area between two curves

[logistic_guy]

Find the area enclosed by the graphs \(\displaystyle y = 3 - x\) and \(\displaystyle y = x^2 - 9\).

 

[khansaheb]

Find the area enclosed by the graphs \(\displaystyle y = 3 - x\) and \(\displaystyle y = x^2 - 9\).

Please show us what you have tried and exactly where you are stuck.

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[logistic_guy]

Find the area enclosed by the graphs \(\displaystyle y = 3 - x\) and \(\displaystyle y = x^2 - 9\).

Let us find the intersection points.

\(\displaystyle 3 - x = x^2 - 9\)

\(\displaystyle x^2 - 9 - 3 + x = 0\)

\(\displaystyle x^2 + x - 12 = 0\)

\(\displaystyle (x + 4)(x - 3) = 0\)

\(\displaystyle x = 3\)
Or
\(\displaystyle x = -4\)

Then, the area is:

\(\displaystyle A = \iint dA\)


\(\displaystyle = \int_{-4}^{3}\int_{x^2 - 9}^{3 - x} dy \ dx\)


\(\displaystyle = \int_{-4}^{3}(3 - x - x^2 + 9) \ dx\)


\(\displaystyle = \int_{-4}^{3}(12 - x - x^2) \ dx\)


\(\displaystyle = \left(12x -\frac{x^2}{2} - \frac{x^3}{3}\right)\bigg|_{-4}^{3}\)


\(\displaystyle = \left(36 -\frac{9}{2} - \frac{27}{3}\right) - \left(-48 - \frac{16}{2} + \frac{64}{3}\right) = \frac{343}{6} = \textcolor{blue}{57.17}\)