What is the area of the region between the graphs of $f(x)=-x^{2}+2 x+12$ and $g(x)=x^{2}-12$ from $x=-3$ to $x=4 ?$

Choose 1 answer:
(A) $\frac{83}{3}$
(B) $\frac{52}{3} \sqrt{13}-1-32 \sqrt{3}$
(c) $\frac{343}{3}$
(D) 7

Question

What is the area of the region between the graphs of $f(x)=-x^{2}+2 x+12$ and $g(x)=x^{2}-12$ from $x=-3$ to $x=4 ?$

Choose 1 answer:
(A) $\frac{83}{3}$
(B) $\frac{52}{3} \sqrt{13}-1-32 \sqrt{3}$
(c) $\frac{343}{3}$
(D) 7

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the area between the graphs of two functions, we integrate the difference of the functions over the given interval. ‖ ‖ step 2 ⋮ The integral of the difference of the functions $f(x) = -x^2 + 2x + 12$ and $g(x) = x^2 - 12$ from $x = -3$ to $x = 4$ is $\int_{-3}^{4} ((-x^2 + 2x + 12) - (x^2 - 12)) \, dx$. ‖ ‖ step 3 ⋮ Simplifying the integrand, we get $\int_{-3}^{4} (-2x^2 + 2x + 24) \, dx$. ‖ ‖ step 4 ⋮ Integrating term by term, we find the definite integral to be $\left[-\frac{2}{3}x^3 + x^2 + 24x\right]_{-3}^{4}$. ‖ ‖ step 5 ⋮ Evaluating the definite integral from $x = -3$ to $x = 4$, we get $\left(-\frac{2}{3}(4)^3 + (4)^2 + 24(4)\right) - \left(-\frac{2}{3}(-3)^3 + (-3)^2 + 24(-3)\right)$. ‖ ‖ step 6 ⋮ Simplifying the expression, we obtain the area as $\frac{343}{3}$. ‖ ***C*** ∻Key Concept∻ ⚹Definite integral of the difference of functions⚹ ∻Explanation∻ ⚹The area between two curves is found by integrating the difference of the functions over the given interval.⚹◊What is the definite integral that represents the area between the two graphs over the given interval?⍭ Generate me a similar question◊

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