The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles.

A = lim n → ∞ [f(x1)Δx + f(x2)Δx + . . . + f(xn)Δx]
Use this definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.
f(x) = 7x cos(7x), 0 ≤ x ≤ π/2

Sure, here's the expression for the area under the graph of f as a limit:

A = lim n → ∞ [f(x1)Δx + f(x2)Δx + . . . + f(xn)Δx]

Let's start by finding the width of each rectangle, Δx. Since we have a continuous function and we want to find the area under the graph from 0 to π/2, we can divide this interval into n equally spaced subintervals. So, each subinterval has a width of (π/2 - 0)/n = π/(2n).

Next, let's find the x-coordinates for the rectangles. We can choose the left endpoint of each subinterval to determine the x-coordinate for each rectangle. Since our interval starts at x = 0, the x-coordinates of the rectangles can be represented as x1 = 0, x2 = π/(2n), x3 = π/n, and so on, up to xn = (n-1)π/(2n).

Now, let's substitute these values into the expression for the area:

A = lim n → ∞ [7x1cos(7x1)Δx + 7x2cos(7x2)Δx + . . . + 7xn cos(7xn)Δx]
= lim n → ∞ [7(0)cos(0)(π/(2n)) + 7(π/(2n))cos(7(π/(2n)))(π/(2n)) + . . . + 7((n-1)π/(2n))cos(7((n-1)π/(2n)))(π/(2n))]

And there you have it! That's the expression for the area under the graph of f as a limit. Remember, we didn't evaluate the limit, so this is just a representation of the area. Feel free to calculate the limit to find the actual area if you want.

To find an expression for the area under the graph of the function f(x) = 7x cos(7x) on the interval [0, π/2] using the definition of the limit of the sum of the areas of approximating rectangles, follow these steps:

1. Divide the interval [0, π/2] into n equal subintervals of width Δx = (π/2)/n.
2. Choose a value for xi in each subinterval, where i ranges from 1 to n.
3. The left endpoint of each subinterval is xi-1 = (i-1)Δx.
4. Substitute these values into the expression for A:
A = lim n → ∞ [f(x1)Δx + f(x2)Δx + . . . + f(xn)Δx]
A = lim n → ∞ [(7x1 cos(7x1))Δx + (7x2 cos(7x2))Δx + . . . + (7xn cos(7xn))Δx]
A = lim n → ∞ [(7(i-1)Δx cos(7(i-1)Δx))Δx + (7iΔx cos(7iΔx))Δx + . . . + (7nΔx cos(7nΔx))Δx]
A = lim n → ∞ [Δx^2 (7(i-1) cos(7(i-1)Δx) + 7i cos(7iΔx) + . . . + 7n cos(7nΔx))]
5. Simplify the expression:
A = lim n → ∞ Δx^2 (7(0) cos(7(0)Δx) + 7(1) cos(7(1)Δx) + . . . + 7(n-1) cos(7(n-1)Δx) + 7n cos(7nΔx))
A = lim n → ∞ Δx^2 (7n cos(7nΔx) + 7(n-1) cos(7(n-1)Δx) + . . . + 7(1) cos(7(1)Δx) + 7(0) cos(7(0)Δx))

This expression represents the area under the graph of f as a limit.

To find an expression for the area under the graph of f(x) = 7x cos(7x) as a limit, we need to use the given definition involving the sum of the areas of approximating rectangles.

First, we need to determine the interval of integration. In this case, it is given that 0 ≤ x ≤ π/2.

Next, we divide the interval [0, π/2] into n subintervals of equal width. We can do this by letting Δx = (π/2 - 0)/n = π/(2n).

Now, we need to determine the left endpoints of each subinterval, denoted as x1, x2, ..., xn. Since we are starting from x = 0 and moving in the positive x direction with equal width subintervals, we can set x1 = 0, x2 = Δx, x3 = 2Δx, and so on up to xn = (n-1)Δx.

With the subintervals and their corresponding left endpoints determined, we can calculate the area of each rectangle. The area of the ith rectangle is given by f(xi) * Δx. In this case, we have f(xi) = 7xi cos(7xi) and Δx = π/(2n). So, the area of the ith rectangle is 7xi cos(7xi) * π/(2n).

Finally, we sum up the areas of all the rectangles by taking the limit as n approaches infinity, according to the definition given:

A = lim n → ∞ [f(x1) Δx + f(x2) Δx + . . . + f(xn) Δx]
= lim n → ∞ [7x1 cos(7x1) * π/(2n) + 7x2 cos(7x2) * π/(2n) + . . . + 7xn cos(7xn) * π/(2n)]

So, the expression for the area under the graph of f(x) as a limit is:

A = lim n → ∞ [7x1 cos(7x1) * π/(2n) + 7x2 cos(7x2) * π/(2n) + . . . + 7xn cos(7xn) * π/(2n)]

It is important to note that this expression represents the formula for the area, but it is not yet evaluated. To find the actual area, we would need to evaluate the limit or use numerical methods to approximate the value.

well, you have the formula, so plug in

xi = 0 + k(pi/2)/n
Δx = (pi/2)/n

sum over k=0..n