What is the total area enclosed between the curve $y=x^2-1$, the x-axis and the lines $x=-2$ and $x=2$?
I tried to find the area by using the integrals $\int_1^2$ and $\int_{-1}^{-2}$ .
$x^2-1$ integrated is $\frac{x^3}{3}-x$
The answer is supposed to be $4$, but by adding up $\int_1^2$ and $\int_{-1}^{-2}$ my area is $\frac{8}{3}$.
What am I doing wrong?
Best Answer
By symmetry, it is twice the area between the $y$-axis and the line $x=2$. The positive $x$-intercept is $x=1$, and on the interval $[0,1]$, the curve is below the $x$-axis, hence the total (geometric) area is $$2\biggl(\int_1^2y(x)\,\mathrm d x-\int_0^1y(x)\,\mathrm d x\biggl).$$