Question

let a be the area of the bounded region enclosed by the parabola y = x² - 48 and the line y = -8x. then a = square units

Explanation:

Step1: Find intersection points

Set $x^2 - 48 = -8x$, rearrange to $x^2 + 8x - 48 = 0$. Factor: $(x+12)(x-4)=0$, so $x=-12$ and $x=4$.

Step2: Set up integral for area

The line $y=-8x$ is above the parabola $y=x^2-48$ between $x=-12$ and $x=4$. Area $A = \int_{-12}^4 [(-8x) - (x^2 - 48)] dx = \int_{-12}^4 (-x^2 -8x +48) dx$.

Step3: Compute the integral

Antiderivative: $-\frac{1}{3}x^3 -4x^2 +48x$. Evaluate at bounds:
At $4$: $-\frac{64}{3} -64 +192 = -\frac{64}{3} +128 = \frac{320}{3}$.
At $-12$: $-\frac{(-1728)}{3} -4(144) +48(-12) = 576 -576 -576 = -576$.
Subtract: $\frac{320}{3} - (-576) = \frac{320}{3} + \frac{1728}{3} = \frac{2048}{3}$.

Answer:

$\frac{2048}{3}$