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
A left Riemann sum, a right Riemann sum, and a trapezoidal sum approximate \(\int_{0}^{1} f(x) \, dx\), each with same number of subintervals. Graph of \( f \) shown. Which sums underestimate \(\int_{0}^{1} f(x) \, dx\)?▶️ Answer/Explanation
Solution
Graph shows \( f(x) \) is concave down (decreasing slope) from 0 to 1.
Left Riemann sum underestimates as rectangles lie below curve.
Right Riemann sum overestimates as rectangles lie above curve.
Trapezoidal sum underestimates as average of left and right, but curve’s concavity pulls it below integral.
Thus, I and III underestimate.
✅ Answer: D
▶️ Answer/Explanation
Solution
For \(\int_{a}^{b} f(x) \, dx\), a Riemann sum is \(\lim_{n \to \infty} \sum_{k=1}^{n} f(x_k) \Delta x\), where \(\Delta x = \frac{b-a}{n}\), \(x_k = a + k \Delta x\).
Here, \(a = 3\), \(b = 5\), \(f(x) = x^4\), so \(\Delta x = \frac{5-3}{n} = \frac{2}{n}\), \(x_k = 3 + k \cdot \frac{2}{n}\).
Sum: \(\lim_{n \to \infty} \sum_{k=1}^{n} \left( 3 + \frac{2k}{n} \right)^4 \frac{2}{n}\).
✅ Answer: D
▶️ Answer/Explanation
Solution
Two inscribed rectangles, width = 1. Use left endpoints: \( x = 0, 1 \). Area = \( e^{0^2} \cdot 1 + e^{1^2} \cdot 1 = 1 + e \approx 3.718 \).
Trapezoidal rule, \( n = 2 \), width = 1. Points: \( x = 0, 1, 2 \). Area = \(\frac{1}{2} (e^{0^2} + 2e^{1^2} + e^{2^2}) = \frac{1 + 2e + e^4}{2} \approx 30.517\).
Difference: \(\frac{1 + 2e + e^4}{2} – (1 + e) = \frac{e^4 – 1}{2} \approx 26.799\).
Rounded: 26.80.
✅ Answer: D