Calculus 2 for Engineers: Riemann sums

[Parker Hays]

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.

 

[LCKurtz]

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.

Try using the right end point of each interval so you have$$
\sum_{i=1}^n \frac{i^3}{n^3}\cdot \frac 1 n$$

 

[Parker Hays]

Try using the right end point of each interval so you have$$
\sum_{i=1}^n \frac{i^3}{n^3}\cdot \frac 1 n$$

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]

Show me what you have done with the sum I gave you. Have you used the given hint?

 

[Parker Hays]

Show me what you have done with the sum I gave you. Have you used the given hint?

I had already completed part a correctly, and you didn't give me a hint for part b.

 

[fresh_42]

I had already completed part a correctly, and you didn't give me a hint for part b.

He did. Use the sum in post #2 with the sum you have for b) and calculate the limit.

 

[Parker Hays]

I figured it out. Thanks for the help, sorry I didn't realize you were giving me another idea I just thought you were restating the answer I already had for part a @LCKurtz.