Evaluate int_{-1}^{1} 5 x^{4} sqrt{x^{5}+1} d | vedclass

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Question

(A) Let $t = x^{5} + 1$. Then,$dt = 5x^{4} dx$.When $x = -1$,$t = (-1)^{5} + 1 = 0$.When $x = 1$,$t = (1)^{5} + 1 = 2$.Therefore,the integral becomes:

Vedclass · Accepted Answer

$\frac{4\sqrt{2}}{3}$

Vedclass · Answer

(A) Let $t = x^{5} + 1$. Then,$dt = 5x^{4} dx$.When $x = -1$,$t = (-1)^{5} + 1 = 0$.When $x = 1$,$t = (1)^{5} + 1 = 2$.Therefore,the integral becomes:$\int_{0}^{2} \sqrt{t} dt = \left[ \frac{2}{3} t^{\frac{3}{2}} \right]_{0}^{2}$$= \frac{2}{3} (2^{\frac{3}{2}} - 0^{\frac{3}{2}})$$= \frac{2}{3} (2\sqrt{2}) = \frac{4\sqrt{2}}{3}$.

Evaluate $\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x$.

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