Question

What is the value of the integral \[\int\limits_0^\pi {x\sin xdx} \]?
A. \[\pi \]
B. 0
C. 1
D. \[{\pi ^2}\]

Answer 1

Hint: Here, a definite integral is given. First, apply the definite integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] and simplify the integral. After that, add this integral with the original integral and solve the integral. In the end, apply the upper and lower limits to get the required answer.

Formula Used: Definite integral rule: \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\]
\[\int {\sin xdx = - \cos x} \]

Complete step by step solution: The given definite integral is \[\int\limits_0^\pi {x\sin xdx} \].
Let consider,
\[I = \int\limits_0^\pi {x\sin xdx} \] \[.....\left( 1 \right)\]

Now apply the definite integral rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] on the above integral.
\[I = \int\limits_0^\pi {\left( {\pi - x} \right)\sin \left( {\pi - x} \right)dx} \]
\[ \Rightarrow I = \int\limits_0^\pi {\left( {\pi - x} \right)\sin xdx} \] \[.....\left( 2 \right)\]
Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[2I = \int\limits_0^\pi {x\sin xdx} + \int\limits_0^\pi {\left( {\pi - x} \right)\sin xdx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\left( {x + \pi - x} \right)\sin xdx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\pi \sin xdx} \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_0^\pi {\sin xdx} \]
Solve the integral.
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - \cos x} \right]_0^\pi \]
Apply the upper and lower limit.
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - \cos \pi + \cos 0} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - \left( { - 1} \right) + 1} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ 2 \right]\]
\[ \Rightarrow I = \pi \]
Therefore, \[\int\limits_0^\pi {x\sin xdx} = \pi \]

Option ‘A’ is correct

Note: students often get confused and solve the integral \[\int {\sin xdx = \cos x} \]. This formula is incorrect. Sometimes they forget to add the negative sign. The correct formula is \[\int {\sin xdx = - \cos x} \].