What is the Area of the Region Bounded by y=0, y=2x+4, and y=x^3?

[karush]

\begin{align*}\displaystyle
R&= y=0 \quad y=2x+4 \quad y=x^3\\
A&=\iint\limits_{R}15x^2\, dA\\
&=\int_{0}^{8} \int_{{(y-4)}/2}^{y^{1/3}} 15x^2\,dx \\
&=\int_{0}^{8}\biggr[5x^3\large\biggr]_{{(y-4)}/2}^{y^{1/3}}\quad dx\\
&=\int_{0}^{8}[5(y^{1/3})^3-5((y-4)/2))^3] \quad dx\\
&= ?
\end{align*}

attempted plot

View attachment 7870

just seeing if I am going in the right direction with this
no book answer?

 

[MarkFL]

It appears to me you're actually computing a volume here, and I would use horizontal strips as follows:

\(\displaystyle V=15\int_0^8\int_{\Large\frac{y-4}{2}}^{y^{\Large\frac{1}{3}}} x^2\,dx\,dy\)

It looks like you are heading in the right direction, but your notation needs a little work. You're missing the outer differential. Next step would be:

\(\displaystyle V=5\int_0^8 \left[x^3\right]_{\Large\frac{y-4}{2}}^{y^{\Large\frac{1}{3}}}\,dy\)

Can you proceed?

 

[karush]

\begin{align*} \displaystyle
V&=5\int_0^8 \biggr[ x^3\biggr]_{(y-4)/2}^{y^{1/3}}\quad dy\\
&=5\int_0^8 [(y^{1/3})^3-((y-4)/2))^3] \quad dy\\
&=5\int_0^8 \biggr [y-\frac{(y-4)^3}{8}\biggr] \quad dy\\
\end{align*}
ok well this expands to via W|A:
\begin{align*} \displaystyle
&=5\int_0^8 -\frac{y^3}{8}+\frac{3y^3}{2}-5y +8 \quad dy\\
&=5\biggr[-\frac{y^4}{32}+\frac{y^3}{2}-\frac{5y^2}{2}+8y \biggr]_0^8\\
&=5 \biggr[-\frac{8^4}{32}+\frac{8^3}{2}-\frac{5(8)^2}{2}+8(8) \biggr]\\
&=160
\end{align*}
View attachment 7872

 

[HOI]

If you were finding an area, where did the "$5x^3$" come from? You seem to have several exercises where you are confusing "find the area of a region" with "integrate a function over the region".

 

[karush]

$5x^3+c$ was the anti derivative of $15x^2$