Question

Prove, in the following two ways, that the signed area under the graph of the function $f\left(x\right)=\sin x{\cos }^{2}x$on an interval $[-a,a]$centered about the origin is always zero:

(a) by calculating a definite integral;

(b) by considering the symmetry of the graph of the function$f\left(x\right)=\sin x{\cos }^{2}x$

Short answer

Hence proved.

Step-by-step solution

Part(a) Step 1. Given information.

The given function and interval is$f\left(x\right)=\sin x{\cos }^{2}xand[-a,a].$

Part (a) Step 2. Explanation.

Using definite integral,

$$\begin{gathered} {\int }_{-a}^a\sin x{\cos }^{2}xdx={\left[-\frac{1}{3}{\cos }^3\left(x\right)\right]}_{-a}^a \\ =\left[-\left[\frac{1}{3}{\cos }^3\left(a\right)-\frac{1}{3}{\cos }^3(-a)\right]\right] \\ =\left[\frac{1}{3}{\cos }^3\left(a\right)-\frac{1}{3}{\cos }^3\left(a\right)\right] \\ =0 \end{gathered}$$

Part (b) Step 1. Explanation.

The graph of the function is ,

Now, from the graph, the function is general sine and cosine function with period whose graph is half above and half below the -axis through origin.