JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Integral Calculus: JEE Main Mathematics Question with Solution

Q: Is the trigonometric substitution necessary, or can we directly combine the integrals in terms of x?

Direct combination in x is even faster: .

\mathrm{I}(\mathrm{m},\mathrm{n})=\int_{0}^{1}\mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1}\mathrm{dx},\; \mathrm{m},\; \mathrm{n}>0

, then

\mathrm{I}(9,14)+\mathrm{I}(10,13)

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\mathrm{I}(9,1)

\mathrm{I}(19,27)

\mathrm{I}(1,13)

\mathrm{I}(9,13)

Step-by-step solutionView

Correct answer

By combining the integrals into a single integrand and factoring out $x^8 (1-x)^{12}$ , the sum simplifies directly to $I(9, 13)$ .

Option analysis

Why each option works or fails

A ·

\mathrm{I}(9,1)

Believing that subtracting indices or canceling exponents yields $I(9, 14-13) = I(9, 1)$ . Factor out common powers from the integrands: $x^8(1-x)^{12}[(1-x) + x] = x^8(1-x)^{12}$ , which corresponds to $I(9, 13)$ , not $I(9, 1)$ .

B ·

\mathrm{I}(19,27)

Believing that the integral operator adds the arguments directly as $I(9+10, 14+13) = I(19, 27)$ . Integrals cannot be combined by adding their parameters; write out the integrands explicitly and simplify algebraically.

C ·

\mathrm{I}(1,13)

Incorrectly simplifying the powers to leave only $(1-x)^{12}$ with power 0 on $x$ , leading to $I(1, 13)$ . Ensure you identify the common factor correctly: both terms share $x^8$ , not $x^0$ , giving parameters $m=9$ and $n=13$ .

D ·

\mathrm{I}(9,13)

None. This is the correct option. Factoring the combined integrand gives $\int_0^1 x^8(1-x)^{12}(1-x+x)\,dx = \int_0^1 x^{9-1}(1-x)^{13-1}\,dx = I(9,13)$ .

Reviewed route

Solution

StepWorking

01given

$I(m,n) = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx$ for $m, n > 0$ . We are required to find $I(9,14) + I(10,13)$ .

02approach

Substitute $x = \sin^2\theta$ , so $dx = 2\sin\theta\cos\theta\,d\theta$ . This converts $I(m,n)$ into $2\int_{0}^{\pi/2} (\sin\theta)^{2m-1}(\cos\theta)^{2n-1} d\theta$ . Then combine the integrals using $\sin^2\theta + \cos^2\theta = 1$ .

03execute

For $I(9,14)$ and $I(10,13)$ : $I(9,14) = 2\int_{0}^{\pi/2} (\sin\theta)^{17}(\cos\theta)^{27}\,d\theta$ $I(10,13) = 2\int_{0}^{\pi/2} (\sin\theta)^{19}(\cos\theta)^{25}\,d\theta$ Summing them: $I(9,14) + I(10,13) = 2\int_{0}^{\pi/2} (\sin\theta)^{17}(\cos\theta)^{25} [\cos^2\theta + \sin^2\theta]\,d\theta$ $= 2\int_{0}^{\pi/2} (\sin\theta)^{17}(\cos\theta)^{25}\,d\theta$ Since $2m-1=17 \implies m=9$ and $2n-1=25 \implies n=13$ , this is $I(9,13)$ .

04verify

Check via algebraic property of Beta function: $I(m,n+1) + I(m+1,n) = \int_0^1 x^{m-1}(1-x)^{n-1}[(1-x)+x]dx = \int_0^1 x^{m-1}(1-x)^{n-1}dx = I(m,n)$ . Setting $m=9, n=13$ gives $I(9,14) + I(10,13) = I(9,13)$ .

05given

$I(9,14) = \int_0^1 x^8(1-x)^{13}\,dx$ and $I(10,13) = \int_0^1 x^9(1-x)^{12}\,dx$ .

06execute

$I(9,14) + I(10,13) = \int_0^1 [x^8(1-x)^{13} + x^9(1-x)^{12}]\,dx = \int_0^1 x^8(1-x)^{12}[(1-x) + x]\,dx = \int_0^1 x^8(1-x)^{12}\,dx = I(9,13)$

✓verify

Since $(1-x)+x = 1$ , the degree in $x$ is 8 and in $(1-x)$ is 12, which matches $I(8+1, 12+1) = I(9,13)$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Is the trigonometric substitution necessary, or can we directly combine the integrals in terms of x?

