The area of the region (x,y):x^2-8x y -x is :

Question

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

Accepted Answer

For the region

${(x,y):{x}^{2}-8x\le y\le -x},$

the lower curve is

$y={x}^{2}-8x$

and the upper curve is

$y=-x.$

To find the area between the curves, we first find their points of intersection.

Set

${x}^{2}-8x=-x$

${x}^{2}-7x=0$

$x(x-7)=0$

So, the curves intersect at

$x=0\, \, x=7.$

Now, on the interval $[0,7]$, the line $y=-x$ lies above the parabola $y={x}^{2}-8x$.

Hence, the required area is

$={\int }_{7}^{0}[(-x)-({x}^{2}-8x)]dx.$

Simplify the integrand:

$(-x)-({x}^{2}-8x)=-x-{x}^{2}+8x=7x-{x}^{2}.$

Therefore,

$={\int }_{7}^{0}(7x-{x}^{2})\, dx.$

Now integrate:

$\int (7x-{x}^{2})\, dx=\frac{7{x}^{2}}{2}-\frac{{x}^{3}}{3}.$

Applying the limits from $0$ to $7$:

$={[\frac{7{x}^{2}}{2}-\frac{{x}^{3}}{3}]}_{7}^{0}$

$=(\frac{7\cdot 49}{2}-\frac{343}{3})-0$

$=\frac{343}{2}-\frac{343}{3}$

Taking LCM $6$:

$=\frac{1029-686}{6}=\frac{343}{6}.$

Hence, the correct answer is

$\frac{343}{6}$

So, the correct option is A.

Solutions · 1