Use the limit process to find the area of the region between the graph of the function and x-axis over the interval

Question

Use the limit process to find the area of the region between the graph of the function  and x-axis over the interval

Quizplus · Accepted Answer

To find the area between the graph of the function and the x-axis, we need to integrate the function with respect to x over the given interval. Using the limit process, we can approximate this area as the sum of the areas of a sequence of rectangles with equal width, which approach zero. Let's divide the interval into n subintervals of equal width Δx = (2-1)/n = 1/n. Then, the x-coordinate of the left endpoint of the k-th subinterval is x_k = 1 + kΔx, and the height of the corresponding rectangle is f(x_k) = 1/(1+x_k) = 1/(1+1+kΔx). The area of this rectangle is A_k = f(x_k)Δx = (1/n)/(1+1+kΔx). Therefore, the total approximate area is:

A ≈ ΣA_k = Σ[(1/n)/(1+1+kΔx)], from k=0 to n-1

This is a Riemann sum, which represents the limit of the approximate areas as n approaches infinity:

A = lim(n → ∞)Σ[(1/n)/(1+1+kΔx)] = ∫[1,2] 1/(1+x) dx

Evaluating this integral, we get:

A = [ln(3) - ln(2)] = ln(3/2) ≈ 0.405

Therefore, the best choice is (B) 0.405. To find the area between the graph and x-axis, we need to integrate the function over the given interval. However, since the function is not integrable in closed form, we need to use the limit process.

We can approximate the area by dividing the interval into n subintervals of equal width, Δx = (8-1)/n. Let xi = 1 + iΔx be the right endpoint of the i-th subinterval, and let yi = 11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11(xi) be the height of the function at that point.

Then, the area of the i-th rectangle is given by Ai = yiΔx, and the total area of all the rectangles is given by A = ΣAi. Evaluating the summation, we get:

A = (11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11(1)Δx) + (11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11(1+Δx)Δx) + ... + (11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11(8-Δx)Δx) + (11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11(8)Δx)

Taking the limit as n approaches infinity, we get:

A = lim[n→∞] [(11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11(1)Δx) + (11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11(1+Δx)Δx) + ... + (11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11(8-Δx)Δx) + (11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11(8)Δx)]

= ∫1^8 11ebbd4d_3edf_d596_8a0a_136093400bbb_TB8202_11 dx

Using numerical integration techniques, we can evaluate this integral to get:

A ≈ 11ebbd4d_3edf_d599_8a0a_2fe9c34441ee_TB8202_11

Therefore, the best choice is B.

Use the Limit Process to Find the Area of the Region