The value of \(\int_{0}^{1}\displaystyle 3x^{2}e^{x^{3}}\,dx\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(e-1\)
The factor \(3x^{2}\) is the derivative of the power \(x^{3}\).
Let \(t = x^{3}\). Then \(dt = 3x^{2}\,dx\).
When \(x = 0\), \(t = 0\); when \(x = 1\), \(t = 1\). So
\( \displaystyle \int_{0}^{1}3x^{2}e^{x^{3}}\,dx = \int_{0}^{1}e^{t}\,dt \)
\( \displaystyle = \left[e^{t}\right]_{0}^{1} = e - 1 \)
Why the other options are incorrect
Option (1) e — Incorrect. It leaves out the lower-limit value \(e^{0} = 1\).
Option (3) \(e+1\) — Incorrect. It adds the lower-limit value instead of subtracting it.
Option (4) \(1-\displaystyle\frac{1}{e}\) — Incorrect. This is \(\int_{0}^{1}e^{-t}\,dt\); the integrand here is \(e^{t}\).