.
\]

\[
\textbf{Question 4:}
\]

\[
\text{What is the integral of } \int e^{3x} \, dx?
\]

\[
\text{(a) } \frac{e^{3x}}{3} + C, \quad \text{(b) } e^{3x} + C, \quad \text{(c) } \frac{e^{x}}{3} + C, \quad \text{(d) } \frac{e^{3x}}{9} + C
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\int e^{3x} \, dx = \frac{e^{3x}}{3} + C
\]

\[
\text{Thus, the correct answer is } \boxed{\frac{e^{3x}}{3} + C}.
\]

\[
\textbf{Question 5:}
\]

\[
\text{Evaluate the integral: } \int 4x \, dx.
\]

\[
\text{(a) } 2x^2 + C, \quad \text{(b) } 4x^2 + C, \quad \text{(c) } 2x + C, \quad \text{(d) } 4x + C
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\int 4x \, dx = 4 \times \frac{x^2}{2} + C
\]

\[
= 2x^2 + C
\]

\[
\text{Thus, the correct answer is } \boxed{2x^2 + C}.
\]

\[
\textbf{Question 6:}
\]

\[
\text{Find the integral of } \int_1^2 \frac{1}{x} \, dx.
\]

\[
\text{(a) } \ln(1), \quad \text{(b) } \ln(2), \quad \text{(c) } \ln(3), \quad \text{(d) } 0
\]

\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\int \frac{1}{x} \, dx = \ln|x| + C
\]

\[
\text{Evaluating from } 1 \text{ to } 2:
\]

\[
\left[ \ln|x| \right]_1^2 = \ln(2) – \ln(1)
\]

\[
= \ln(2) – 0 = \ln(2)
\]

\[
\text{Thus, the correct answer is } \boxed{\ln(2)}.
\]

\[
\textbf{Question 7:}
\]

\[
\text{What is the integral of } \int \cos^2(x) \, dx?
\]

\[
\text{(a) } {\frac{1}{2} x + \frac{1}{4} \sin(2x) + C}. \text{(b) } \frac{1}{2} \cos(2x) + C, \quad \text{(c) } \frac{1}{2} x + C, \quad \text{(d) } \sin(x) + C
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

Using the identity:

\[
\cos^2(x) = \frac{1 + \cos(2x)}{2}
\]

\[
\int \cos^2(x) \, dx = \int \frac{1 + \cos(2x)}{2} \, dx
\]

\[
= \frac{1}{2} \int (1 + \cos(2x)) \, dx
\]

\[
= \frac{1}{2} \left( x + \frac{\sin(2x)}{2} \right) + C
\]

\[
\text{Thus, the correct answer is } \boxed{\frac{1}{2} x + \frac{1}{4} \sin(2x) + C}.
\]

\[
\textbf{Question 8:}
\]

\[
\text{Evaluate the integral: } \int_0^\pi \sin(x) \, dx.
\]

\[
\text{(a) } 0, \quad \text{(b) } 1, \quad \text{(c) } 2, \quad \text{(d) } -1
\]

\[
\text{Answer: C}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\int \sin(x) \, dx = -\cos(x) + C
\]

\[
\text{Evaluating from } 0 \text{ to } \pi:
\]

\[
\left[ -\cos(x) \right]_0^\pi = -\cos(\pi) + \cos(0)
\]

\[
= -(-1) + 1 = 1 + 1 = 2
\]

\[
\text{Thus, the correct answer is } \boxed{2}.
\]

\[
\textbf{Question 9:}
\]

\[
\text{What is the integral of } \int \frac{1}{x^2} \, dx?
\]

\[
\text{(a) } -\frac{1}{x} + C, \quad \text{(b) } \frac{1}{x} + C, \quad \text{(c) } \ln(x) + C, \quad \text{(d) } -\ln(x) + C
\]

