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Question Number 32688 by Rio Mike last updated on 31/Mar/18

$$\mathrm{Evaluate}$$$$1){\int }_{-1}^0({\mathrm{x}}^2+\mathrm{x}1)\mathrm{dx}$$$$2){\int }_1^5(\mathrm{x}-1+\frac{1}{{\mathrm{x}}^2})\mathrm{dx}$$$$$$

Answered by Joel578 last updated on 31/Mar/18

$$\left(1\right)$$$$I={\int }_{-1}^0{x}^2+x+1dx$$$$={[\frac{1}{3}{x}^3+\frac{1}{2}{x}^2+x]}_{-1}^0$$$$=\left(0\right)-(-\frac{1}{3}+\frac{1}{2}-1)$$$$=...$$$$$$$$\left(2\right)$$$$I={\int }_1^{5}x-1+\frac{1}{x^2}dx$$$$={[\frac{1}{2}{x}^2-x-\frac{1}{x}]}_1^5$$$$=(\frac{25}{2}-5-\frac{1}{5})-(\frac{1}{2}-1-1)$$$$=...$$

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