Applications – Area Between Curves I

Problem 244: Find the area of the region that is enclosed between $y = 12-x^2$ and $y = x^2 - 6$ .

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) $\begin{equation*} 12-x^2 = x^2 - 6 \implies 18 - 2x^2 = 0 \implies x = \pm 3.\end{equation*}$

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

(2) $\begin{align*} A & = \int_{a}^{b} \text{Top/Right Function-Bottom/Left Function} \ dx \\ & = \int_{-3}^{3} 12-x^2-(x^2-6) \ dx \\ & = \int_{-3}^{3} -2x^2 + 18 \ dx \\ & = -\frac{2}{3}x^3 + 18x \bigg|_{-3}^{3} \\ & = 72. \end{align*}$

That is, the area of the region enclosed between the two given functions is 72.

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Applications – Area Between Curves I

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