Question

What is the value of the definite integral \[\int\limits_0^{2\pi } {\left| {\sin x} \right|} dx\]?
A. 0
B. 1
C. 2
D. 4

Answer 1

Hint: Here, a definite integral is given. First, simplify the given integral by applying the integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\]. Then, check the values of the given absolute function in that interval. After that, apply the integration formula \[\int\limits_a^b {\sin xdx = \left[ { - \cos x} \right]} _a^b\]and solve it. In the end, apply the upper and lower limits to get the required answer.



Formula Used:\[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\]
\[\int\limits_a^b {\sin xdx = \left[ { - \cos x} \right]} _a^b\]



Complete step by step solution:The given definite integral is \[\int\limits_0^{2\pi } {\left| {\sin x} \right|} dx\].
Let consider,
\[I = \int\limits_0^{2\pi } {\left| {\sin x} \right|} dx\]
Simplify the integral by applying the integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\].
\[I = \int\limits_0^\pi {\left| {\sin x} \right|} dx + \int\limits_\pi ^{2\pi } {\left| {\sin x} \right|} dx\]
Simplify the absolute function.
We know that, values of \[\sin x\] are positive in the interval \[\left[ {0,\pi } \right]\] and negative in the interval \[\left[ {\pi ,2\pi } \right]\].
So, we get
\[I = \int\limits_0^\pi {\sin x} dx + \int\limits_\pi ^{2\pi } { - \sin x} dx\]
\[ \Rightarrow I = \int\limits_0^\pi {\sin x} dx - \int\limits_\pi ^{2\pi } {\sin x} dx\]
Solve the integrals by applying the integration formula \[\int\limits_a^b {\sin xdx = \left[ { - \cos x} \right]} _a^b\].
\[ \Rightarrow I = \left[ { - \cos x} \right]_0^\pi - \left[ { - \cos x} \right]_\pi ^{2\pi }\]
Apply the upper and lower limits.
\[ \Rightarrow I = \left[ { - \cos \pi - \left( { - \cos 0} \right)} \right] - \left[ { - \cos 2\pi - \left( { - \cos \pi } \right)} \right]\]
\[ \Rightarrow I = \left[ { - \cos \pi + \cos 0} \right] - \left[ { - \cos 2\pi + \cos \pi } \right]\]
\[ \Rightarrow I = \left[ { - \left( { - 1} \right) + 1} \right] - \left[ { - 1 + \left( { - 1} \right)} \right]\]
\[ \Rightarrow I = 1 + 1 + 1 + 1\]
\[ \Rightarrow I = 4\]
Therefore, \[\int\limits_0^{2\pi } {\left| {\sin x} \right|} dx = 4\].



Option ‘D’ is correct



Note: Students often do mistake to integrating \[\int\limits_a^b {\sin x} dx\] . They apply the formula \[\int\limits_a^b {\sin x} dx = \left[ {\cos x} \right]_a^b\] which is an incorrect formula. They get confused because \[\dfrac{d}{{dx}}\sin x = \cos x\] . The correct formula is \[\int\limits_a^b {\sin x} dx = \left[ { - \cos x} \right]_a^b\].