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} $ .