Integral Calculus Notes for GATE CS

Integral Calculus notes for GATE CS: 28 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

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.

Choosing u and dv: The ILATE Rule

When the integrand is a product of two different function types, use the ILATE priority to select $u$ :

Priority	Function Type	Example
I	Inverse Trigonometric	$sin^{- 1} (x)$
L	Logarithmic	$ln (x)$
A	Algebraic	$x, x^{2}$
T	Trigonometric	$sin (x)$
E	Exponential	$e^{x}$

Rule: The function appearing earlier in the ILATE list is chosen as $u$ . The remaining part becomes $d v$ .

Example: For $\int x ln (x) d x$ , Logarithmic (L) comes before Algebraic (A). Therefore, $u = ln (x)$ and $d v = x d x$ .

Standard Application: Algebraic times Exponential

Standard Application

Evaluate:

\int x e^{x} d x

Step 1: Assign $u$ and $d v$ using ILATE

Algebraic (A) > Exponential (E)

u = x ⟹ d u = d x

d v = e^{x} d x ⟹ v = e^{x}

Step 2: Apply the formula

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

\int x e^{x} d x = (x) (e^{x}) - \int e^{x} d x

Step 3: Evaluate the remaining integral

= x e^{x} - e^{x} + C

= e^{x} (x - 1) + C

Integral Calculus: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Engineering Mathematics MCQ

For any $a, b, c > 0$ , let $I = \int_{- a}^{a} \int_{- b}^{b} \int_{- c}^{c} (x^{2} y z^{2} + x y^{2} z^{3}) d z d y d x$ . What is the minimum value of $I$ ?

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

Correct Answer:

Step-by-Step Solution

Key idea: In a symmetric box, if any term in the integrand is odd in at least one variable, that entire term integrates to zero.

Step 1: Check the limits. The region is a box with limits $[- a, a] \times [- b, b] \times [- c, c]$ , all symmetric about zero.

Step 2: Split the integral into two terms.

$I = ∭ x^{2} y z^{2} d V + ∭ x y^{2} z^{3} d V$

Step 3: Analyze the first term $x^{2} y z^{2}$ .

The power of $y$ is $1$ (odd). Since the $y$ -limits are symmetric, this term integrates to $0$ .

Step 4: Analyze the second term $x y^{2} z^{3}$ .

The power of $x$ is $1$ (odd) and the power of $z$ is $3$ (odd). Since the $x$ and $z$ limits are symmetric, this term also integrates to $0$ .

Step 5: Combine.

$I = 0 + 0 = 0$ . Since $I$ is always exactly $0$ for any $a, b, c > 0$ , its minimum value is $0$ .

Answer: A

Question 2 · Engineering Mathematics MCQ

Which of the following statements about symmetry properties of definite integrals is TRUE?

A.
The Even-Odd rule applies to integrals over $[0, a]$ and King's Rule applies to integrals over $[- a, a]$
B.
King's Rule states that $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (x - a) d x$
C.
The Even-Odd rule applies to integrals over $[- a, a]$ and King's Rule applies to integrals over $[0, a]$
D.
For an odd function, $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$

Correct Answer:

Step-by-Step Solution

Key idea: The Even-Odd rule and King's Rule apply to different interval types. The Even-Odd rule requires a symmetric interval $[- a, a]$ , while King's Rule works on $[0, a]$ .

Step 1: Check option A. It claims Even-Odd applies to $[0, a]$ and King's to $[- a, a]$ . This is backwards. The Even-Odd rule needs the interval to be symmetric about the origin ( $[- a, a]$ ), not $[0, a]$ . FALSE.

Step 2: Check option B. King's Rule states $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ , not $f (x - a)$ . The argument is $(a - x)$ , not $(x - a)$ . FALSE.

Step 3: Check option C. Even-Odd rule → symmetric interval $[- a, a]$ . King's Rule → interval $[0, a]$ . This matches the correct definitions. TRUE.

Step 4: Check option D. For an odd function, $\int_{- a}^{a} f (x) d x = 0$ , not $2 \int_{0}^{a} f (x) d x$ . The doubling formula applies to even functions. FALSE.

Answer: C

Question 3 · Engineering Mathematics MCQ

When proving the Reflection Property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ using the substitution $u = a - x$ , which of the following statements correctly describes the transformation of the differential and the limits?

A.
$d x = d u$ , limits $[0, a] \to [a, 0]$
B.
$d x = - d u$ , limits $[0, a] \to [a, 0]$
C.
$d x = - d u$ , limits $[0, a] \to [0, a]$
D.
$d x = d u$ , limits $[0, a] \to [0, a]$

Correct Answer:

Step-by-Step Solution

Key idea: The substitution $u = a - x$ requires applying the chain rule to the differential and updating the limits of integration.

