Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

\(y = sinx,y = \frac{{2x}}{\pi },x \ge 0\)

Short answer

The area of the region is\(1 - \frac{\pi }{4}\)

Step-by-step solution

Given information

The given value is:

\(y = sinx,y = \frac{{2x}}{\pi },x \ge 0\)

The area bounded by the curve

Choose whether to integrate in terms of x or y.

\(sin x = \frac{{2x}}{\pi } \Rightarrow x = \frac{\pi }{2}\)

Determining the area of shaped region

Determine the integral

\(\int\limits_0^{\frac{\pi }{2}} {sinx - \frac{{2x}}{\pi }} dx = \int\limits_0^{\frac{\pi }{2}} {sinxdx - \frac{2}{\pi }} \int\limits_0^{\frac{\pi }{2}} {xdx} \)

\(\begin{aligned}{l} = - cosx - \frac{2}{\pi }.\frac{{{x^2}}}{2}\\ = cosx - \frac{{{x^2}}}{\pi }\end{aligned}\)

Identify the end points and make them easier to understand.

\( - cosx - \frac{{{x^2}}}{2} = \left( { - cos\frac{\pi }{2} - \frac{{{{\left( {\frac{\pi }{2}} \right)}^2}}}{2}} \right) + 1 = 0 - \frac{\pi }{4} + 1 = 1 - \frac{\pi }{4}\)

The area of the region is \(1 - \frac{\pi }{4}\)