If f x = ∫ 5 x 8 + 7 x 6 x 2 + 1 + 2 x 7 2 d x ,   x ≥ 0 ,   f 0 = 0 and f 1 = 1 K ,…

If fx=5x8+7x6x2+1+2x72dx, x0, f0=0 and f1=1K, then the value of K is

Solution

fx=5x8+7x6dxx14x-5+x-7+22

Let x-5+x-7+2=t

-5x-6-7x-8dx=dt

fx=-dtt2=1t+c

fx=x7x2+1+2x7

f1=14

Asked in: JEE Main 2021 (18 Mar Shift 1)

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