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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 =lim⁡n→∞(∑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],…,[xn−1,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 [xk−1,xk}\left[x_{k-1}, x_{k}\right\}, where 1≤k≤n1 \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. lim⁡n→∞(∑k=1nf(xk)Δx)=\lim _{n \rightarrow \infty}\left(\sum_{k=1}^{n} f\left(x_{k}\right) \Delta x\right)= □\square

Studdy Solution

STEP 1

1. We are calculating the area between the curve f(x)=6x3 f(x) = 6x^3 and the x x -axis over the interval [0,2][0, 2].
2. We will use the limit of right-endpoint Riemann sums to find this area.
3. The interval [0,2][0, 2] is divided into n n equal subintervals.

STEP 2

1. Determine the width of each subinterval, Δx\Delta x.
2. Find the right endpoints x1,x2,x3x_1, x_2, x_3 of the first three subintervals.
3. Develop a general expression for the right endpoint xkx_k of the kk-th subinterval.
4. Express f(xk)f(x_k) in terms of kk and nn.
5. Calculate f(xk)Δxf(x_k) \Delta x.
6. Evaluate the Riemann sum ∑k=1nf(xk)Δx\sum_{k=1}^{n} f(x_k) \Delta x.
7. Compute the limit of the Riemann sum as n→∞n \to \infty.

STEP 3

The width of each subinterval Δx\Delta x is given by the length of the interval divided by the number of subintervals:
Δx=2−0n=2n\Delta x = \frac{2 - 0}{n} = \frac{2}{n}

STEP 4

The right endpoints of the first three subintervals are calculated as follows:
x1=0+2n=2nx_1 = 0 + \frac{2}{n} = \frac{2}{n} x2=0+2⋅2n=4nx_2 = 0 + 2 \cdot \frac{2}{n} = \frac{4}{n} x3=0+3⋅2n=6nx_3 = 0 + 3 \cdot \frac{2}{n} = \frac{6}{n}
Thus, the right endpoints are:
x1,x2,x3=2n,4n,6nx_1, x_2, x_3 = \frac{2}{n}, \frac{4}{n}, \frac{6}{n}

STEP 5

The general expression for the right endpoint xkx_k of the kk-th subinterval is:
xk=0+k⋅2n=2knx_k = 0 + k \cdot \frac{2}{n} = \frac{2k}{n}

STEP 6

Substitute xk=2knx_k = \frac{2k}{n} into the function f(x)=6x3f(x) = 6x^3:
f(xk)=6(2kn)3=6⋅8k3n3=48k3n3f(x_k) = 6\left(\frac{2k}{n}\right)^3 = 6 \cdot \frac{8k^3}{n^3} = \frac{48k^3}{n^3}

STEP 7

Calculate f(xk)Δxf(x_k) \Delta x:
f(xk)Δx=48k3n3⋅2n=96k3n4f(x_k) \Delta x = \frac{48k^3}{n^3} \cdot \frac{2}{n} = \frac{96k^3}{n^4}

STEP 8

Evaluate the Riemann sum:
∑k=1nf(xk)Δx=∑k=1n96k3n4=96n4∑k=1nk3\sum_{k=1}^{n} f(x_k) \Delta x = \sum_{k=1}^{n} \frac{96k^3}{n^4} = \frac{96}{n^4} \sum_{k=1}^{n} k^3
Using the formula for the sum of cubes, ∑k=1nk3=(n(n+1)2)2\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2, we have:
∑k=1nf(xk)Δx=96n4(n(n+1)2)2=96n4⋅n2(n+1)24\sum_{k=1}^{n} f(x_k) \Delta x = \frac{96}{n^4} \left(\frac{n(n+1)}{2}\right)^2 = \frac{96}{n^4} \cdot \frac{n^2(n+1)^2}{4}
=24n2(n+1)2n4=24(n+1)2n2= \frac{24n^2(n+1)^2}{n^4} = \frac{24(n+1)^2}{n^2}

STEP 9

Compute the limit of the Riemann sum as n→∞n \to \infty:
lim⁡n→∞24(n+1)2n2=lim⁡n→∞24(n2+2n+1)n2=lim⁡n→∞(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}

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