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Integral Calculus: JEE Main Mathematics Question with Solution

Let

f_n = \int_{0}^{\frac{\pi}{2}} \left(\sum_{k=1}^{n} \sin^{k-1} x\right) \left(\sum_{k=1}^{n} (2k - 1) \sin^{k-1} x\right) \cos x \, dx, n \in \mathbb{N}

. Then

f_{21} - f_{20}

is equal to

\underline{\quad\quad\quad}

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Step-by-step solutionView

Correct answer

The value of $f_{21} - f_{20}$ is equal to 41.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

$f_n = \int_{0}^{\frac{\pi}{2}} \left(\sum_{k=1}^{n} \sin^{k-1} x\right) \left(\sum_{k=1}^{n} (2k - 1) \sin^{k-1} x\right) \cos x \, dx$ , where $n \in \mathbb{N}$ .

02approach

Perform the substitution $u = \sin x$ , which converts the integral into $\int_0^1 \left(\sum_{k=1}^{n} u^{k-1}\right) \left(\sum_{k=1}^{n} (2k - 1) u^{k-1}\right) du$ . Multiply the first term by $\sqrt{u}$ and divide the differential by $\sqrt{u}$ , noting that the derivative of $u^{k - 1/2}$ produces $(k - 1/2) u^{k - 3/2} = \frac{1}{2}(2k-1)u^{k-1}/\sqrt{u}$ .

03execute

Let $t = \sum_{k=1}^n u^{k - 1/2} = u^{1/2} + u^{3/2} + \dots + u^{n - 1/2}$ . Then $dt = \sum_{k=1}^n \left(k - \frac{1}{2}\right) u^{k - 3/2} du = \frac{1}{2\sqrt{u}} \left(\sum_{k=1}^n (2k - 1) u^{k-1}\right) du$ . Thus, $\left(\sum_{k=1}^n (2k - 1) u^{k-1}\right) \frac{du}{\sqrt{u}} = 2 \, dt$ . At $u = 0$ , $t = 0$ . At $u = 1$ , $t = n$ . The integral becomes: $f_n = \int_0^n 2t \, dt = \left[ t^2 \right]_0^n = n^2.$

04verify

For $n=1$ : $f_1 = \int_0^{\pi/2} 1 \cdot 1 \cdot \cos x \, dx = [\sin x]_0^{\pi/2} = 1 = 1^2$ , which matches $f_n = n^2$ . Thus, $f_{21} - f_{20} = 21^2 - 20^2 = 441 - 400 = 41$ .

05given

Let $P_n(u) = \sum_{k=1}^n u^{k-1}$ and $Q_n(u) = \sum_{k=1}^n (2k-1)u^{k-1}$ , so $f_n = \int_0^1 P_n(u) Q_n(u) du$ .

06approach

Note that $P_n(u) = \frac{1-u^n}{1-u}$ , and $Q_n(u) = 2u P_n'(u) + P_n(u)$ . Therefore, $\frac{d}{du}\left[ u (P_n(u))^2 \right] = (P_n(u))^2 + 2u P_n(u) P_n'(u) = P_n(u) Q_n(u)$ .

07execute

$f_n = \int_0^1 \frac{d}{du}\left[ u (P_n(u))^2 \right] du = \left[ u (P_n(u))^2 \right]_0^1 = 1 \cdot (P_n(1))^2 - 0 = n^2.$ Therefore, $f_{21} - f_{20} = 21^2 - 20^2 = (21 - 20)(21 + 20) = 41$ .

✓verify

$21^2 - 20^2 = 441 - 400 = 41$ . Consistent across both methods.

✓ Source and academic review

Question type: Numerical
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

Why is the upper limit for t equal to n when u = 1?

Because t is the sum of n terms, each of the form u^(k - 1/2). When u = 1, each term is 1^(k - 1/2) = 1, so t = 1 + 1 + ... + 1 (n times) = n.

Answer

The value of $f_{21} - f_{20}$ is equal to 41.

Why each option works or fails

Step-by-step solution

given: $f_n = \int_{0}^{\frac{\pi}{2}} \left(\sum_{k=1}^{n} \sin^{k-1} x\right) \left(\sum_{k=1}^{n} (2k - 1) \sin^{k-1} x\right) \cos x \, dx$ , where $n \in \mathbb{N}$ .
approach: Perform the substitution $u = \sin x$ , which converts the integral into $\int_0^1 \left(\sum_{k=1}^{n} u^{k-1}\right) \left(\sum_{k=1}^{n} (2k - 1) u^{k-1}\right) du$ . Multiply the first term by $\sqrt{u}$ and divide the differential by $\sqrt{u}$ , noting that the derivative of $u^{k - 1/2}$ produces $(k - 1/2) u^{k - 3/2} = \frac{1}{2}(2k-1)u^{k-1}/\sqrt{u}$ .
execute: Let $t = \sum_{k=1}^n u^{k - 1/2} = u^{1/2} + u^{3/2} + \dots + u^{n - 1/2}$ . Then $dt = \sum_{k=1}^n \left(k - \frac{1}{2}\right) u^{k - 3/2} du = \frac{1}{2\sqrt{u}} \left(\sum_{k=1}^n (2k - 1) u^{k-1}\right) du$ . Thus, $\left(\sum_{k=1}^n (2k - 1) u^{k-1}\right) \frac{du}{\sqrt{u}} = 2 \, dt$ . At $u = 0$ , $t = 0$ . At $u = 1$ , $t = n$ . The integral becomes: $f_n = \int_0^n 2t \, dt = \left[ t^2 \right]_0^n = n^2.$
verify: For $n=1$ : $f_1 = \int_0^{\pi/2} 1 \cdot 1 \cdot \cos x \, dx = [\sin x]_0^{\pi/2} = 1 = 1^2$ , which matches $f_n = n^2$ . Thus, $f_{21} - f_{20} = 21^2 - 20^2 = 441 - 400 = 41$ .
given: Let $P_n(u) = \sum_{k=1}^n u^{k-1}$ and $Q_n(u) = \sum_{k=1}^n (2k-1)u^{k-1}$ , so $f_n = \int_0^1 P_n(u) Q_n(u) du$ .
approach: Note that $P_n(u) = \frac{1-u^n}{1-u}$ , and $Q_n(u) = 2u P_n'(u) + P_n(u)$ . Therefore, $\frac{d}{du}\left[ u (P_n(u))^2 \right] = (P_n(u))^2 + 2u P_n(u) P_n'(u) = P_n(u) Q_n(u)$ .
execute: $f_n = \int_0^1 \frac{d}{du}\left[ u (P_n(u))^2 \right] du = \left[ u (P_n(u))^2 \right]_0^1 = 1 \cdot (P_n(1))^2 - 0 = n^2.$ Therefore, $f_{21} - f_{20} = 21^2 - 20^2 = (21 - 20)(21 + 20) = 41$ .
verify: $21^2 - 20^2 = 441 - 400 = 41$ . Consistent across both methods.

Integral Calculus: JEE Main Mathematics Question with Solution

The value of f21−f20f_{21} - f_{20}f21​−f20​ is equal to 41.

01given

02approach

03execute

04verify

05given

06approach

07execute

✓verify

Students also ask

The value of $f_{21} - f_{20}$ is equal to 41.