Mathematics - Definite Integration Question with Solution | TestHub

If $f(x)$ is odd $\Rightarrow f^{\prime }(x)$ is even, but the converse is not true. For example, if $f^{\prime }(x)=x\sin x$, then $f(x)=\sin x-x\cos x+C$.

Then $f(-x)=-\sin x+x\cos x+C$.

So, $f(x)+f(-x)$ is a constant, which need not be zero.

For S-1: $f(x)={\int }_0^x\sqrt{1+t^2}dt$; $g(x)=\sqrt{1+x^2}$.

$$\begin{array}{rl}f(-x)& ={\int }_0^{-x}\sqrt{1+t^2}dt\\ Let& t=-y\\ dt& =-dy\\ When& t=0,y=0\\ When& t=-x,y=x\\ f(-x)& ={\int }_0^x\sqrt{1+(-y{)}^2}(-dy)\\ f(-x)& =-{\int }_0^x\sqrt{1+y^2}dy\\ \therefore f(x)+f(-x)& =0\end{array}$$

This implies that $f$ is odd, and $g$ is obviously even.