Find the area, A, of the region between the curve y=(2x)/(1+x^(2)) and the interval -2 <= x <= 2 of the x-axis. The area is A=◻. (Type an exact answer.)

Determine Integral: Determine the integral to find the area under the curve y = (2x)/(1 + x^2) from x = -2 to x = 2. Set up Integral: Set up the integral for the area A: A = ∫ from -2 to 2 of (2x)/(1 + x^2) dx. Function Symmetry: Notice the function (2x)/(1 + x^2) is an odd function, and the interval [-2, 2] is symmetric about the y-axis. Area Calculation: Since the integral of an odd function over a symmetric interval around zero is zero, the area under the curve from -2 to 2 is zero.

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Find the area, A, of the region between the curve
y=(2x)/(1+x^(2)) and the interval
-2 <= x <= 2 of the
x-axis.
The area is
A=◻.
(Type an exact answer.)

Find the area, $A$ , of the region between the curve $y=\frac{2x}{1+x^{2}}$ and the interval $-2 \leq x \leq 2$ of the $x$ -axis. The area is $A=\square$ . $\newline$ (Type an exact answer.)

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Q. Find the area,

A

, of the region between the curve

y=\frac{2x}{1+x^{2}}

and the interval

-2 \leq x \leq 2

of the

x

-axis. The area is

A=\square

\newline

(Type an exact answer.)

Determine Integral: Determine the integral to find the area under the curve $y = \frac{2x}{1 + x^2}$ from $x = -2$ to $x = 2$ .
Set up Integral: Set up the integral for the area $A$ : $A = \int_{-2}^{2} \frac{2x}{1 + x^2} \, dx$ .
Function Symmetry: Notice the function $(2x)/(1 + x^2)$ is an odd function, and the interval $[-2, 2]$ is symmetric about the y-axis.
Area Calculation: Since the integral of an odd function over a symmetric interval around zero is zero, the area under the curve from $-2$ to $2$ is zero.