Integral Calculus PYQs for GATE CS

Solve 4+ Integral Calculus previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Integral Calculus

Chapter Roadmap

Integration by Parts and Integral Equations

Current Topic: Core techniques for products of functions and unknown functions under integrals.

Symmetry Properties of Definite Integrals

Exploiting even and odd functions and interval shifts to simplify evaluations.

Multiple Integrals and Symmetry

Double and triple integrals, changing order of integration, and 3D symmetry.

Integration by Parts: The Intuition

The product rule for differentiation states:

\frac{d}{d x} [u (x) v (x)] = u (x) v^{'} (x) + v (x) u^{'} (x)

Integrating both sides with respect to $x$ and rearranging yields the Integration by Parts formula:

\int u d v = uv - \int v d u

Core Idea: We transform a difficult integral $\int u d v$ into a potentially easier integral $\int v d u$ . The goal is to choose $u$ and $d v$ such that the new integral is simpler than the original one.

Integral Calculus: Solved Questions with Step-by-Step Explanations (4 Problems)

Question 1 · Engineering Mathematics · 2026_Set2 MSQ

For a real number $a$ , let $I (a) = \int_{- 1}^{1} (3 x^{2} - a x + 1) d x$ . Which of the following statements is/are true?

A.
The value of $I (a)$ is independent of the value of $a$
B.
The value of $I (a)$ can vary with the value of $a$
C.
There exists $a \in (- \infty, + \infty)$ such that $I (a)$ is a positive real number
D.
There exists $a \in (- \infty, + \infty)$ such that $I (a)$ is a negative real number

Correct Answer:

["A","C"]

Step-by-Step Solution

Key idea: Split the integral using linearity and apply the Even-Odd function rules over the symmetric limits $[- 1, 1]$ to eliminate the parameter-dependent term.

Step 1: Write the integral as $I (a) = \int_{- 1}^{1} (3 x^{2} - a x + 1) d x$ .

Step 2: Use the linearity of integration to split it into three separate integrals:

$I (a) = \int_{- 1}^{1} 3 x^{2} d x - \int_{- 1}^{1} a x d x + \int_{- 1}^{1} 1 d x$ .

Step 3: Evaluate the middle term. The function $g (x) = a x$ is an odd function because $g (- x) = a (- x) = - a x = - g (x)$ . The integral of any odd function over symmetric limits $[- 1, 1]$ is exactly 0. So, $\int_{- 1}^{1} a x d x = 0$ .

Step 4: Evaluate the remaining terms. Both $3 x^{2}$ and $1$ are even functions.

$\int_{- 1}^{1} 3 x^{2} d x = 2 \int_{0}^{1} 3 x^{2} d x = 2 [x^{3}]_{0}^{1} = 2 (1 - 0) = 2$ .

$\int_{- 1}^{1} 1 d x = 2 \int_{0}^{1} 1 d x = 2 [x]_{0}^{1} = 2 (1 - 0) = 2$ .

Step 5: Combine the results: $I (a) = 2 - 0 + 2 = 4$ .

Step 6: Analyze the options based on $I (a) = 4$ .

The value is always 4, regardless of $a$ . Thus, it is independent of $a$ (Option A is true, Option B is false).
Since 4 is a positive real number, there exists an $a$ (in fact, all real $a$ ) that makes $I (a)$ positive (Option C is true, Option D is false).

Answer: Options A and C.

Question 2 · Engineering Mathematics · 2025_Set2 MCQ

The value of $x$ such that $x > 1$ , satisfying the equation $\int_{1}^{x} t ln t d t = \frac{1}{4}$ is

A.
$e$
B.
$e$
C.
$e^{2}$
D.
$e - 1$

Correct Answer:

Step-by-Step Solution

Key idea: Use Integration by Parts to evaluate the definite integral, then solve the resulting algebraic equation for $x$ .

Step 1: Set up Integration by Parts for the indefinite integral $\int t ln t d t$ . Following the ILATE rule, choose $u = ln t$ (Logarithmic) and $d v = t d t$ (Algebraic).

Step 2: Differentiate $u$ and integrate $d v$ :

$d u = \frac{1}{t} d t$ and $v = \frac{t ^{2}}{2}$ .

Step 3: Apply the Integration by Parts formula $\int u d v = uv - \int v d u$ :

$\int t ln t d t = (ln t) (\frac{t ^{2}}{2}) - \int (\frac{t ^{2}}{2}) (\frac{1}{t}) d t$

$= \frac{t ^{2}}{2} ln t - \frac{1}{2} \int t d t = \frac{t ^{2}}{2} ln t - \frac{t ^{2}}{4}$ .

Step 4: Evaluate the definite integral from 1 to $x$ :

$[\frac{t ^{2}}{2} ln t - \frac{t ^{2}}{4}]_{1}^{x} = (\frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4}) - (\frac{1 ^{2}}{2} ln 1 - \frac{1 ^{2}}{4})$ .