Step 1: Differentiate the substitution.

$u = a - x ⟹ d u = - d x ⟹ d x = - d u$ .

Step 2: Update the limits of integration.

When $x = 0$ , $u = a - 0 = a$ .

When $x = a$ , $u = a - a = 0$ .

So the limits change from $[0, a]$ to $[a, 0]$ .

Step 3: Combine the transformations.

The differential becomes $- d u$ and the limits become $[a, 0]$ .

(Note: The negative sign from $- d u$ and the flipped limits $[a, 0]$ cancel out to give the final positive integral $\int_{0}^{a} f (u) d u$ ).

Answer: B

Question 4 · Engineering Mathematics MCQ

What is the value of the iterated integral $\int_{- 1}^{1} \int_{- 1}^{1} x y d x d y$ ?

A.
$0$
B.
$1$
C.
$2$
D.
$4$

Correct Answer:

Step-by-Step Solution

Key idea: This is a symmetric-box integral where the integrand is odd in both variables.

Step 1: Check the limits. Both $x$ and $y$ range from $- 1$ to $1$ , which are symmetric about zero.

Step 2: Check the parity of the integrand $f (x, y) = x y$ .

The function is odd in $x$ (since $(- x) y = - x y$ ) and odd in $y$ .

Step 3: Apply the rule. Since the integrand is odd in $x$ and the $x$ -limits are symmetric, the inner integral $\int_{- 1}^{1} x y d x = 0$ for any $y$ .

Step 4: The outer integral of $0$ is $0$ .

Answer: A

Question 5 · Engineering Mathematics MCQ

Let $I (a) = \int_{- a}^{a} (x^{3} + 2 x^{2}) d x$ . What is the maximum value of $I (a)$ for $a \in [0, 2]$ ?

A.
$10.67$
B.
$4.00$
C.
$0.00$
D.
$8.00$

Correct Answer:

Step-by-Step Solution

Key idea: Use the single-variable symmetry rules to simplify the integral, then maximize the resulting function of $a$ .

Step 1: Split the integral using linearity.

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

Step 2: Apply the odd-even rules.

The limits are symmetric ( $[- a, a]$ ). $x^{3}$ is odd, so its integral is $0$ .

$2 x^{2}$ is even, so its integral is $2 \int_{0}^{a} 2 x^{2} d x = 4 \int_{0}^{a} x^{2} d x$ .

Step 3: Evaluate the remaining integral.

$I (a) = 4 [\frac{x ^{3}}{3}]_{0}^{a} = \frac{4 a ^{3}}{3}$

Step 4: Maximize $I (a)$ for $a \in [0, 2]$ .

Since $I (a)$ is strictly increasing for $a \geq 0$ , the maximum occurs at the boundary $a = 2$ .

$I (2) = \frac{4 ( 2 ) ^{3}}{3} = \frac{32}{3} \approx 10.67$ .

Answer: A

Question 6 · Engineering Mathematics MCQ

Consider the following two equations involving an unknown function $f (x)$ :

(P) $f (x) = x + \int_{0}^{x} f (t) d t$

(Q) $f^{'} (x) + 2 f (x) = 0$

Which of the following correctly classifies each equation?

A.
P is an integral equation; Q is a differential equation
B.
P is a differential equation; Q is an integral equation
C.
Both P and Q are integral equations
D.
P is an algebraic equation; Q is a differential equation

Correct Answer:

Step-by-Step Solution

Key idea: An integral equation is one in which the unknown function appears under an integral sign. A differential equation involves derivatives of the unknown function.

Step 1: Examine equation (P). The unknown $f (x)$ appears both outside and inside the integral $\int_{0}^{x} f (t) d t$ . Since $f$ is under the integral sign, (P) is an integral equation.

Step 2: Examine equation (Q). The unknown $f (x)$ appears through its derivative $f^{'} (x)$ and as $f (x)$ itself, but there is no integral sign. This is a differential equation.

Step 3: Match the classification. P → Integral Equation, Q → Differential Equation.

Answer: A

Question 7 · Engineering Mathematics NAT

The number of real values of $x > 1$ satisfying

$\int_{1}^{x} 2 t d t = 8$

Correct Answer:

1.00

Step-by-Step Solution

Key idea: Evaluate the definite integral directly using the Fundamental Theorem of Calculus, then solve the resulting algebraic equation for $x$ .

Step 1: Evaluate the integral.

$\int_{1}^{x} 2 t d t = [t^{2}]_{1}^{x} = x^{2} - 1$

Step 2: Set equal to 8 and solve.

