For the intergral of -(x^2)/3 +6 interval [0,3]
Rewrite it as a function of n using the right hand endpoint without any summation sign.
can anyone help me start this problem out?
I noticed somebody else had that post before, perhaps it was you.
I am not sure exactly what you are asking, I am not familiar with the terminology you are using.
do you want the integral of -1/3(x^2) + 6 from 0 to 3?
the integral would be -1/9(x^3) + 6x + c
which when evaluated is 15
i know how to evaluate the equation but how do i rewrite it as a function of n using the right hand endpoint.
i got delta x =3/n and x*k=3k/n
what do i do next?
To rewrite the integral as a function of n using the right hand endpoint, you need to express the integral in terms of n instead of x. Let's break it down step by step.
First, you correctly identified that Δx = 3/n, which represents the width of each subinterval.
Next, you can express the right-hand endpoint of each subinterval. Since there are n subintervals, the right-hand endpoint of the k-th subinterval can be written as xk = 3k/n. This represents the value of x at the right end of the k-th subinterval.
Now, we can rewrite the integral using these new terms. Instead of integrating with respect to x, we will integrate with respect to k (since k represents the subintervals).
The integral becomes ∫[-(x^2)/3 + 6] dx from x = 0 to x = 3.
Using the substitution x = 3k/n, we can rewrite the integral as ∫[-(9k^2)/(n^2) + 6] dx from k = 0 to k = n.
The Δx becomes Δk = 1 because we are now dealing with k as the variable of integration.
Therefore, the rewritten integral is ∫[-(9k^2)/(n^2) + 6] dx from k = 0 to k = n, where Δk = 1.
Hope this helps! Let me know if you have any further questions.