Question

Which of the following inequalities is/are TRUE?

• A. $\underset{0}{\overset{1}{\int }}x\cos xdx\ge \frac{3}{8}$
• B. $\underset{0}{\overset{1}{\int }}x\sin xdx\ge \frac{3}{10}$
• C. $\underset{0}{\overset{1}{\int }}{x}^2\cos xdx\ge \frac{1}{2}$
• D. $\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge \frac{2}{9}$

Answer 1

Answer: (A), (B), (D)

$\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge \frac{2}{9}$

$\left(A\right)$
Expansion of $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}\cdots$
$\therefore \cos x\ge 1-\frac{x^2}{2!}$
Multiply $x$ both side
$x\cos x\ge x-\frac{x^2}{2!}$
Integrating both side
$\underset{0}{\overset{1}{\int }}x\cos xdx\ge \underset{0}{\overset{1}{\int }}(x-\frac{x^3}{2})dx$
$\underset{0}{\overset{1}{\int }}x\cos xdx\ge {(\frac{x^2}{2}-\frac{x^4}{8})}_0^1$
$\underset{0}{\overset{1}{\int }}x\cos xdx\ge \frac{1}{2}-\frac{1}{8}$
$\underset{0}{\overset{1}{\int }}x\cos xdx\ge \frac{3}{8}$
Similarly
$\left(B\right)$ Expansion of $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$
$\sin x\ge x-\frac{x^3}{3!}$
Multiply $x$ both side
$x\sin x\ge x^2-\frac{x^4}{6}$
Integrating both side
$\underset{0}{\overset{1}{\int }}x\sin xdx\ge \underset{0}{\overset{1}{\int }}(x^2-\frac{x^4}{6})dx$
$\underset{0}{\overset{1}{\int }}x\sin xdx\ge {(\frac{x^3}{3}-\frac{x^5}{6\cdot 5})}_0^1$
$\underset{0}{\overset{1}{\int }}x\sin xdx\ge {(\frac{x^3}{3}-\frac{x^5}{30})}_0^1$
$\underset{0}{\overset{1}{\int }}x\sin xdx\ge \frac{1}{3}-\frac{1}{30}$
$\underset{0}{\overset{1}{\int }}x\sin xdx\ge \frac{3}{10}$
$\left(C\right)\cos x<1$
Multiply $x^2$ both side
$x^2\cos x<x^2$
Integrating both side
$\underset{0}{\overset{1}{\int }}{x}^2\cos xdx<\underset{0}{\overset{1}{\int }}{x}^{2}dx$
$\underset{0}{\overset{1}{\int }}{x}^2\cos xdx<\frac{1}{3}$
$\left(D\right)$
$\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge \underset{0}{\overset{1}{\int }}{x}^2(x-\frac{x^3}{6})dx$
​​​​​​​$\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge {(\frac{x^4}{4}-\frac{x^6}{36})}_0^1$
​​​​​​​$\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge \frac{1}{4}-\frac{1}{36}$
​​​​​​​$\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge \frac{2}{9}$