Try using the right end point of each interval so you have$$Parker Hays said:Homework Statement
a. Write down a Riemann sum for the integral ∫x3dx from 0 to 1.
b. Given the following identity 13+23+33...+N3=(N(N+1)/2)2, show that the Riemann sums for ∫x3dx from 0 to 1 converge to 1/4.
The Attempt at a Solution
I believe I have gotten part a. I got ∑i^3/N^4 from i=0 to N-1. Part b I have attempted to solve from multiple angles for the past hour but none of the ways I've tried have yielded any positive results.
Yes, I got part a correct. However, part b still does not make sense to me. I have tried solving for N^4 and various other strategies but I don't see how I can plug it in or do anything to get that it converges to 1/4.LCKurtz said:Try using the right end point of each interval so you have$$
\sum_{i=1}^n \frac{i^3}{n^3}\cdot \frac 1 n$$
I had already completed part a correctly, and you didn't give me a hint for part b.LCKurtz said:Show me what you have done with the sum I gave you. Have you used the given hint?
He did. Use the sum in post #2 with the sum you have for b) and calculate the limit.Parker Hays said:I had already completed part a correctly, and you didn't give me a hint for part b.
Riemann sums are used to approximate the area under a curve by dividing the region into smaller rectangles and calculating the sum of their areas. In Calculus 2 for Engineers, Riemann sums are used to solve problems involving continuous functions and to find the volume of solids of revolution.
Riemann sums are calculated by dividing the interval of the function into smaller subintervals and evaluating the function at specific points within each subinterval. The sum of these function values multiplied by the width of each subinterval gives an approximation of the area under the curve.
A left Riemann sum uses the left endpoint of each subinterval to evaluate the function, a right Riemann sum uses the right endpoint, and a midpoint Riemann sum uses the midpoint of each subinterval. The accuracy of the approximation depends on the choice of Riemann sum, with the midpoint Riemann sum generally being the most accurate.
Riemann sums are used to approximate the value of a definite integral. As the number of subintervals increases, the Riemann sum becomes a more accurate approximation of the definite integral. In the limit, as the number of subintervals approaches infinity, the Riemann sum becomes equal to the definite integral.
Riemann sums can be used to approximate the area under a curve for any continuous function. However, the accuracy of the approximation may vary depending on the characteristics of the function, such as its shape and behavior. In some cases, other methods such as the Trapezoidal Rule or Simpson's Rule may provide a more accurate approximation.