Home / AP Calculus AB 8.4 Finding the Area Between Curves Expressed as Functions of x – MCQs

AP Calculus AB 8.4 Finding the Area Between Curves Expressed as Functions of x - MCQs - Exam Style Questions

AP Calculus AB-MCQs and Answers - All Topics

No-Calc Question

Let \(R\) be the region bounded by the graphs of \(y=2x\) and \(y=4x-x^{2}\). What is the area of \(R\) ?
(A) \(\tfrac{2}{3}\)
(B) \(\tfrac{4}{3}\)
(C) \(\tfrac{16}{3}\)
(D) \(\tfrac{28}{3}\)
▶️ Answer/Explanation

Find intersections: \(2x=4x-x^{2}\Rightarrow x^{2}-2x=0\Rightarrow x=0,2\).
Area \(=\displaystyle\int_{0}^{2}\big((4x-x^{2})-2x\big)\,dx=\int_{0}^{2}(2x-x^{2})\,dx\).
Antiderivative: \(x^{2}-\tfrac{x^{3}}{3}\).
Evaluate \(0\to 2:\ 4-\tfrac{8}{3}=\tfrac{4}{3}\).

✅ Answer: (B)

No-Calc Question

What is the area of the region enclosed by the graphs of \(f(x)=x-2x^{2}\) and \(g(x)=-5x\)?

(A) \(\tfrac{7}{3}\)
(B) \(\tfrac{16}{3}\)
(C) \(\tfrac{20}{3}\)
(D) \(9\)
(E) \(36\)

▶️ Answer/Explanation

Find intersections: \(x-2x^{2}=-5x \Rightarrow -2x^{2}+6x=0 \Rightarrow -2x(x-3)=0\).
So \(x=0\) and \(x=3\).

On \(0\le x\le 3\), \(f(x)\) is above \(g(x)\) (e.g., \(f(1)=-1\), \(g(1)=-5\)).
Area \(=\displaystyle \int_{0}^{3}\big(f(x)-g(x)\big)\,dx\) \(=\int_{0}^{3}\big((x-2x^{2})-(-5x)\big)\,dx\) \(=\int_{0}^{3}(6x-2x^{2})\,dx\).

Antiderivative: \(\displaystyle \int(6x-2x^{2})\,dx=3x^{2}-\tfrac{2}{3}x^{3}\).
Evaluate: \(\big[3x^{2}-\tfrac{2}{3}x^{3}\big]_{0}^{3} =3\times 3^{2}-\tfrac{2}{3}\times 3^{3} =3\times 9-\tfrac{2}{3}\times 27 =27-18=9\).
✅ Answer: (D)

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