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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, AA, of the region between the curve y=2x1+x2y=\frac{2x}{1+x^{2}} and the interval −2≤x≤2-2 \leq x \leq 2 of the xx-axis. The area is A=□A=\square. (Type an exact answer.)

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Full solution

Q. Find the area, AA, of the region between the curve y=2x1+x2y=\frac{2x}{1+x^{2}} and the interval −2≤x≤2-2 \leq x \leq 2 of the xx-axis. The area is A=□A=\square. (Type an exact answer.)
  1. Determine Integral: Determine the integral to find the area under the curve y=2x1+x2y = \frac{2x}{1 + x^2} from x=−2x = -2 to x=2x = 2.
  2. Set up Integral: Set up the integral for the area AA: A=∫−222x1+x2 dxA = \int_{-2}^{2} \frac{2x}{1 + x^2} \, dx.
  3. Function Symmetry: Notice the function (2x)/(1+x2)(2x)/(1 + x^2) is an odd function, and the interval [−2,2][-2, 2] is symmetric about the y-axis.
  4. Area Calculation: Since the integral of an odd function over a symmetric interval around zero is zero, the area under the curve from −2-2 to 22 is zero.

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