Evaluate $\iint(x^2+4y) \,dA$, where $R$ is bounded by $y=x$, $y=x^3$, $x\geq 0$

Question

First I set the two $y$ values equal to each other. I got $x=x^3$ so $x$ must equal $0$ or $1$ to be true.

Next I did the double integral evaluated from $0$ to $1$ as the outer integral and $x$ to $x^3$ as my inner integral, of $x^2+4y \,dy\,dx$.

I integrated from the first integral to get $x^2+2y^2$ evaluated from $x$ to $x^3$, to get $x^6+2x^6-(x^2+2x^2)$ and then simplified to get $3x^6-3x^2$.

Next I evaluated the integral from $0$ to $1$ of $3x^6-3x^2\,dx$ to get $\dfrac{3}{7}x^7-x^3$ evaluated from $0$ to $1$ and got $-\dfrac{4}{7}$.

This is not the correct answer. Also sorry this is messy.

Answer 1

Alternatively, use change of order for the region bounded by $y=x$, $y=x^3$ & $x\ge 0$ by considering a horizontal infinitesimal small rectangular slab & integrate with proper limits as follows $$\int\int (x^2+4y)dA=\int_0^1\int_{x=y}^{x=y^{1/3}} (x^2+4y)dxdy$$ $$=\int_0^1\left(\frac{x^3}{3}+4xy\right)_{y}^{y^{1/3}}dy$$ $$=\int_0^1\left(\frac{y}{3}+4y^{4/3}-\frac{y^3}{3}-4y^2\right)dy$$ $$=\left(\frac{y^2}{6}+\frac{12}{7}y^{7/3}-\frac{y^4}{12}-\frac{4}{3}y^3\right)_0^1$$ $$=\frac16+\frac{12}{7}-\frac1{12}-\frac43$$ $$=\color{blue}{\frac{13}{28}}$$

Answer 2

Your domaine is defined by

$$R=\{(x,y)\in\Bbb R^2\;:\; 0\le x\le 1 \; and\; x^3\le y\le x\}$$

the integral becomes

$$\int_0^1(\int_{x^3}^x(x^2+4y)dy)dx=$$

$$\int_0^1\Bigl[x^2y+2y^2\Bigr]_{x^3}^xdx=$$

$$\int_0^1(x^3+2x^2-x^5-2x^6)dx=$$ $$\frac 14+\frac 23-\frac 16-\frac 27=\frac{13}{28}$$