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Question Number 6748 by Leila Akram last updated on 21/Jul/16

Answered by Yozzii last updated on 21/Jul/16

$$I={\int }_{-1}^{0}t\sqrt{t+2}dx$$$$u=t+2\Rightarrow du=dt$$$$t=u-2$$$$Att=0,u=2;Att=-1,u=1.$$$$I={\int }_1^2(u-2)\sqrt{u}du$$$$I={\int }_1^2(u^{3/2}-2u^{1/2})du$$$$I=\frac{2}{5}{u}^{5/2}-\frac{2}{3}\times 2\times u^{3/2}{\mid }_1^2$$$$I=2(\frac{1}{5}{2}^{5/2}-\frac{2}{3}{2}^{3/2}-\frac{1}{5}+\frac{2}{3})$$$$I=2(\frac{3-5}{15}{2}^{5/2}+\frac{2\times 5-3}{15})$$$$I=2(\frac{7}{15}-\frac{2}{15}{2}^{5/2})$$$${\int }_{-1}^{0}t\sqrt{t+2}dt=\frac{2}{15}(7-8\sqrt{2})$$$$$$

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