Math  /  Calculus

QuestionIn this problem you will calculate the area between f(x)=6x3f(x)=6 x^{3} and the xx-axis over the interval [0,2][0,2] using a limit of right-endpoint Riemann sums:  Area =limn(k=1nf(xk)Δx)\text { Area }=\lim _{n \rightarrow \infty}\left(\sum_{k=1}^{n} f\left(x_{k}\right) \Delta x\right)
Express the following quantities in terms of nn, the number of rectangles in the Riemann sum, and kk, the index for the rectangles in the Riemann sum. a. We start by subdividing [0,2][0,2] into nn equal width subintervals [x0,x1],[x1,x2],,[xn1,xn]\left[x_{0}, x_{1}\right],\left[x_{1}, x_{2}\right], \ldots,\left[x_{n-1}, x_{n}\right] each of width Δx\Delta x. Express the width of each subinterval Δx\Delta x in terms of the number of subintervals nn. Δx=\Delta x=\square b. Find the right endpoints x1,x2,x3x_{1}, x_{2}, x_{3} of the first, second, and third subintervals [x0,x1],[x1,x2],[x2,x3]\left[x_{0}, x_{1}\right],\left[x_{1}, x_{2}\right],\left[x_{2}, x_{3}\right] and express your answers in terms of nn. x1,x2,x3= (Enter a comma separated list.) x_{1}, x_{2}, x_{3}=\square \text { (Enter a comma separated list.) } c. Find a general expression for the right endpoint xkx_{k} of the kk th subinterval [xk1,xk}\left[x_{k-1}, x_{k}\right\}, where 1kn1 \leq k \leq n. Express your answer in terms of kk and nn. xk=x_{k}= \square d. Find f(xk)f\left(x_{k}\right) in terms of kk and nn. f(xk)=f\left(x_{k}\right)= \square e. Find f(xk)Δxf\left(x_{k}\right) \Delta x in terms of kk and nn. f(xk)Δx=f\left(x_{k}\right) \Delta x= \square f. Find the value of the right-endpoint Riemann sum in terms of nn. k=1nf(xk)Δx=\sum_{k=1}^{n} f\left(x_{k}\right) \Delta x= \square g. Find the limit of the right-endpoint Riemann sum. limn(k=1nf(xk)Δx)=\lim _{n \rightarrow \infty}\left(\sum_{k=1}^{n} f\left(x_{k}\right) \Delta x\right)= \square

Studdy Solution
Compute the limit of the Riemann sum as nn \to \infty:
limn24(n+1)2n2=limn24(n2+2n+1)n2=limn(24+48n+24n2)=24\lim_{n \to \infty} \frac{24(n+1)^2}{n^2} = \lim_{n \to \infty} \frac{24(n^2 + 2n + 1)}{n^2} = \lim_{n \to \infty} \left(24 + \frac{48}{n} + \frac{24}{n^2}\right) = 24
Thus, the area is:
24\boxed{24}

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord