Question

Evaluate the double integral over the specified region.

${\iint }_{\Omega }x{e}^{x^3}dA$where $\Omega$is the triangular region in the first quadrant bounded below by the x-axis, bounded above $y=m{x}_{,}\text{where}m>0$and bounded on the

right by the line with equation$x=1$.

Short answer

The required integral is$\frac{m}{3}(e-1)$.

Step-by-step solution

Given Information

We are given following double integral as ${\iint }_{\Omega }x{e}^{x^3}dA$

$\Omega$is triangular region in first quadrant bounded by $y=mx$, slope $m>0$

bounded on right by $x=1$

Simplification

Region of integration is region between line $y=mx,y=0,x=1$

It can be seen that for type I integral, $0\le x\le 1,0\le y\le mx$

Therefore

${\iint }_{\Omega }x{e}^{x^3}dA={\int }_0^1{\int }_0^{mx}x{e}^{x^3}dydx$

${\int }_0^1{\int }_0^{mx}x{e}^{x^3}dydx={\int }_0^1[y{]}_0^{mx}x{e}^{x^3}dx$

$=m{\int }_0^1{x}^2{e}^{x^3}dx$

Using $x^3=t$

$\Rightarrow 3x^{2}dx=dt$

Limits are

$t=0^3=0,t=1^3=1$

${\int }_0^1{\int }_0^{mx}x{e}^{x^3}dydx=\frac{m}{3}{\int }_0^1{e}^{t}dt$

$=\frac{m}{3}{\left[e^t\right]}_0^1$

$=\frac{m}{3}(e-1)$