f(x)=∫x14(x−5+x−7+2)2(5x8+7x6)dx
Let x−5+x−7+2=t
(−5x−6−7x−8)dx=dt
⇒f(x)=∫−t2dt=t1+c
f(x)=x2+1+2x7x7
f(1)=41
If f(x)=∫(x2+1+2x7)25x8+7x6dx,(x≥0),f(0)=0 and f(1)=K1, then the value of K is
Held on 18 Mar 2021 · Verified 6 Jul 2026.
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