We have, I=∫x5e−4x3dx
Let, −4x3=t⇒−12x2dx=dt
⇒I=∫(−4t)et(−12dt)=481∫tetdt
=481[tet−∫etdt]=48(t−1)et+C=48(−4x3−1)e−4x3+C,
where C is the constant of integration.
Hence, f(x)=−4x3−1
JEE Main 2019 — Mathematics Calculus
If ∫x5e−4x3dx=481e−4x3f(x)+C, where C is a constant of integration, then f(x) is equal to
Held on 10 Jan 2019 · Verified 6 Jul 2026.
−4x3−1
−2x3+1
−2x3−1
4x3+1
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