Question Number 169852 by Shrinava last updated on 10/May/22
$$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2} \\ $$$$\mathrm{y}\:=\:-\:\mathrm{x} \\ $$$$\mathrm{x}\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{bounded}\:\mathrm{by} \\ $$$$\mathrm{lines} \\ $$
Answered by mahdipoor last updated on 11/May/22
![∫_0 ^( 1) (x^2 +2−(−x))dx=[(x^3 /3)+2x+(x^2 /2)]_0 ^1 =((17)/6)](https://www.tinkutara.com/question/Q169853.png)
$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{2}−\left(−{x}\right)\right){dx}=\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{17}}{\mathrm{6}} \\ $$