# What is the total area enclosed between the curve $y=x^2-1$, the x-axis and the lines $x=-2$ and $x=2$?

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?

You forgot the area under the x-axis. The signed area of this is:

$$\int_{-1}^1 x^2-1 \ dx \$$

Can you continue?

• Thanks! I calculated the area under the x-axis. It gives me the area $\frac{-4}{3}$. Do I add it to the other areas even though it's negative? – Eris Tyenns Aug 11 '19 at 12:29
• If $-1 < x < 1$, then $x^2 - 1 < 0$. – N. F. Taussig Aug 11 '19 at 12:30
• No, since areas cannot be negative (in this case). You should add the absolute value of all the areas since this will always give a positive answer. – Toby Mak Aug 11 '19 at 12:31
• Thanks! Now the area is 4. – Eris Tyenns Aug 11 '19 at 12:36

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 $$2\biggl(\int_1^2y(x)\,\mathrm d x-\int_0^1y(x)\,\mathrm d x\biggl).$$

It is an even function Required Area will be , $$Area = 2\;(\int_0^2| x^2\;-1| \,dx)$$ $$=2(\;\int_0^1(1-x^2)\,dx\;+\int_1^2(x^2-1)\,dx\;)$$$$=2( \, \frac23\;+\;\frac43\,)=\;4$$