\(\int_{-1}^{1}\displaystyle 5x^{4}\sqrt{x^{5}+1}\,dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{4\sqrt{2}}{3}\)
Let \(t = x^{5}+1\), so \(dt = 5x^{4}\,dx\).
When \(x = -1\), \(t = 0\); when \(x = 1\), \(t = 2\). By substitution with new limits,
\( \displaystyle \int_{-1}^{1} 5x^{4}\sqrt{x^{5}+1}\,dx = \int_{0}^{2}\sqrt{t}\,dt \)
\( \displaystyle = \left[\frac{2}{3}t^{3/2}\right]_{0}^{2} \)
\( \displaystyle = \frac{2}{3}\cdot 2\sqrt{2} = \frac{4\sqrt{2}}{3} \)
Why the other options are incorrect
Option (1) 0 — Incorrect. The integrand is not odd: at \(x = 1\) it is \(5\sqrt{2}\), but at \(x = -1\) it is 0.
Option (3) \(\displaystyle\frac{2\sqrt{2}}{3}\) — Incorrect. It takes \(2^{3/2}\) as \(\sqrt{2}\); in fact \(2^{3/2} = 2\sqrt{2}\).
Option (4) \(\displaystyle\frac{10}{3}\) — Incorrect. The value is \(\frac{4\sqrt{2}}{3} \approx 1.89\), not \(3.33\).