Home / AP Calculus BC 6.4 The Fundamental Theorem of Calculus and Accumulation Functions – Exam Style Questions – MCQs

AP Calculus BC 6.4 The Fundamental Theorem of Calculus and Accumulation Functions - Exam Style Questions - MCQs - New Syllabus

AP Calculus BC-MCQs and Answers - All Topics

Question


The figure shows the graph of a function \(f\) (a line with positive slope). If \( \displaystyle g(x)=\int_{1}^{x} f(t)\,dt \), which of the following could be the graph of \(g\)?

▶️ Answer/Explanation
Detailed solution

By the Fundamental Theorem of Calculus (Part 1), \[ g'(x)=f(x), \qquad g(1)=\int_{1}^{1}f(t)\,dt=0. \] Because the graph of \(f\) is a line with positive slope, we have \[ g”(x)=f'(x) > 0 \quad \text{for all }x, \] so \(g\) is concave up everywhere. Also \(g'(1)=f(1)>0\), hence the graph of \(g\) crosses the \(x\)-axis at \(x=1\) with a positive slope. The only option that is concave up, satisfies \(g(1)=0\), and has \(g'(1)>0\) is: ✅ Answer: (D)

Question

If \( f \) is defined by \( f(x)=\displaystyle\int_{5}^{x^{3}}\tan(t)\,dt \), then \( f'(x)= \)

(A) \( \sec^{2}\!\big(x^{3}\big) \)
(B) \( \tan\!\big(x^{3}\big) \)
(C) \( 3x^{2}\sec^{2}\!\big(x^{3}\big) \)
(D) \( 3x^{2}\tan\!\big(x^{3}\big) \)
▶️ Answer/Explanation

Detailed solution
By FTC Part 1 with a variable upper limit \(g(x)=x^{3}\): \( \dfrac{d}{dx}\!\left[\int_{5}^{g(x)} \tan(t)\,dt\right] = \tan\!\big(g(x)\big)\cdot g'(x) \).
Here \( g'(x)=3x^{2} \). Therefore \( f'(x)=\tan\!\big(x^{3}\big)\cdot 3x^{2}=3x^{2}\tan\!\big(x^{3}\big) \).

✅ Answer: (D)

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