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Find the area under the curve f(x)=x^{2} on the interval [-1,1].

The main definition given for an integral in many undergraduate calculus textbooks today is one using a type of Riemann sums, that is summing up rectangles while making them smaller and smaller while adding in more and more of them until you have filled in all of the space and get the exact area under the curve. Really, all that is saying is an integral gives you the area between a function and the x-axis on a given interval. So, to find the area between this curve and the x-axis, all we need to do is take the integral on the interval.

A=\int _{{-1}}^{{1}}x^{2}\,dx\,

The first step you can use to simplify this integral a tiny bit is to recognize that x^{2} is an even function. An even function is one for which it is true that f(-x)=f(x) for all x. Since this is true, it has a symmetry about x=0. Thus, the value of this integral on [-1,1] is simply twice that of the integral on [0,1]. In some cases, this is a great help; in others, not so much.

A=\int _{{-1}}^{{1}}x^{2}\,dx=2\int _{{0}}^{{1}}x^{2}\,dx={\frac  {2}{3}}x^{3}{\bigg |}_{{0}}^{{1}}={\frac  {2}{3}}\,

Thus, the area between the function and the x-axis, on the interval [-1,1] is {\frac  {2}{3}}.

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