Need help with limit for integration

Your answer to part c is correct, but your answers to parts d and e are incorrrect. In part d, you will need to use the formula for the sum of
1^2 + 2^2 + ...+ n^2 = (n(n + 1)(2n + 1))/6. Once you create a sigma (summation) for your answer to part c, you can factor out both the 4/n and the (4/n)^2 in front of the sigma since each is constant as k changes from 1 to n. Also for (k*4/n)^2 + 6), you can compute the sum for the 6 term separately. If you write out some work for this and submit it, someone can check it.
 
Prof said:
Your answer to part c is correct, but your answers to parts d and e are incorrrect. In part d, you will need to use the formula for the sum of
1^2 + 2^2 + ...+ n^2 = (n(n + 1)(2n + 1))/6. Once you create a sigma (summation) for your answer to part c, you can factor out both the 4/n and the (4/n)^2 in front of the sigma since each is constant as k changes from 1 to n. Also for (k*4/n)^2 + 6), you can compute the sum for the 6 term separately. If you write out some work for this and submit it, someone can check it.
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Thanks. I figured it out myself.
 
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