AP Calculus AB : 6.6 Applying Properties of  Definite Integrals- Exam Style questions with Answer- MCQ

Question

\(\int_{1}^{2}(4x^3-6x)dx\)

(A) 2
(B) 4
(C) 6
(D) 36
(E) 42

▶️Answer/Explanation

Ans:C

\(\int_{1}^{2}(4x^3-6x)dx=(x^4-3x^2)|_{1}^{2}=(16-12)-(1-3)=6\)

Question

If \(\int_{-1}^{1}e^{-x^{2}}dx=k\),then\(\int_{-1}^{0}e^{-x^{2}}dx\)=

(A) −2k                      (B) −k                          (C) \(− \frac{2}{k}\)             (D)\(\frac{2}{k}\)                                         (E) 2k

▶️Answer/Explanation

Ans:D

since \(e^{-x^{2}}\) is even \(\int_{-1}^{0}e^{-x^{2}}dx=\frac{1}{2}\int_{-1}^{1}e^{-x^{2}}dx=\frac{1}{2}k\)

Question

Let g be the function given by  g(x)=’\(\int ^{x}_{3}(t^{2}-5t-14)dt\) . What is the x-coordinate of the point of inflection of the graph of g?
A -2
B\(\frac{5}{2}\)
C3
D7

▶️Answer/Explanation

Ans:B
To find the point of inflection of the graph of g,determine where g″ changes sign.
g′(x)\(=x^{2}-5x-14\)
g′′(x)=2x−5
Then g′′(x)=0 at \(x=\frac{5}{2}\). Since g′′(x)<0 for \(x<\frac{5}{2}\) and g′′(x)>0 for \(x>\frac{5}{2}\), the graph of g changes concavity at \(x=\frac{5}{2}\) and therefore, the graph of g has a point of inflection at \(x=\frac{5}{2}\).

Question

 The graph of a piecewise-linear function f , for \(−1 ≤ x  ≤4 \), is shown above. What is the value of \(\int_{-1}^{4}f(x)dx\)?

(A) 1                                         (B) 2.5                              (C) 4                               (D) 5.5                                        (E) 8

▶️Answer/Explanation

Ans:B

\(\int_{-1}^{4}f(x)dx=\int_{-1}^{2}f(x)dx+\int_{2}^{4}f(x)dx\)
= Area of trapezoid(1) – Area of trapezoid(2) = 4 -1.5= 2.5 

Scroll to Top