Question

[-/0.83 Points] SCALCET8 6.4.018. of a hole in the bucket at a rate of 0.25lb/s. Find the work done in pulling the bucket to the top of the well. Show how to approximate the required work by a Riemann sum. (Let x be the height in feet above the bottom of the well. Enter x

i
​

∗
as x
i
​
.) lim
n→[infinity]
​
∑
i=1
n
​
()Δx Express the work as an integral. ∫
0
1
​
(1)dx Evaluate the integral. ft−lb

Answer 1

Explanation:

To express the given sum as an integral, we need to consider the limit as n approaches infinity. The sum can be rewritten using the sigma notation:

lim(n→∞) ∑(i=1 to n) ()Δx

where ()Δx represents the function being summed and Δx represents the width of each subinterval.

In this case, the sum can be expressed as:

lim(n→∞) ∑(i=1 to n) [f(xi) Δx]

where xi represents the x-value within each subinterval.

To find the integral form, we need to express the sum as an Riemann sum. Considering the interval [0, 1] and dividing it into n subintervals of equal width, we have:

Δx = (1 - 0)/n = 1/n

xi = 0 + i(Δx) = i/n

The sum can now be written as:

lim(n→∞) ∑(i=1 to n) [f(xi) (1/n)]

To express it as an integral, we rewrite it as:

lim(n→∞) [1/n] ∑(i=1 to n) f(i/n)

Taking the limit as n approaches infinity, we have:

∫(0 to 1) f(x) dx

Therefore, the given sum is equivalent to the integral ∫(0 to 1) f(x) dx.

The second part of the question asks to evaluate the integral ∫(0 to 1) (1) dx.

Since the integrand is a constant function, integrating it with respect to x gives:

∫(0 to 1) (1) dx = x ∣(0 to 1) = 1 - 0 = 1.

Therefore, the value of the integral ∫(0 to 1) (1) dx is 1.

The unit of the work or the ft-lb mentioned in the question suggests it represents a quantity of work and is not directly related to the integral evaluation.