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mcastillo356

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- Been searching backwards in my textbook, but can't find it; I mean, the reason. I will provide a solved exercise

**EXAMPLE 4**Find the area of the region ##R## lying above the line ##y=1## and below the curve ##y=5/(x^2+1)##.

*The region ##R## is shaded in Figure 5.24. To find the intersections of ##y=1## and ##y=5/(x^2+1)##, we must solve these equations simultaneously:*

**Solution**##1=\frac{5}{x^2+1}##

so ##x^2+1=5##,##x^2=4##, and ##x=\pm 2##

The area of the region ##R## is the area under the curve ##y=5/(x^2+1)## and above the ##x##-axis between ##x=-2## and ##x=2##, minus the area of a rectangle of width 4 and height 1. Since ##\tan^{-1}x## is an antiderivative of ##1/(x^2+1)##,

##A=\int_{-2}^2 \frac{5}{x^2+1}\ ,dx-4=2\int_0^2\frac{5}{x^2+1}\ ,dx-4=10\tan^{-1} x\left. \right |_0^2-4=10\tan{-1} 2-4## square units.

Observe the use of even symmetry (Theorem 3(h) of Section 5,4) to replace the lower limit of integration by 0. It si easier to susbstitute 0 into the antiderivative than -2.

**THEOREM 3**Let ##f## and ##g## be integrable on an interval containing the points ##a##, ##b## and ##c##. Then

(...)

(h) The integral of an even function over an interval symmetric about zero is twice the integral over the positive half of the interval. If ##f## is an even function (i.e., ##f(-x)=f(x))##, then

##\int_{-a}^a f(x)\ ,dx=\int_0^a f(x)\ ,dx##

Attempt: It is not explicited the unit