Definite Integrals Part 1

Given, f(0)=1 . . . (i)
f(2)‌‌=e2 . . . (ii)
f′(x)‌‌=f′(2−x)
Integrating w.r.t. x,
f(x)=−f(2−x)+C
Put x=0
‌f(0)‌‌=−f(2)+C
⇒‌1‌‌=−e2+C [from Eqs. (i) and (ii)]
⇒‌C‌‌=1+e2
‌∴‌f(x)‌‌=−f(2−x)+1+e2
‌ or ‌f(x)+f(2−x)=1+e2 . . . (iii)
‌ Let ‌I=

2

∫

0

f(x)dx . . . (iv)
‌ Also, ‌I=

2

∫

0

f(2−x)dx. . . (v)
Now, adding Eqs. (iv) and (v),
2I‌‌=

2

∫

0

[f(x)+f(2−x)]dx
2I‌‌=

2

∫

0

(1+e2)dx [from Eq. (iii)]
2I‌‌=2(1+e2)
∴‌‌I‌‌=(1+e2)