Mathematics - Definite Integration Question with Solution | TestHub

We are given the functions $P(x)=a{x}^2+bx+c$ and $f(x)={\sin }^{3}x$.

We need to evaluate the integral:

$$I={\int }_0^{2\pi }\underset{I}{\underbrace{P(x)}}\cdot \underset{II}{\underbrace{f^{\prime \prime }(x)}}dx$$

Using Integration by Parts

$$\begin{array}{rl}I& ={P(x)\cdot f^{\prime }(x)|}_0^{2\pi }-{\int }_0^{2\pi }{P}^{\prime }(x)\cdot f^{\prime }(x)dx\\ & =\underset{zero}{\underbrace{{P(x)\cdot f^{\prime }(x)|}_0^{2\pi }}}-{\int }_0^{2\pi }\underset{I}{\underbrace{P^{\prime }(x)}}\cdot \underset{II}{\underbrace{f^{\prime }(x)}}dx\\ & =-\left[{P^{\prime }(x)\cdot f(x)|}_0^{2\pi }-{\int }_0^{2\pi }{P}^{\prime \prime }(x)\cdot f(x)dx\right]\\ & ={\int }_0^{2\pi }{P}^{\prime \prime }(x)\cdot f(x)dx\end{array}$$

Since $P(x)=a{x}^2+bx+c$, then $P^{\prime }(x)=2ax+b$ and $P^{\prime \prime }(x)=2a$.

Substituting $P^{\prime \prime }(x)$ into the integral:

$$I={\int }_0^{2\pi }2a\cdot {\sin }^{3}xdx=2a{\int }_0^{2\pi }{\sin }^{3}xdx$$

We know that ${\int }_0^{2\pi }{\sin }^{3}xdx=0$. This is because ${\sin }^{3}x$ is an odd function about $x=\pi$ and the integral over a full period of an odd power of sine or cosine is zero.

Therefore, $I=2a\cdot 0=0$.