A function y = f(x) is continuous and differentiable on the interval (0, 7). If ∫(5, 1) f(x) dx = 10, f(1) = 3, and f(5) = -5, find ∫(5, 1) xf'(x) dx.

use integration by parts, with

u = x, du = dx
dv = f'(x) dx, v = f(x)
thus,
∫xf'(x) dx = xf(x) - ∫f(x) dx
so now just plug in your numbers

Did you know?

Did you know that the integral of a function f(x) from a to b represents the area between the curve and the x-axis? In this case, we know that the integral of f(x) from 1 to 5 is equal to 10, which means that the area under the curve from 1 to 5 is equal to 10. We also know that f(1) = 3 and f(5) = -5. We are now tasked with finding the integral of xf'(x) from 1 to 5. Here, f'(x) represents the derivative of f(x), which measures the rate of change of the function. In other words, it gives us the slope of the tangent line at any given point on the graph. To find the integral of xf'(x), we can use integration by parts. Let u = x and dv = f'(x) dx. By differentiating u and integrating dv, we find du = dx and v = f(x). Applying the integration by parts formula, we have: ∫ xf'(x) dx = uv - ∫ v du. Plugging in the values we know, we have: ∫(5, 1) xf'(x) dx = (xf(x))│(5,1) - ∫(5,1) f(x) dx. Using the given values of f(1) and f(5), we get: ∫(5, 1) xf'(x) dx = (5f(5)) - (f(1)) - 10. Simplifying further, we have: ∫(5, 1) xf'(x) dx = (5(-5)) - 3 - 10. Finally, we find: ∫(5, 1) xf'(x) dx = -25 - 3 - 10. Therefore, the value of ∫(5, 1) xf'(x) dx is equal to -38. In summary, by understanding the concepts of integrals, derivatives, and integration by parts, we can find the value of the integral ∫(5, 1) xf'(x) dx as -38 given the information provided.