JEE Main 2014MathematicsDefinite IntegrationHardMCQ

JEE Main 2014 — Definite Integration Question with Solution

JEE Main 2014 (19 Apr Online)

Question

Let, the function F be defined as Fx=∫1xettdt, x>0, then the value of the integral ∫1xett+adt, where a>0, is

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Show full solutionCorrect option: D
Correct answer
De-aFx+a-F1+a

Step-by-step explanation

Given Fx=∫1xettdt

Let, I=∫1xett+adt

Let, t+a=y   ⇒dt=dy

Also, t=1 ⇒y=1+a and t=x ⇒y=x+a

∴  I=∫1+ax+aey-aydy

⇒I=e-a∫1+ax+aeyydy

Using ∫abfxdx=∫abftdt, we get

I=e-a∫1+ax+aettdt

⇒I=e-a∫11+aettdt+∫1+ax+aettdt-∫11+aettdt

Using ∫abfxdx+∫bcfxdx=∫acfxdx, a<c<b,

⇒I=e-a∫1x+aettdt-∫11+aettdt

Using, the given relation, we get

I=e-aFx+a-F1+a.

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About this question

This is a previous-year question from JEE Main 2014, 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.