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Mathematics Question on Relations and Functions

Let f:R→Rf:R→R be defined by f(x)={x if x≤1−x+2 if x>1x \text{ if } x≤1 -x+2 \text{ if } x>1}.Then ∫02f(x)dx∫_0^2f(x)dx=?

A

=2(tan−1e−π4)=2(tan^{-1}e-\dfrac{\pi}{4})

B

=(tan−1e+π2)=(tan^{-1}e+\dfrac{\pi}{2})

C

=4(tan−1e+π4)=4(tan^{-1}e+\dfrac{\pi}{4})

D

=(tan−1e−π2)=(tan^{-1}e-\dfrac{\pi}{2})

E

=(tan−1e−π4)=(tan^{-1}e-\dfrac{\pi}{4})

Answer

=2(tan−1e−π4)=2(tan^{-1}e-\dfrac{\pi}{4})

Explanation

Solution

Given that:

∫012ex1+e2xdx∫_0^1 \dfrac{2e^x}{1+e^{2x}} dx

Take, u=exu = e^x 

     $du=e^xdx$

Then corresponding limits will be 1,e1,e

∴\therefore 2∫01t1+t2dx2∫_0^1 \dfrac{t}{1+t^2} dx

=2[tan−1t]1e2[tan^{-1}t]_1^e

=2(tan−1e−tan−1(1))=2(tan^{-1}e-tan^{-1}(1))

=2(tan−1e−π4)=2(tan^{-1}e-\dfrac{\pi}{4}) (_Ans)