AP Calculus BC 8.5 Finding the Area Between Curves Expressed as Functions of y - MCQs - Exam Style Questions
No-Calc Question
Let \(R\) be the region bounded by the graph of \(y=\ln x\), the vertical line \(x=e^{2}\), and the \(x\)-axis, as shown. Which of the following integrals are equal to the area of \(R\)?
I. \(\displaystyle \int_{1}^{e^{2}}\ln x\,dx\)
II. \(\displaystyle \int_{0}^{2}e^{y}\,dy\)
III. \(\displaystyle \int_{0}^{2}\big(e^{2}-e^{y}\big)\,dy\)
II. \(\displaystyle \int_{0}^{2}e^{y}\,dy\)
III. \(\displaystyle \int_{0}^{2}\big(e^{2}-e^{y}\big)\,dy\)
(A) I only
(B) III only
(C) I and II only
(D) I and III only
(B) III only
(C) I and II only
(D) I and III only
▶️ Answer/Explanation
Detailed solution
Vertical slices (with \(x\) as variable) give the area as \(\displaystyle \int_{1}^{e^{2}}\ln x\,dx\) since \(\ln x\ge0\) on \([1,e^{2}]\). This is I.
Horizontal slices from \(y=0\) to \(y=2\) run from \(x=e^{y}\) to \(x=e^{2}\); thus area \(=\displaystyle \int_{0}^{2}\big(e^{2}-e^{y}\big)\,dy\). This is III.
II is just the area under \(x=e^{y}\) from \(y=0\) to \(2\) and omits the rectangle \(x=e^{2}\); therefore it is not the region \(R\).
✅ Answer: (D)