Applications: Area Between Curves II

Problem 245: Find the area of the region that is enclosed between $y = x$ and $y = \sqrt[3]{x}$ .

Solution: In order to find the area enclosed by the two functions, we need to find the limits of integration. That is, since we have two functions equal to $y$ , we can make it equal to each other.

(1) $\begin{equation*} x = \sqrt[3]{x} \implies x^3 = x \implies x^3 - x =0 \implies x =0, \pm 1.\end{equation*}$

Let us now find the area of the region. A graph is always helpful when doing these type of problems.

(2) $\begin{align*} A & = \int_{a}^{b} \text{Top/Right Function-Bottom/Left Function} \ dx \\ & = \int_{-1}^{0} x- \sqrt[3]{x} \ dx + \int_{0}^{1} \sqrt[3]{x}-x \ dx \\ & = \int_{-1}^{0} x - x^{\frac{1}{3}} \ dx + \int_{0}^{1} x^{\frac{1}{3}} - x \ dx \\ & = \frac{x^2}{2} -\frac{3}{4}x^{\frac{4}{3}}\bigg|_{-1}^{0} + \frac{3}{4}x^{\frac{4}{3}}-\frac{x^2}{2}\bigg|_{0}^{1}\\ & = 0.25 + 0.25 \\ & = 0.50 . \end{align*}$

That is, the area of the region enclosed between the two given functions is 1/2.

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