The area of the region {(x, y) : x^2 - 8x le | vedclass

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Question

(A) To find the area of the region,first determine the intersection points of the curves $y = x^2 - 8x$ and $y = -x$.Setting the equations equal: $x^2

Vedclass · Accepted Answer

$\frac{343}{6}$

Vedclass · Answer

(A) To find the area of the region,first determine the intersection points of the curves $y = x^2 - 8x$ and $y = -x$.Setting the equations equal: $x^2 - 8x = -x \implies x^2 - 7x = 0 \implies x(x - 7) = 0$.Thus,the intersection points are $x = 0$ and $x = 7$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = 0$ to $x = 7$:$A = \int_0^7 (-x - (x^2 - 8x)) dx = \int_0^7 (7x - x^2) dx$.Evaluating the integral:$A = [\frac{7x^2}{2} - \frac{x^3}{3}]_0^7 = (\frac{7(49)}{2} - \frac{343}{3}) - 0 = \frac{343}{2} - \frac{343}{3}$.$A = \frac{1029 - 686}{6} = \frac{343}{6}$.

The area of the region ${(x, y) : x^2 - 8x \le y \le -x}$ is :

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