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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?

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

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

Can you continue?

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  • $\begingroup$ 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? $\endgroup$ – Eris Tyenns Aug 11 '19 at 12:29
  • $\begingroup$ If $-1 < x < 1$, then $x^2 - 1 < 0$. $\endgroup$ – N. F. Taussig Aug 11 '19 at 12:30
  • $\begingroup$ 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. $\endgroup$ – Toby Mak Aug 11 '19 at 12:31
  • $\begingroup$ Thanks! Now the area is 4. $\endgroup$ – Eris Tyenns Aug 11 '19 at 12:36
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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 $$2\biggl(\int_1^2y(x)\,\mathrm d x-\int_0^1y(x)\,\mathrm d x\biggl).$$

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

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