Let \left( 2^{1 - a} + 2^{1 + a} \right),f(a),\left( 3^{a} + 3^{- a} \right) be in A.P. and \alpha be the minimum value of f(a) . Then the value of the integral \int_{\log_{e}(\alpha - 1)}^{\log_{e}(\alpha)}\frac{dx}{\left( e^{2x} - e^{- 2x} \right)} is :

\frac{1}{4}\log_{e}\left( \frac{4}{3} \right)

\frac{1}{2}\log_{e}\left( \frac{4}{3} \right)

\frac{1}{2}\log_{e}\left( \frac{8}{5} \right)

\frac{1}{4}\log_{e}\left( \frac{8}{5} \right)

Let [.] denote the greatest integer function. Then the value of \int_{0}^{3}\left( \frac{e^{x} + e^{- x}}{\lbrack x\rbrack!} \right)dx is :

\frac{1}{2}\left( e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}} \right)

e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}}

e^{2} + e^{3} - \frac{1}{2e^{2}} - \frac{1}{2e^{3}}

\frac{1}{2}(e^{2} + e^{3}) - \frac{1}{e^{2}} - \frac{1}{e^{3}}

JEE Main · Mathematics · Chapter 17

Definite Integrals JEE Main PYQs (2020–2026)

259 questions from Definite Integrals have appeared in JEE Main between 2020 and 2026, including 183 in the last five papers (2022–2026). The most-asked topic is Fundamental Theorem of Calculus & Standard Definite Integrals (83 questions). Practise them all with instant answers and step-by-step solutions.

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Solved Definite Integrals PYQs

3 recent questions with answers. The full set of 259 is free to practise.

JEE Main 2026EasyFundamental Theorem of Calculus & Standard Definite Integrals
Q1
Let $(2^{1 - a} + 2^{1 + a}), f (a), (3^{a} + 3^{- a})$ be in A.P. and $α$ be the minimum value of $f (a)$ . Then the value of the integral $\int_{l o g_{e} (α - 1)}^{l o g_{e} (α)} \frac{d x}{( e ^{2 x} - e ^{- 2 x} )}$ is :
- A $\frac{1}{2} lo g_{e} (\frac{4}{3})$
- B $\frac{1}{4} lo g_{e} (\frac{4}{3})$
- C $\frac{1}{2} lo g_{e} (\frac{8}{5})$
- D $\frac{1}{4} lo g_{e} (\frac{8}{5})$
Show answer & solution
Correct answer: (B)
$f (a) = \frac{1}{2} (2 (2^{a} + 2^{- a}) + (3^{a} + 3^{- a})) a = \frac{1}{2} (2 \times 2 + 2) = 3 I = \int_{l n (3 - 1)}^{l n 3} \frac{e ^{2 x}}{e ^{4 x} - 1} d x put e^{2 x} = t I = \frac{1}{2} \int_{4}^{9} \frac{d t}{t ^{2} - 1} I = \frac{1}{4} (ln (\frac{t - 1}{t + 1}))_{4}^{9} I = \frac{1}{4} ln (\frac{4}{3})$
JEE Main 2026EasyDefinite Integration involving Modulus and Step Functions
Q2
Let [.] denote the greatest integer function. Then the value of $\int_{0}^{3} (\frac{e ^{x} + e ^{- x}}{[ x ]!}) d x$ is :
- A $e^{2} + e^{3} - \frac{1}{e ^{2}} - \frac{1}{e ^{3}}$
- B $\frac{1}{2} (e^{2} + e^{3} - \frac{1}{e ^{2}} - \frac{1}{e ^{3}})$
- C $e^{2} + e^{3} - \frac{1}{2 e ^{2}} - \frac{1}{2 e ^{3}}$
- D $\frac{1}{2} (e^{2} + e^{3}) - \frac{1}{e ^{2}} - \frac{1}{e ^{3}}$
Show answer & solution
Correct answer: (B)
$\int_{0}^{1} \frac{e ^{x} + e ^{- x}}{1} d x + \int_{1}^{2} \frac{e ^{x} + e ^{- x}}{1} d x + \int_{2}^{3} \frac{e ^{x} + e ^{- x}}{2} d x$

= $\int_{0}^{2} e^{x} d x + \int_{0}^{2} e^{- x} d x + \frac{1}{2} \int_{2}^{3} (e^{x} + e^{- x}) d x$

= $(e^{2} - 1) + \frac{e ^{- 2} - 1}{- 1} + \frac{1}{2} ((e^{3} - e^{2}) - (e^{- 3} - e^{- 2}))$

= $\frac{e ^{3} + e ^{2}}{2} + 1 - 1 - \frac{e ^{- 2}}{2} - \frac{e ^{- 3}}{2}$

= $\frac{e ^{3} + e ^{2} - e ^{- 2} - e ^{- 3}}{2}$
JEE Main 2026EasyLeibniz Rule of Differentiation under Integral Sign
Q3
Let $f : [1, \infty) \to R$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is :
- A4
- B3
- C-4
- D-3
Show answer & solution
Correct answer: (B)
$6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{2} - 4 By using leibnz rule 6 f (x) = 3 f (x) + 3 x f^{'} (x) + 3 x^{2} \frac{f ( x ) - x f ^{'} ( x )}{x ^{2}} = 1 \frac{d}{d x} (\frac{f ( x )}{x}) = - 1 f (x) = - x^{2} + k x 1 = - 1 + k (since f (1) = 1) f (x) = - x^{2} + 2 x f (2) - f (3) = 3$