AP Calculus BC

Conducting Body: College Board
For: High school students seeking college credit
Syllabus: Calculus I & II topics, including series and polar coordinates
Question Types: Multiple Choice and Free Response
Duration: 3 hours 15 minutes
Calculator Use: Permitted for parts of the exam
Frequency: Annually in May
Qualifies For: Potential college credit or advanced placement

The AP Calculus BC exam is a standardized test from the College Board for high school students who have completed a college-level calculus course. It covers a wider range of material than the AP Calculus AB exam, roughly equivalent to a full year of university calculus. Students take the exam to demonstrate mastery of the subject, which can earn them credit or advanced placement at many colleges and universities.

The exam has two sections of equal weight: multiple choice and free response. Each of these sections is divided into two parts, one where a graphing calculator is permitted and another where it is not. The questions assess everything from computational skill to the application of calculus concepts in various problem-solving scenarios.

The difficulty of the Calculus BC exam comes from its extensive syllabus. It includes all topics from Calculus AB, plus additional units on parametric equations, polar coordinates, and infinite series. The free response questions are especially challenging because they require students to write out complete, logical solutions, not just find a final answer. Managing time effectively while handling complex, multi-step problems is essential for a high score.

How a question gets worked here

From a mathematics paper rather than AP Calculus BC, worked the way every question on this site is. AP Calculus BC questions are being added.

JEE Advanced 2013

The area enclosed by the curves $y = \sin x + \cos x$ and $y = |\cos x − \sin x|$ over the interval [0, π/2] is $(1) 4(√2 − 1) (2) 2√2(√2 − 1) (3) 2(√2 + 1) (4) 2√2(√2 + 1)$

Applications of Integrals The area enclosed by the curves

y = \sin x + \cos x

and

y = |\cos x − \sin x|

over the interval [0, π/2] is

(1) 4(√2 − 1) (2) 2√2(√2 − 1) (3) 2(√2 + 1) (4) 2√2(√2 + 1)

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$y_1 = \sin x + \cos x, \qquad y_2 = |\cos x - \sin x|$ $x \in \left[0, \tfrac{\pi}{2}\right]$ $\text{Remove the modulus, find upper/lower, then } A = \int_0^{\pi/2} (\text{upper} - \text{lower})\,dx$ $|\cos x - \sin x| = \begin{cases} \cos x - \sin x & x \in [0,\tfrac{\pi}{4}] \\[2pt] \sin x - \cos x & x \in [\tfrac{\pi}{4},\tfrac{\pi}{2}] \end{cases}$ $y_1 - y_2 = 2\sin x \geq 0 \quad \left(0 \leq x \leq \tfrac{\pi}{4}\right)$ $y_1 - y_2 = 2\cos x \geq 0 \quad \left(\tfrac{\pi}{4} \leq x \leq \tfrac{\pi}{2}\right)$ $\Rightarrow y_1 \text{ is on top throughout}$ $A = \int_0^{\pi/4} 2\sin x\,dx + \int_{\pi/4}^{\pi/2} 2\cos x\,dx$ $\int_0^{\pi/4} 2\sin x\,dx = [-2\cos x]_0^{\pi/4} = -\tfrac{2}{\sqrt{2}} + 2 = 2\left(1 - \tfrac{1}{\sqrt{2}}\right)$ $\int_{\pi/4}^{\pi/2} 2\cos x\,dx = [2\sin x]_{\pi/4}^{\pi/2} = 2 - \tfrac{2}{\sqrt{2}} = 2\left(1 - \tfrac{1}{\sqrt{2}}\right)$ $A = 4\left(1 - \tfrac{1}{\sqrt{2}}\right) = 4 - 2\sqrt{2} = 2\sqrt{2}(\sqrt{2} - 1)$ $\Rightarrow \text{option (2)}$ $A = 2\sqrt{2}(\sqrt{2} - 1) \text{ square units}$

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