Step 5: Simplify using the fact that $ln 1 = 0$ :

$= \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} - (0 - \frac{1}{4}) = \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} + \frac{1}{4}$ .

Step 6: Set this result equal to the given value $\frac{1}{4}$ :

$\frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} + \frac{1}{4} = \frac{1}{4}$ .

Step 7: Subtract $\frac{1}{4}$ from both sides and factor out $\frac{x ^{2}}{4}$ :

$\frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} = 0 ⟹ \frac{x ^{2}}{4} (2 ln x - 1) = 0$ .

Step 8: Since the problem states $x > 1$ , we know $\frac{x ^{2}}{4} \neq = 0$ . Therefore, we can divide by it, leaving:

$2 ln x - 1 = 0 ⟹ ln x = \frac{1}{2} ⟹ x = e^{1/2} = e$ .

Answer: $e$ (Option A)

Question 3 · Engineering Mathematics · 2024_Set2 MCQ

Let $f (x)$ be a continuous function from $R$ to $R$ such that

$f (x) = 1 - f (2 - x)$

Which one of the following options is the CORRECT value of $\int_{0}^{2} f (x) d x$ ?

A.
0
B.
1
C.
2
D.
$- 1$

Correct Answer:

Step-by-Step Solution

Key idea: Use the Reflection Property (King's Rule) of definite integrals to create a solvable algebraic equation for the unknown integral.

Step 1: Let the required integral be $I = \int_{0}^{2} f (x) d x$ .

Step 2: Apply King's Rule, which states $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ . For $a = 2$ , this gives $I = \int_{0}^{2} f (2 - x) d x$ .

Step 3: Use the given functional equation $f (x) = 1 - f (2 - x)$ . Rearranging this yields $f (2 - x) = 1 - f (x)$ .

Step 4: Substitute this expression into the integral from Step 2:

$I = \int_{0}^{2} (1 - f (x)) d x$ .

Step 5: Use linearity to split the integral:

$I = \int_{0}^{2} 1 d x - \int_{0}^{2} f (x) d x$ .

Step 6: Evaluate the first part and substitute $I$ for the second part:

$I = [x]_{0}^{2} - I = 2 - I$ .

Step 7: Solve the algebraic equation for $I$ :

$2 I = 2 ⟹ I = 1$ .

Answer: 1 (Option B)

Question 4 · Engineering Mathematics · 2023 NAT

The value of the definite integral

\int_{- 3}^{3} \int_{- 2}^{2} \int_{- 1}^{1} (4 x^{2} y - z^{3}) d z d y d x

is __________. (Rounded off to the nearest integer)

Correct Answer:

0.00

Step-by-Step Solution

Insight: Every limit is symmetric about zero and the integrand is a sum of monomials — check parity before integrating. Any term odd in any symmetric variable vanishes.

Exam route (under 30 seconds):

Limits: $x \in [- 3, 3]$ , $y \in [- 2, 2]$ , $z \in [- 1, 1]$ — all symmetric.
Split: $I = ∭ 4 x^{2} y d V - ∭ z^{3} d V$ .
Term $4 x^{2} y$ : power of $y$ is $1$ (odd), $y$ -limits symmetric $\Rightarrow$ this term is $0$ .
Term $z^{3}$ : power of $z$ is $3$ (odd), $z$ -limits symmetric $\Rightarrow$ this term is $0$ .
$I = 0 - 0 = 0$ .

Learning route (full derivation):

Integrate inside out.

Inner: $\int_{- 1}^{1} (4 x^{2} y - z^{3}) d z = [4 x^{2} y z - z^{4} /4]_{- 1}^{1} = (4 x^{2} y - 1/4) - (- 4 x^{2} y - 1/4) = 8 x^{2} y$ .

Middle: $\int_{- 2}^{2} 8 x^{2} y d y = [4 x^{2} y^{2}]_{- 2}^{2} = 4 x^{2} (4) - 4 x^{2} (4) = 0$ .

Outer: $\int_{- 3}^{3} 0 d x = 0$ .

Common trap: brute-forcing the integral without noticing symmetry. Even if you do compute, you must still get $0$ — but you waste a minute and risk arithmetic errors. Always inspect parity first on any multiple integral over a symmetric box.

Verification: substitute the result back — a triple integral of an odd-in- $y$ function over a symmetric $y$ -interval must be $0$ by the one-variable odd-function rule, and the same holds for the $z^{3}$ term. Both routes agree: $0$ .

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Integral Calculus PYQs for GATE CS

Chapter Roadmap: Integral Calculus

Chapter Roadmap

Integration by Parts: The Intuition

Integration by Parts: The Intuition

Integral Calculus: Solved Questions with Step-by-Step Explanations (4 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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