JEE Main 2024MathematicsDefinite IntegrationEasyNumerical

JEE Main 2024 — Definite Integration Question with Solution

JEE Main 2024 (29 Jan Shift 2)

Question

If ∫π6π31-sin2xdx=α+β2+γ3, where α,β and γ are rational numbers, then 3α+4β-γ is equal to _____.

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Show full solutionCorrect answer: 6
Correct answer
6

Step-by-step explanation

Given,

∫π6π31-sin2xdx=α+β2+γ3

Now, let I=∫π6π31-sin2xdx

⇒I=∫π6π3|sinx-cosx|dx

⇒I=∫π6π4(cosx-sinx)dx+∫π4π3(sinx-cosx)dx

⇒I=-1+22-3

Now, comparing with 

I=α+β2+γ3

We get, α=-1,β=2,γ=-1

Hence, 3α+4β-γ=-3+8+1=6

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About this question

This is a previous-year question from JEE Main 2024, 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.