$x^{2} - 1 = 8 ⟹ x^{2} = 9 ⟹ x = 3$ or $x = - 3$

Step 3: Apply the constraint $x > 1$ .

Only $x = 3$ satisfies $x > 1$ . The value $x = - 3$ is rejected.

Step 4: Count the valid solutions.

There is exactly 1 real value.

Answer: 1.00

Question 8 · Engineering Mathematics MCQ

For which of the following integrands $h (x)$ is it IMPOSSIBLE to evaluate $\int_{0}^{π /2} h (x) d x$ as $\frac{π}{4}$ using the standard trigonometric symmetry shortcut $\int_{0}^{π /2} \frac{g ( x )}{g ( x ) + g ( π /2 - x )} d x$ ?

A.
$\frac{s i n ^{2} x}{s i n ^{2} x + c o s ^{2} x}$
B.
$\frac{t a n x}{t a n x + c o t x}$
C.
$\frac{s i n x}{s i n x + c o s ^{2} x}$
D.
$\frac{e ^{s i n x}}{e ^{s i n x} + e ^{c o s x}}$

Correct Answer:

Step-by-Step Solution

Key idea: The trigonometric symmetry shortcut requires the denominator to be exactly $g (x) + g (\frac{π}{2} - x)$ for the same function $g$ .

Step 1: Check each option to see if it fits the pattern.

Option A: Let $g (x) = sin^{2} x$ . Then $g (\frac{π}{2} - x) = cos^{2} x$ . The denominator is $sin^{2} x + cos^{2} x$ , which matches $g (x) + g (\frac{π}{2} - x)$ . (Fits)

Option B: Let $g (x) = tan x$ . Then $g (\frac{π}{2} - x) = cot x$ . The denominator is $tan x + cot x$ , which matches. (Fits)

Option C: If we try $g (x) = sin x$ , then $g (\frac{π}{2} - x) = cos x$ . The denominator would need to be $sin x + cos x$ , but it is $sin x + cos^{2} x$ . This does not match the form for any single function $g$ . (Impossible)

Option D: Let $g (x) = e^{s i n x}$ . Then $g (\frac{π}{2} - x) = e^{c o s x}$ . The denominator matches. (Fits)

Answer: C

Question 9 · Engineering Mathematics NAT

Consider the iterated integral $\int_{0}^{2} \int_{0}^{1} \int_{- 2}^{2} x y z^{2} d z d y d x$ . The innermost definite integral (with respect to $z$ ) evaluates to $k x y$ . What is the numerical value of $k$ (rounded to two decimal places)?

Correct Answer:

5.33

Step-by-Step Solution

Key idea: Evaluate the iterated integral from the inside out, applying the correct power rule and limits.

Step 1: Identify the innermost integral. The rightmost differential is $d z$ , so we integrate with respect to $z$ first, treating $x$ and $y$ as constants.

Step 2: Compute the innermost integral.

$\int_{- 2}^{2} x y z^{2} d z = x y [\frac{z ^{3}}{3}]_{- 2}^{2}$

Step 3: Evaluate the limits.

$(\frac{2 ^{3}}{3}) - (\frac{( - 2 ) ^{3}}{3}) = \frac{8}{3} - (- \frac{8}{3}) = \frac{16}{3}$

Step 4: Find $k$ .

The result is $\frac{16}{3} x y$ , so $k = \frac{16}{3} \approx 5.33$ .

Answer: 5.33

Question 10 · Engineering Mathematics MCQ

When evaluating $\int e^{x} cos x d x$ using the cyclic integration by parts method, the original integral reappears after how many applications of integration by parts?

A.
1
B.
2
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: For integrals of the form $\int e^{a x} cos (b x) d x$ or $\int e^{a x} sin (b x) d x$ , applying integration by parts twice brings back the original integral, allowing you to solve for it algebraically.

Step 1: First application of parts. Let $u = e^{x}$ , $d v = cos x d x$ . Then $d u = e^{x} d x$ , $v = sin x$ .

$I = e^{x} sin x - \int e^{x} sin x d x$

Step 2: Second application of parts on $\int e^{x} sin x d x$ . Let $u = e^{x}$ , $d v = sin x d x$ . Then $d u = e^{x} d x$ , $v = - cos x$ .

$\int e^{x} sin x d x = - e^{x} cos x + \int e^{x} cos x d x = - e^{x} cos x + I$

Step 3: Substitute back.

$I = e^{x} sin x - (- e^{x} cos x + I) = e^{x} sin x + e^{x} cos x - I$

The original integral $I$ reappears after exactly 2 applications.

Answer: B