Question

Evaluate the integral $\int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx$
A) $20$
B) $8$
C) $10$
D) $18$

Answer 1

We will first check the periodicity of the given integrand. Then we will use the periodic property to simplify the integral. We will then use the definition of the absolute value to simplify it further. We will apply the integration formula of the trigonometric function to find the required value.

Formula used:
For a periodic function $f(x)$ with period $a$, $\int\limits_{ma}^{na} {f(x)dx = (n - m)\int\limits_0^a {f(x)dx} }$

Complete step by step solution:
We have to evaluate the integral $\int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx$.
Here, the lower limit is $\pi$ and the upper limit is $10\pi$. Also, $f(x) = \left| {\sin x} \right|$.
Let us check the periodicity of the function $f(x) = \left| {\sin x} \right|$.
We know that $\sin (\pi + x) = - \sin x$.
Taking modulus on both sides, we get
$\left| {\sin (\pi + x)} \right| = \left| { - \sin x} \right| = \left| {\sin x} \right|$
This means that the function $f(x) = \left| {\sin x} \right|$ is periodic with period $\pi$.
Now, let us apply the property $\int\limits_{ma}^{na} {f(x)dx = (n - m)\int\limits_0^a {f(x)dx} }$, where $f(x)$ is a periodic function with period $a$.
Comparing the given integral $\int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx$ with $\int\limits_{ma}^{na} {f(x)dx}$, we see that $m = 1,n = 10,a = \pi$ and $f(x) = \left| {\sin x} \right|$.
Hence, applying the property to the given integral, we get
$\int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx = (10 - 1)\int\limits_0^\pi {\left| {\sin x} \right|} dx$
$\Rightarrow \int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx = 9\int\limits_0^\pi {\left| {\sin x} \right|} dx$ ……….$(1)$
In equation $(1)$, the integral to be evaluated on the RHS is $\int\limits_0^\pi {\left| {\sin x} \right|} dx$.
We know that for an absolute value function, \left| x \right| = \left\\{ \begin{array}{l}x{\text{ if }}x > 0\\\0{\text{ if }}x = 0\\\ - x{\text{ if }}x < 0\end{array} \right.
Using this definition of absolute value function, we have
\left| {\sin x} \right| = \left\\{ \begin{array}{l}0{\text{ if }}x = 0\\\\\sin x{\text{ if }}0 < x < \pi \\\ - \sin x{\text{ if }}\pi < x < 2\pi \end{array} \right.
We observe from the integral $\int\limits_0^\pi {\left| {\sin x} \right|} dx$ that $x$ lies between $0$and $\pi$.
So, from the definition of the function $\left| {\sin x} \right|$, for the interval $0 < x < \pi$, $\left| {\sin x} \right| = \sin x$.
Therefore, $\int\limits_0^\pi {\left| {\sin x} \right|} dx = \int\limits_0^\pi {\sin xdx}$
Substituting this in equation $(1)$, we have
$\int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx = 9\int\limits_0^\pi {\sin xdx}$ ……….$(2)$
We know that $\int {\sin x} = - \cos x + c$.
Using this in equation $(2)$, we get
$\Rightarrow \int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx = 9 \times \left[ { - \cos x} \right]_0^\pi$ …………$(3)$
We know that $\int\limits_a^b {f(x)dx} = f(b) - f(a)$.
Using this in equation $(3)$, we have
$\Rightarrow \int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx = 9 \times \left( { - \cos \pi - ( - \cos 0)} \right)$ ……….$(4)$
Now, $\cos \pi = - 1$ and $\cos 0 = 1$. Hence, equation $(4)$ becomes
$\Rightarrow \int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx = 9 \times \left( 2 \right) = 18$

Therefore, $\int\limits_\pi ^{10\pi } {\left| {\sin x} \right|} dx = 18$ and so the correct option is D.

Note:
Trigonometry is a branch of mathematics that helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers to make maps. It is also used by the aviation and naval industries.
A periodic function is a function that repeats its values at regular intervals. In the above problem, $\left| {\sin x} \right|$ is a periodic function with period $\pi$, which means that $\left| {\sin x} \right|$ repeats its values after every $\pi$ radians.