JEE Main 2023MathematicsDefinite IntegrationEasyMCQ

JEE Main 2023 — Definite Integration Question with Solution

JEE Main 2023 (10 Apr Shift 2)

Question

Let f be a continuous function satisfying ∫0t2f(x)+x2dx=43t3,∀t>0 . Then fπ24 is equal to

Choose an option

Show full solutionCorrect option: C
Correct answer
Cπ1-π316

Step-by-step explanation

Given equation is ∫0t2f(x)+x2dx=43t3,∀t>0

According to Newton Leibnitz theorem we haveddx∫uxvxftdt=fvx×v'x-fux×u'x

Apply Newtons Leibnitz theorem in the given equation.

⇒ft2+t42t-0=4t2

⇒ft2+t4=2t

⇒fx2=-x4+2x

⇒f(x)=-x2+2x

⇒fπ24=-π442+2×π2

=-π416+π

=π1-π316

Hence this is the correct option.

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About this question

This is a previous-year question from JEE Main 2023, covering the Definite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.