The curves $x=7 y+18$ and $x=2 y^{2}+19 y+18$ are graphed.

Which expression represents the area bounded by the curves?
Choose 1 answer:
(A) $\int_{-6}^{0}\left(-2 y^{2}-12 y\right) d y$
(B) $\int_{-6}^{0}\left(2 y^{2}+26 y+36\right) d y$
(c) $\int_{-4.75}^{18}\left(-2 y^{2}-12 y\right) d y$
() $\int_{-4.75}^{18}\left(2 y^{2}+26 y+36\right) d y$

Question

The curves $x=7 y+18$ and $x=2 y^{2}+19 y+18$ are graphed.

Which expression represents the area bounded by the curves?
Choose 1 answer:
(A) $\int_{-6}^{0}\left(-2 y^{2}-12 y\right) d y$
(B) $\int_{-6}^{0}\left(2 y^{2}+26 y+36\right) d y$
(c) $\int_{-4.75}^{18}\left(-2 y^{2}-12 y\right) d y$
() $\int_{-4.75}^{18}\left(2 y^{2}+26 y+36\right) d y$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the area between the curves $x = 7y + 18$ and $x = 2y^2 + 19y + 18$, we first need to set the equations equal to each other to find the points of intersection. ‖ ‖ step 2 ⋮ Setting $7y + 18 = 2y^2 + 19y + 18$, we get $2y^2 + 12y = 0$. ‖ ‖ step 3 ⋮ Solving $2y^2 + 12y = 0$ gives $y(y + 6) = 0$, so $y = 0$ or $y = -6$. ‖ ‖ step 4 ⋮ The area between the curves from $y = -6$ to $y = 0$ is given by the integral $\int_{-6}^{0} \left[(7y + 18) - (2y^2 + 19y + 18)\right] dy$. ‖ ‖ step 5 ⋮ Simplifying the integrand, we get $\int_{-6}^{0} \left(-2y^2 - 12y\right) dy$. ‖ ‖ step 6 ⋮ Therefore, the correct expression for the area is $\int_{-6}^{0} \left(-2y^2 - 12y\right) dy$. ‖ ***A*** ∻Key Concept∻ ⚹Area between curves⚹ ∻Explanation∻ ⚹To find the area between two curves, you need to integrate the difference of the functions over the interval determined by their points of intersection.⚹◊What is the process for finding the area between two curves on a coordinate plane?⍭ Generate me a similar question◊

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