Question

The area bounded by the curve $y = x\sin x$ and $x$ -axis between $x = 0$ and $x = 2\pi$ is:
A. $2\pi \;sq.\;unit$
B. $3\pi \;sq.\;unit$
C. $4\pi \;sq.\;unit$
D. $5\pi \;sq.\;unit$

Answer 1

Hint : Use the integration of the given curve with the limits according to the constraints given in the question to get the area bounded by the curve and $x$ -axis. Make use of properties of definite integral accordingly.

Complete step-by-step answer :
The area under the curve $y = f\left( x \right)$ bounded with $x$ -axis and with the condition that $x$ is in between $a$ and $b$ is equal to $\int\limits_a^b {f\left( x \right)dx}$ .
The area bounded by the curve $y = x\sin x$ and $x$ -axis between $x = 0$ and $x = 2\pi$ is given by the integration $\int\limits_0^{2\pi } {x\sin xdx}$ .
The integration of product of two functions $f\left( x \right)$ and $g\left( x \right)$ is given by the integral $\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right) \times \int {g\left( x \right)dx} - \int {\left( {\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) \times \left( {\int {g\left( x \right)dx} } \right)} \right)dx}$ .
Simplify the integration:
$\int {x\sin xdx} = x\int {\sin xdx} - \int {\dfrac{d}{{dx}}\left( x \right)\int {\sin xdx} dx} \\\ = x\left( { - \cos x} \right) - \int {1\left( { - \cos x} \right)dx} \\\ = - x\cos x + \int {\cos xdx} \\\ = - x\cos x + \sin x \;$
Now put the limits in the integration.
$\int\limits_0^{2\pi } {x\sin xdx} = \left. { - x\cos x + \sin x} \right|_0^{2\pi } \\\ = - 2\pi \cos \left( {2\pi } \right) + \sin \left( {2\pi } \right) - \left( { - 0 \times \cos 0 + \sin 0} \right) \\\ = - 2\pi \left( 1 \right) + 0 - 0 \\\ = - 2\pi \;$
The area is always positive. So, take the modulus value of the area under the curve as calculated above.
So, the area bounded by the curve $y = x\sin x$ and $x$ -axis between $x = 0$ and $x = 2\pi$ is $2\pi \;sq.\;unit$ .
So, the correct answer is “Option A”.

Note : Use the integration by parts formula to solve the integration of the curve $y = x\sin x$ with respect to the variable $x$ and the constraints given in the question are the limits of the integration. The integration of product of two functions $f\left( x \right)$ and $g\left( x \right)$ is given by the integral $\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right) \times \int {g\left( x \right)dx} - \int {\left( {\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) \times \left( {\int {g\left( x \right)dx} } \right)} \right)dx}$ . The area under the curve $y = f\left( x \right)$ bounded with $x$ -axis and with the condition that $x$ is in between $a$ and $b$ is equal to $\int\limits_a^b {f\left( x \right)dx}$ .