# Enclosed Areas and Integration

Find the area enclosed between the curve $y=x^3$, the $x$-axis and the line $y=-3x+4$.

So there intersection point is $(1,1)$, and I understand the area is below this point to the $x$-axis, but how do I find the area? The answer is approximately $0.42$.

Thank you.

• How do you think the integral for the area should be set up? Feel free to edit your post to include what you think the integral should be. This will make it easier for people to help you. – Mike Pierce Feb 8 '15 at 4:40

Below shows the shaded region, and you want to find the total area of it. You are right about the intersection of $(1,1)$.

It is suggested that you compute the following:

$$A=\int_a^b x_{\text{right}}-x_{\text{left}} \, dy=\int_0^1 \frac 13(4-y) - y^{1/3} \, dy.$$

Note: The $x=\frac 13(4-y)$ comes from solving for $x$ in the given $y=-3x+4$. The $x= y^{1/3}$ comes from solving for $x$ in the given $y=x^3$.

• Why do you have to solve for $x$? The integral is pretty straightforward with respect to $x$... – bartgol Feb 8 '15 at 5:54
• @bartgol This is only one way of evaluating the integral. If one evaluates this way with respect to $y$, then he/she can avoid having to separate into two integrals. Of course, the OP is free to choose however she wants to compute the area. – Cookie Feb 8 '15 at 6:21

We need to find area under $y=x^3$ from $(0,0)$ till $(1,1)$ And $y=-3x+4$ from $(1,1)$ till $(\frac43,0)$.

$$\int_0^1x^3dx + \int_1^{\frac43}(-3x+4)dx$$
$$=[\frac{x^4}4]_0^1+[\frac{-3x^2}2]_1^{\frac43}+[4x]_1^{\frac43}$$
$$=\frac14-\frac83+\frac32+\frac43$$
$$=\frac5{12}=0.41666...$$
$x^3 = -3x+4 \to x^3+3x-4=0 \to (x-1)(x^2 + x +4) = 0 \to x = 1$ is the only intersection of the two curves. This leads us to: $A=\displaystyle \int_{0}^1 \int_{\sqrt[3]{y}}^{\frac{4-y}{3}} 1dxdy$. I think you can take it from here.