The area of the region enclosed by the parabo | vedclass

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(C) To find the area of the region enclosed by the parabolas $y=x^2-5x$ and $y=7x-x^2$,we first find their points of intersection by setting the equat

Vedclass · Accepted Answer

$72$

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(C) To find the area of the region enclosed by the parabolas $y=x^2-5x$ and $y=7x-x^2$,we first find their points of intersection by setting the equations equal to each other:$x^2-5x = 7x-x^2$$2x^2-12x = 0$$2x(x-6) = 0$Thus,the points of intersection are $x=0$ and $x=6$.In the interval $[0, 6]$,the parabola $g(x) = 7x-x^2$ lies above $f(x) = x^2-5x$.The area $A$ is given by the integral:$A = \int_0^6 (g(x) - f(x)) dx$$A = \int_0^6 ((7x-x^2) - (x^2-5x)) dx$$A = \int_0^6 (12x - 2x^2) dx$$A = [12 \frac{x^2}{2} - \frac{2x^3}{3}]_0^6$$A = [6x^2 - \frac{2}{3}x^3]_0^6$$A = (6(6)^2 - \frac{2}{3}(6)^3) - (0)$$A = 216 - \frac{2}{3}(216)$$A = 216 - 144 = 72 	ext{ unit}^2$

The area of the region enclosed by the parabolas $y=x^2-5x$ and $y=7x-x^2$ is:

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