Question
The expression \((\frac{1}{50}\left ( \sqrt{\frac{1}{50}+}\sqrt{\frac{2}{50}}+\sqrt{\frac{3}{50}}+…+\sqrt{\frac{50}{50}} \right )\) is a Riemann sum approximation for
(A)\(\int_{0}^{1}\sqrt{\frac{x}{50}}dx\)
(B)\(\int_{0}^{1}\sqrt{X}dx\)
(C)\(\frac{1}{50}\int_{0}^{1}\sqrt{\frac{x}{50}}dx \)
(D)\(\frac{1}{50}\int_{0}^{1}\sqrt{x}dx\)
(E)\(\frac{1}{50}\int_{0}^{50}\sqrt{x}dx\)
▶️Answer/Explanation
Ans:B
Question
A left Riemann sum, a right Riemann sum, and a trapezoidal sum are used to approximate the value of \(\int_{0}^{1}f(x)dx\), each using the same number of subintervals. The graph of the function f is shown in the figure above. Which of the sums give an underestimate of the value of \(\int_{0}^{1}f(x)dx,\)
I. Left sum
II. Right sum
III. Trapezoidal sum
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
▶️Answer/Explanation
Ans:D
Question
Which of the following limits is equal to \(\int_{3}^{5}x^{4}dx?\)
(A) \(\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left ( 3+\frac{k}{n} \right )^{4}\frac{1}{n}\)
(B)\( \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left ( 3+\frac{k}{n} \right )^{4}\frac{2}{n}\)
(C) \(\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left ( 3+\frac{2k}{n} \right )^{4}\frac{1}{n}\)
(D) \(\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left ( 3+\frac{2k}{n} \right )^{4}\frac{2}{n}\)
▶️Answer/Explanation
Ans:D
Question
If the definite integral \(\int_{0}^{2}e^{x^{2}} dx\) is first approximated by using two inscribed rectangles of equal
width and then approximated by using the trapezoidal rule with n = 2 , the difference between the
two approximations is
(A) 53.60 (B) 30.51 (C) 27.80 (D) 26.80 (E) 12.78
▶️Answer/Explanation
Ans:D
Rectangle approximation \(= e^{0}+e^{1}=1+e\) Trapezoid approximation. = \((1+2e+e^{4})/2.\) Difference = \((e^{4}-1)/2 = 26.799.\)