Question

Draw the region and evaluate the following integral

(a) ${\int }_0^1{\int }_1^2(y+2x)dydx$

 (b) ${\int }_0^1{\int }_{x^2}^{x}xydydx$ (c) ${\int }_0^{\pi }{\int }_y^{\pi }\frac{\sin x}{x}dxdy$ 

Answer 1

Solution : Draw the region and evaluate the following integral

(a) ${\int }_0^1{\int }_1^2\left(y+2x\right)dydx={\int }_0^1\left(2x+\frac{1}{2}{y}^2\right){|}_1^{2}dx={\int }_0^1\left(2x+\frac{3}{2}\right)dx=\left(x^2+\frac{3}{2}x\right){|}_0^1=\frac{5}{2}$

 Therefore, ${\int }_0^1{\int }_1^2(y+2x)dydx=\frac{5}{2}■$ 

(b) ${\int }_0^1{\int }_{x^2}^{x}xydydx={\int }_0^1\frac{1}{2}x{y}^2{|}_{x^2}^{x}dx={\int }_0^1\left(\frac{1}{2}{x}^3-\frac{1}{2}{x}^5\right)dx=\left(\frac{1}{8}{x}^4-\frac{1}{12}{x}^6\right){|}_0^1=\frac{1}{24}$

 Therefore, ${\int }_0^1{\int }_{x^2}^{x}xydydx=\frac{1}{24}■$ 

(c) ${\int }_0^{\pi }{\int }_y^{\pi }\frac{\sin x}{x}dxdy\Rightarrow {\int }_0^{\pi }{\int }_0^x\frac{\sin x}{x}dydx={\int }_0^{\pi }\sin xdx=\left(-\cos x\right){|}_0^{\pi }=2$

 Therefore, ${\int }_0^{\pi }{\int }_y^{\pi }\frac{\sin x}{x}dxdy=2■$