Direct combination in x is even faster: $x^8(1-x)^{13} + x^9(1-x)^{12} = x^8(1-x)^{12}[(1-x) + x] = x^8(1-x)^{12}$ .

Answer

By combining the integrals into a single integrand and factoring out $x^8 (1-x)^{12}$ , the sum simplifies directly to $I(9, 13)$ .

Why each option works or fails

A: $\mathrm{I}(9,1)$ - Believing that subtracting indices or canceling exponents yields $I(9, 14-13) = I(9, 1)$ . Factor out common powers from the integrands: $x^8(1-x)^{12}[(1-x) + x] = x^8(1-x)^{12}$ , which corresponds to $I(9, 13)$ , not $I(9, 1)$ .
B: $\mathrm{I}(19,27)$ - Believing that the integral operator adds the arguments directly as $I(9+10, 14+13) = I(19, 27)$ . Integrals cannot be combined by adding their parameters; write out the integrands explicitly and simplify algebraically.
C: $\mathrm{I}(1,13)$ - Incorrectly simplifying the powers to leave only $(1-x)^{12}$ with power 0 on $x$ , leading to $I(1, 13)$ . Ensure you identify the common factor correctly: both terms share $x^8$ , not $x^0$ , giving parameters $m=9$ and $n=13$ .
D · correct: $\mathrm{I}(9,13)$ - None. This is the correct option. Factoring the combined integrand gives $\int_0^1 x^8(1-x)^{12}(1-x+x)\,dx = \int_0^1 x^{9-1}(1-x)^{13-1}\,dx = I(9,13)$ .

Step-by-step solution

given: $I(m,n) = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx$ for $m, n > 0$ . We are required to find $I(9,14) + I(10,13)$ .
approach: Substitute $x = \sin^2\theta$ , so $dx = 2\sin\theta\cos\theta\,d\theta$ . This converts $I(m,n)$ into $2\int_{0}^{\pi/2} (\sin\theta)^{2m-1}(\cos\theta)^{2n-1} d\theta$ . Then combine the integrals using $\sin^2\theta + \cos^2\theta = 1$ .
execute: For $I(9,14)$ and $I(10,13)$ : $I(9,14) = 2\int_{0}^{\pi/2} (\sin\theta)^{17}(\cos\theta)^{27}\,d\theta$ $I(10,13) = 2\int_{0}^{\pi/2} (\sin\theta)^{19}(\cos\theta)^{25}\,d\theta$ Summing them: $I(9,14) + I(10,13) = 2\int_{0}^{\pi/2} (\sin\theta)^{17}(\cos\theta)^{25} [\cos^2\theta + \sin^2\theta]\,d\theta$ $= 2\int_{0}^{\pi/2} (\sin\theta)^{17}(\cos\theta)^{25}\,d\theta$ Since $2m-1=17 \implies m=9$ and $2n-1=25 \implies n=13$ , this is $I(9,13)$ .
verify: Check via algebraic property of Beta function: $I(m,n+1) + I(m+1,n) = \int_0^1 x^{m-1}(1-x)^{n-1}[(1-x)+x]dx = \int_0^1 x^{m-1}(1-x)^{n-1}dx = I(m,n)$ . Setting $m=9, n=13$ gives $I(9,14) + I(10,13) = I(9,13)$ .
given: $I(9,14) = \int_0^1 x^8(1-x)^{13}\,dx$ and $I(10,13) = \int_0^1 x^9(1-x)^{12}\,dx$ .
execute: $I(9,14) + I(10,13) = \int_0^1 [x^8(1-x)^{13} + x^9(1-x)^{12}]\,dx = \int_0^1 x^8(1-x)^{12}[(1-x) + x]\,dx = \int_0^1 x^8(1-x)^{12}\,dx = I(9,13)$
verify: Since $(1-x)+x = 1$ , the degree in $x$ is 8 and in $(1-x)$ is 12, which matches $I(8+1, 12+1) = I(9,13)$ .

Integral Calculus: JEE Main Mathematics Question with Solution

By combining the integrals into a single integrand and factoring out x8(1−x)12x^8 (1-x)^{12}x8(1−x)12, the sum simplifies directly to I(9,13)I(9, 13)I(9,13).

01given

02approach

03execute

04verify

05given

06execute

✓verify

Students also ask

By combining the integrals into a single integrand and factoring out $x^8 (1-x)^{12}$ , the sum simplifies directly to $I(9, 13)$ .