Find the area of the region between y = x sin x and y = x for 0 ≤ x ≤ pi divided by 2
You will need to integrate the function
f(x)=xsin(x)-x
=x(sin(x)-1)
between 0 and π/2.
Since we are looking for the area, we have to make sure that the function f(x) remains positive within the limits of integration.
Since f(0)=0, f(π/2)=0, and f(x) is positive between the two limits, we just have to evaluate the integral:
∫x(sin(x)-1)dx between 0 and π/2.
This can be done by integration by parts:
∫x(sin(x)-1)dx
= x(-cos(x)-x) + ∫1*(cos(x)+x)dx
= sin(x)+x²/2 - xcos(x) - x² + C
= sin(x) - xcos(x) - x²/2 + C
Evaluate the integral between the given limits (ignore the constant C).
I get (π²/8)-1.
Scott, if you cannot wait and repost, i suggest you indicate that you are reposting. If teachers repeat the same work, the end result is everyone gets slower responses.
To find the area of the region between the two curves, we first need to determine the points of intersection. These occur when y = x sin x is equal to y = x.
Setting x sin x = x, we can simplify the equation to sin x = 1. Rearranging further, we have x = pi/2.
Now, we can find the area of the region by integrating y = x sin x - y = x with respect to x over the interval [0, pi/2].
The area A is given by the integral:
A = ∫[0, pi/2] (x sin x - x) dx
Evaluating the integral, we get:
A = [1/2 x^2 (sin x - 2) - cos x] evaluated from 0 to pi/2
A = [(1/2 (pi/2)^2 (sin (pi/2) - 2) - cos (pi/2)] - [1/2 (0)^2 (sin (0) - 2) - cos (0)]
A = [(pi^2/8 - 1) - (-1/2)] - [0]
A = pi^2/8 - 1 + 1/2
A = pi^2/8 + 1/2
Therefore, the area of the region between y = x sin x and y = x for 0 ≤ x ≤ pi/2 is pi^2/8 + 1/2.