Question
The region bounded by the x-axis and the part of the graph of y= cos x between \(x=-\frac{\pi }{2} \)and \(x=\frac{\pi }{2}\) is separated into two regions by the line x = k . If the area of the region for three times the area of the region for \(-\frac{\pi }{2}\leqslant x\leqslant\) k is three times the area of the region for \(k\leq x\leq \frac{\pi }{2}\) then k=
(A) arcsin\(\left ( \frac{1}{4} \right ) \) (B) arcsin\(\left ( \frac{1}{3} \right )\) (C) \(\frac{\pi }{6} \) (D)\(\frac{π}{4}\) (E) \(\frac{π}{3}\)
▶️Answer/Explanation
Ans:C
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 \(\frac{2}{3}\)
B \(\frac{4}{3}\)
C \(\frac{16}{3}\)
D \(\frac{28}{3}\)
▶️Answer/Explanation
Ans:A
The graphs of y=2x and \(y=4x-x^{2}\) intersect when x=0 and x=2. The graph of \(y=4x-x^{2}\) lies above the graph y=2x on the interval 0≤x≤2. (One way to see this is to sketch a graph of the parabola and the line, observing that the graph of \(y=4x-x^{2}\) has a slope of 4 at x=0, while the graph of y=2x has a slope of 2) The area of the region bounded by the two graphs is therefore \(\int ^{2}_{0}(4x-x^{2}-2x)dx=\int _{0}^{2}(2-x^{2})dx=[x^{2}-\frac{x^{3}}{3}]|_{0}^{2}=4-\frac{8}{3}=\frac{4}{3}\)
Question
Calculate the approximate area of the shaded region in the figure by the trapezoidal rule, using divisions at x = \(\frac{4}{3}\) and x=\(\frac{5}{3}\)
(A)\(\frac{50}{27}\) (B)\(\frac{251}{108}\) (C)\(\frac{7}{3}\) (D)\(\frac{127}{54}\) (E)\(\frac{77}{27}\)
▶️Answer/Explanation
Ans:D
Question
The area of the region in the first quadrant that is enclosed by the graphs of \(y=x^{3}+8\) and y=x+8 is
(A) \(\frac{1}{4}\) (B) \(\frac{1}{2}\) (C) \(\frac{3}{4}\) (D) 1 (E) \(\frac{65}{4}\)
▶️Answer/Explanation
Ans:C
