# Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 339: 83

The area of the region is $56/15$.

#### Work Step by Step

$x+4y^2=4$ and $x+y^4=1$ for $x\ge0$ - For the curve $x+4y^2=4$, we can rewrite into $$x=-4y^2+4$$ - For the curve $x+y^4=1$, we can rewrite into $$x=-y^4+1$$ 1) Draft the graph: The graph is drafted below. We would use integration with respect to $y$ in this exercise. 2) Find limits of integration: To find the limits of integration, we will find the intersection points with respect to $y$. $$-4y^2+4=-y^4+1$$ $$y^4-4y^2+3=0$$ $$(y^2-1)(y^2+3)=0$$ $$y=-1\hspace{1cm}\text{or}\hspace{1cm}y=1$$ Therefore, the upper limit is $1$ and the lower one is $-1$. 3) Find the area: The region from $y=-1$ to $y=1$ is bounded on the right by $x=-4y^2+4$ and on the left by $x=-y^4+1$. So according to definition, the area of the region is $$A=\int^1_{-1}(-4y^2+4+y^4-1)dy=\int^1_{-1}(y^4-4y^2+3)dy$$ $$A=\frac{y^5}{5}-\frac{4y^3}{3}+3y\Big]^1_{-1}$$ $$A=\Big(\frac{1}{5}-\frac{4}{3}+3\Big)-\Big(-\frac{1}{5}+\frac{4}{3}-3\Big)$$ $$A=2\Big(\frac{1}{5}-\frac{4}{3}+3\Big)$$ $$A=2\times\frac{28}{15}=\frac{56}{16}$$

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