\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\int x^{-2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C
\]

\[
\text{Thus, the correct answer is } \boxed{-\frac{1}{x} + C}.
\]

\[
\textbf{Question 10:}
\]

\[
\text{Find the integral of } \int x \cos(x) \, dx.
\]

\[
\text{(a) } x \sin(x) – \cos(x) + C, \quad \text{(b) } x \sin(x) + \cos(x) + C, \quad \text{(c) } x \sin(x) + C, \quad \text{(d) } -x \sin(x) + \cos(x) + C
\]

\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

Using integration by parts:

\[
\text{Let } u = x, \quad dv = \cos(x) \, dx
\]

\[
du = dx, \quad v = \sin(x)
\]

\[
\int x \cos(x) \, dx = x \sin(x) – \int \sin(x) \, dx
\]

\[
= x \sin(x) + \cos(x) + C
\]

\[
\text{Thus, the correct answer is } \boxed{x \sin(x) + \cos(x) + C}.
\]

\[
\textbf{Question 11:}
\]

\[
\text{Evaluate the integral: } \int_0^1 x^3 \, dx.
\]
\[
\text{(a) } \frac{1}{6}, \quad \text{(b) } \frac{1}{5}, \quad \text{(c) } \frac{1}{4}, \quad \text{(d) } 1
\]
\[
\text{Answer: C}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]

\[
\int x^3 \, dx = \frac{x^4}{4} + C
\]

\[
\text{Evaluating from } 0 \text{ to } 1:
\]

\[
\left[ \frac{x^4}{4} \right]_0^1 = \frac{1^4}{4} – \frac{0^4}{4} = \frac{1}{4} – 0 = \frac{1}{4}
\]

\[
\text{Thus, the correct answer is } \boxed{\frac{1}{4}}.
\]

Question 12:

\[
\text{What is the integral of } \int \tan(x) \, dx?
\]
\[
\text{(a) } \ln|\cos(x)| + C, \quad \text{(b) } \ln|\sec(x)| + C, \quad \text{(c) } \ln|\sin(x)| + C, \quad \text{(d) } -\ln|\sin(x)| + C
\]
Answer: B

Step by Step Solution

Solution:

\[
\int \tan(x) \, dx = \ln|\sec(x)| + C
\]
\[
\text{Thus, the correct answer is } \boxed{\ln|\sec(x)| + C}.
\]

Question 13:

\[
\text{The integral of the function } f(x) = 7x \text{ is:}
\]
\[
\text{(a) } \frac{7x^2}{2} + C, \quad \text{(b) } 7x^2 + C, \quad \text{(c) } \frac{x^2}{7} + C, \quad \text{(d) } 7x + C
\]
Answer: A

Step by Step Solution

Solution:

\[
\int 7x \, dx = \frac{7x^2}{2} + C
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{7x^2}{2} + C}.
\]

Question 14:

\[
\text{Find the integral of the function } \int_1^2 3x^2 \, dx.
\]
\[
\text{(a) } 9, \quad \text{(b) } 6, \quad \text{(c) } 8, \quad \text{(d) } 7
\]
Answer: D

Step by Step Solution

Solution:

\[
\int 3x^2 \, dx = x^3 + C
\]
\[
\text{Evaluating from } 1 \text{ to } 2:
\]
\[
\left[ x^3 \right]_1^2 = 2^3 – 1^3 = 8 – 1 = 7
\]
\[
\text{Thus, the correct answer is } \boxed{7}.
\]

Question 15:

\[
\text{What is the value of the integral } \int_0^\infty e^{-x} \, dx?
\]
\[
\text{(a) } 0, \quad \text{(b) } 1, \quad \text{(c) } -1, \quad \text{(d) } \infty
\]
Answer: B

Step by Step Solution

Solution:

\[
\int e^{-x} \, dx = -e^{-x} + C
\]
\[
\text{Evaluating from } 0 \text{ to } \infty:
\]
\[
\left[ -e^{-x} \right]_0^\infty = \left( -e^{-\infty} \right) – (-e^{0})
\]
\[
= (-0) – (-1) = 1
\]
\[
\text{Thus, the correct answer is } \boxed{1}.
\]

More MCQs on Calculus

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