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Question: Answered & Verified by Expert
Let exx2dx=fxex+C (where, C is the constant of integration). The range of fx as xR is a,. The value of a4 is
MathematicsIndefinite IntegrationJEE Main
Solution:
2535 Upvotes Verified Answer
The correct answer is: 0.25
As we know, exfx+f'xdx=exfx+C,
Thus, exx2dx=exx2+2x-2x-2+2dx
=exx2-2x+2+ex2x-2dx
=exx2-2x+2+C
i.e. fx=x2-2x+2=x-12+1
Hence, range is 1,a=1
a4=0.25

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