Question:

Evaluate the integral: \(\int^2_0 x\sqrt{x+2}\)

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Put \(x+2=t^2\)
Updated On: Oct 7, 2023
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Solution and Explanation

\(\int_{0}^{2} \sqrt{x+2} \,dx \)

Let x+2=t2⧠dx=2tdt

When x=0,t=√2 and,when x=2,t=2

\(\int_{0}^{2} \sqrt{x+2} \,dx \) =∫2√2(t2-2)√t22tdt

=2∫2√2(t2-2)t2dt

=2∫2√2(t4-2t2)dt

=2[t5/5-2t3/3]2√2

=2[\(\frac{32}{5}-\frac{16}{3}-\frac{4√2}{5}+\frac{4√2}{3}\)]

=2[\(\frac{96-80-12√2+20√2}{15}\)]

=2[16+8√2/15]

=\(\frac{16(2+√2)}{15}\)

=\(\frac{16√2(√2+1)}{15}\)

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