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