Question

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

Answer 1

Hint: Here, a definite integral is given. First, simplify the integral by applying the integration formula \[\int\limits_a^b {f\left( x \right)dx = } \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx} } \]. Then, simplify the integrals by calculating the period of the functions in the corresponding integrals. After that, solve the integral by applying the integration formula \[\int\limits_a^b {\sin xdx = \left[ { - \cos x} \right]} _a^b\]. In the end, apply the limits and solve the equation to get the required answer.



Formula Used:Integration rule: \[\int\limits_a^b {f\left( x \right)dx = } \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx} } \]
\[\int\limits_a^{a + nT} {f\left( x \right)dx = } n\int\limits_0^T {f\left( x \right)dx} \], where \[T\] is the period of the function.
\[\int\limits_a^b {\sin xdx = \left[ { - \cos x} \right]} _a^b\]



Complete step by step solution:The given definite integral is \[\int\limits_0^{n\pi + v} {\left| {\sin x} \right|} dx\].
Let consider,
\[I = \int\limits_0^{n\pi + v} {\left| {\sin x} \right|} dx\]
Simplify the integral by applying the integration rule \[\int\limits_a^b {f\left( x \right)dx = } \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx} } \].
\[I = \int\limits_0^v {\left| {\sin x} \right|} dx + \int\limits_v^{n\pi + v} {\left| {\sin x} \right|} dx\]
We know that, \[\sin x\] is positive in the first and the second quadrant.
Also, period of the function \[\sin x\] is \[\pi \]. So, apply the integration rule \[\int\limits_a^{a + nT} {f\left( x \right)dx = } n\int\limits_0^T {f\left( x \right)dx} \], where \[T\] is the period of the function.
We get,
\[I = \int\limits_0^v {\sin x} dx + n\int\limits_0^\pi {\sin x} dx\]
Now solve the integral by applying the integration formula \[\int\limits_a^b {\sin xdx = \left[ { - \cos x} \right]} _a^b\].
\[I = \left[ { - \cos x} \right]_0^v + n\left[ { - \cos x} \right]_0^\pi \]
Apply the upper and lower limits.
\[I = \left[ { - \cos v - \left( { - \cos 0} \right)} \right] + n\left[ { - \cos \pi - \left( { - \cos 0} \right)} \right]\]
\[ \Rightarrow I = \left[ { - \cos v - \left( { - 1} \right)} \right] + n\left[ { - \left( { - 1} \right) - \left( { - 1} \right)} \right]\]
\[ \Rightarrow I = 1 - \cos v + n\left[ {1 + 1} \right]\]
\[ \Rightarrow I = 1 - \cos v + 2n\]
\[ \Rightarrow I = 2n + 1 - \cos v\]
Thus, \[\int\limits_0^{n\pi + v} {\left| {\sin x} \right|} dx = 2n + 1 - \cos v\].



Option ‘B’ 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 \[\int\limits_a^b {\sin x} dx = \left[ { - \cos x} \right]_a^b\].