Mathematics - Definite Integration Question with Solution | TestHub

We know that $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \dots \dots \dots$

$\Rightarrow \cos x\ge 1-\frac{x^2}{2}$

$\therefore {\int }_0^{1}x\cos xdx\ge {\int }_0^{1}x\left(1-\frac{x^2}{2}\right)dx={\left(\frac{x^2}{2}-\frac{x^4}{8}\right)}_0^1=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}$

So, option $\left(A\right)$ is correct.

Now, we also know that $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots \dots \dots \dots \dots$

$\Rightarrow \sin x\ge x-\frac{x^3}{3!}, x\in \left[0, 1\right]$

$\therefore {\int }_0^{1}x\sin xdx\ge {\int }_0^{1}x\left(x-\frac{x^3}{6}\right)dx={\left(\frac{x^3}{3}-\frac{x^5}{6\times 5}\right)}_0^1=\frac{1}{3}-\frac{1}{30}=\frac{9}{30}=\frac{3}{10}$

So, option $\left(B\right)$ is correct.

Now, ${\int }_0^1{x}^2\cos xdx\ge {\int }_0^1{x}^2\left(1-\frac{x^2}{2}\right)dx={\left(\frac{x^3}{3}-\frac{x^5}{2\times 5}\right)}_0^1=\frac{1}{3}-\frac{1}{10}=\frac{7}{30}$

Also, $x^2\cos x\le x^2$

$\Rightarrow {\int }_0^1{x}^2\cos xdx\le {\int }_0^1{x}^{2}dx$

$\Rightarrow {\int }_0^1{x}^2\cos xdx\le {\left(\frac{x^3}{3}\right)}_0^1=\frac{1}{3}$

So, $\frac{7}{30}\le {\int }_0^1{x}^2\cos xdx\le \frac{1}{3}$

So, option $\left(C\right)$ is incorrect.

Now, ${\int }_0^1{x}^2\sin xdx\ge {\int }_0^1{x}^2\left(x-\frac{x^3}{3!}\right)dx={\left(\frac{x^4}{4}-\frac{x^6}{6\times 6}\right)}_0^1=\frac{1}{4}-\frac{1}{36}=\frac{8}{36}=\frac{2}{9}$

So, option $\left(D\right)$ is correct.