Question

Find the area of the region bounded by the graphs of the equations. $$ y=\cos ^{2} x, \quad y=\sin x \cos x, \quad x=-\pi / 2, \quad x=\pi / 4 $$

Short answer

The area of the region bounded by the graphs of the equations can be found by evaluating the definite integral \[\int_{-\pi/2}^{\pi/4}\left(\frac{1+\cos 2x}{2}-\frac{1}{2}\sin 2x\right) dx\]. The resulting value represents the area.

Step-by-step solution

Rewrite the function

Rewrite the functions using trigonometric identities: \(y=\cos ^{2} x=\frac{1+\cos 2x}{2}\) and \(y=\sin x \cos x = \frac{1}{2}\sin 2x\). The new functions are easier to integrate and simplify.

Find intersects

Set the two functions equal to each other to find x-values where the functions intersect within the range of \(x=[-π/2,π/4]\): \(\frac{1+\cos 2x}{2}=\frac{1}{2}\sin 2x\). Solve for x to find the intersection points.

Integrate functions

Integrate the functions from the left intersection to the right intersection found in Step 2. Subtract the integral of the lower function (\(\frac{1}{2}\sin 2x\)) from the integral of the upper function (\(\frac{1+\cos 2x}{2}\)) to find the area between the curves: \[\int_{-\pi/2}^{\pi/4}\left(\frac{1+\cos 2x}{2}-\frac{1}{2}\sin 2x\right) dx\]

Evaluate the integral

Evaluate the definite integral found in step 3, applying the fundamental theorem of calculus.