Test Index

Definite Integration

Section: Mathematics

Question : 16 of 46

Marks: +1, -0

Solution:

We know that

\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\dots

So,

\cos x>1-\frac{x^{2}}{2}

for all

x \in (0,1)

\Rightarrow x \cos x > x - \frac{x^{3}}{2}

\Rightarrow \int\limits_{0}^{1} x \cos x\,dx > \int\limits_{0}^{1} \left( x - \frac{x^{3}}{2} \right) dx

\Rightarrow \int\limits_{0}^{1} x \cos x\,dx > \frac{3}{8}

(Option

A

is correct)

\because x^{2} \cos x < x

for all

x \in (0,1)

\Rightarrow \int\limits_{0}^{1} x^{2} \cos x\,dx < \int\limits_{0}^{1} x\,dx

\Rightarrow \int\limits_{0}^{1} x^{2} \cos x\,dx < \frac{1}{2}

(Option

C

is incorrect)

Again we know that,

\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots

So,

\sin x > x - \frac{x^{3}}{6}

for all

x \in (0,1)

\Rightarrow x \sin x > x^{2} - \frac{x^{4}}{6}

\int\limits_{0}^{1} x \sin x\,dx > \int\limits_{0}^{1} \left( x^{2} - \frac{x^{4}}{6} \right) dx

\Rightarrow \int\limits_{0}^{1} x \sin d x > \frac{3}{10}

(Option

B

is correct)

\because x^{2} \sin x > x^{3} - \frac{x^{5}}{6}

\Rightarrow \int\limits_{0}^{1} x^{2} \sin x\,dx > \frac{2}{9}

(Option

D

is correct)

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