- Recall that the formula for finding the area under the curve using the left endpoints is
[y 0 + y 1 + y 2 +...+ yn+1]. This formula assumes that the x-values are evenly spaced, but theyaren’t here, so we will replace the values of with the appropriate widths of eachrectangle. The width of the first rectangle is 1 − 0 = 1; the second width is 3 − 1 = 2; the thirdis 5 − 3 = 2; the fourth is 9 − 5 = 4; and the fifth is 14 − 9 = 5. We find the height of eachrectangle by evaluating g(x) at the appropriate value of x, the left endpoint of each interval onthe x-axis. Here y 0 = 10, y 1 = 8, y 2 = 11, y 3 = 17, and y 4 = 20. Therefore, we can approximatethe integral with: g(x) dx = (1)(10) + (2)(8) + (2)(11) + (4)(17) + (5)(20) = 216.SOLUTIONS TO PRACTICE PROBLEM SET 21
1.
We find the average value of the function, f(x), on the interval [a, b] using the formula f(c) = f(x) dx. Here we are looking for the average value of f(x) = 4x cos x^2 on the interval . Using the formula, we need to find =
.
We will need to use u-substitution to evaluate the integral. Let u = x^2 and du = 2x dx. We needto substitute for 4x dx, so we multiply by 2 to get 2 du = 4x dx. Now we can substitute into theintegral: ∫(4x cos x^2 ) dx = 2 ∫cos u du = 2 sin u. If we substitute back, we get 2 sin x^2 . Now
we can evaluate the integral: = =