In this problem you will calculate ∫305x3dx by using the formal definition of the definite integral: ∫baf(x)dx=limn→[infinity][∑k=1nf(x∗ k)Δx]. (a) The interval [0,3] is divided into n equal subintervals of length Δx. What is Δx (in terms of n)? Δx = (b) The right-hand endpoint of the kth subinterval is denoted x∗k. What is x∗k (in terms of k and n)? x∗k = (c) Using these choices for x∗k and Δx, the definition tells us that ∫305x3dx=limn→[infinity][∑k=1nf(x∗k )Δx]. What is f(x∗k)Δx (in terms of k and n)? f(x∗k)Δx = (d) Express ∑k=1nf(x∗k)Δx in closed form. (Your answer will be in terms of n.) ∑k=1nf(x∗k)Δx = (e) Finally, complete the problem by taking the limit as n→[infinity] of the expression that you found in the previous part. ∫305x3dx=limn→[infinity][∑k=1nf(x∗k )Δx] =