Question

Given $f'\left( x \right) = \dfrac{{\cos x}}{x}$, $f\left( {\dfrac{\pi }{2}} \right) = a$, and $f\left( {\dfrac{{3\pi }}{2}} \right) = b$, find the value of $\int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx$.

Answer 1

Here, we need to find the value of the definite integral. We will integrate the given function by parts. Then, we will simplify the expression using the given values. We will again substitute the limits into the expressions and simplify the expression to obtain the required value of the integral.
Formula Used: Using integration by parts, the integral of the product of two differentiable functions of $x$ can be written as $\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx}$, where $u$ and $v$ are the differentiable functions of $x$.

Complete step by step solution:
We will integrate the given definite integral using integration by parts.
Using integration by parts, the integral of the product of two differentiable functions of $x$ can be written as $\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx}$, where $u$ and $v$ are the differentiable functions of $x$.
Rewriting the given integral, we get
$\int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right) \times 1} dx$
Let $u$ be $f\left( x \right)$ and $v$ be $1$.
Therefore, by integrating $\int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx$ by parts, we get
$\begin{array}{l}\int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} \times 1dx = \left. {\left[ {f\left( x \right)\int {\left( 1 \right)} dx - \int {\left( {\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} \times \int {\left( 1 \right)} dx} \right)} dx} \right]} \right|_{\pi /2}^{3\pi /2}\\\ \Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \left. {\left[ {f\left( x \right)\int {\left( 1 \right)} dx - \int {\left( {\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} \times \int {\left( 1 \right)} dx} \right)} dx} \right]} \right|_{\pi /2}^{3\pi /2}\end{array}$
We know that the derivative of the function $f\left( x \right)$ is $f'\left( x \right)$.
Also, we know that the integral of a constant $\int {\left( 1 \right)} dx$ is $x$.
Therefore, we can simplify the integral as
$\Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \left. {\left[ {f\left( x \right)x - \int {\left( {f'\left( x \right) \times x} \right)} dx} \right]} \right|_{\pi /2}^{3\pi /2}$
Substitute $f'\left( x \right) = \dfrac{{\cos x}}{x}$ in the expression, we get
$\begin{array}{l} \Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \left. {\left[ {f\left( x \right)x - \int {\left( {\dfrac{{\cos x}}{x} \times x} \right)} dx} \right]} \right|_{\pi /2}^{3\pi /2}\\\ \Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \left. {\left[ {xf\left( x \right) - \int {\cos x} dx} \right]} \right|_{\pi /2}^{3\pi /2}\end{array}$
Integrating the function $\cos x$, we get
$\Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \left. {\left[ {xf\left( x \right) - \sin x} \right]} \right|_{\pi /2}^{3\pi /2}$
Now, we will substitute the limits into the expressions.
Therefore, we get
$\Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \left( {\dfrac{{3\pi }}{2}f\left( {\dfrac{{3\pi }}{2}} \right) - \sin \dfrac{{3\pi }}{2}} \right) - \left( {\dfrac{\pi }{2}f\left( {\dfrac{\pi }{2}} \right) - \sin \dfrac{\pi }{2}} \right)$
We know that $\sin \dfrac{\pi }{2} = 1$ and $\sin \dfrac{{3\pi }}{2} = - 1$.
Thus, the integral becomes
$\begin{array}{l} \Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \left( {\dfrac{{3\pi }}{2}f\left( {\dfrac{{3\pi }}{2}} \right) - \left( { - 1} \right)} \right) - \left( {\dfrac{\pi }{2}f\left( {\dfrac{\pi }{2}} \right) - 1} \right)\\\ \Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \left( {\dfrac{{3\pi }}{2}f\left( {\dfrac{{3\pi }}{2}} \right) + 1} \right) - \left( {\dfrac{\pi }{2}f\left( {\dfrac{\pi }{2}} \right) - 1} \right)\\\ \Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \dfrac{{3\pi }}{2}f\left( {\dfrac{{3\pi }}{2}} \right) + 1 - \dfrac{\pi }{2}f\left( {\dfrac{\pi }{2}} \right) + 1\\\ \Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \dfrac{{3\pi }}{2}f\left( {\dfrac{{3\pi }}{2}} \right) - \dfrac{\pi }{2}f\left( {\dfrac{\pi }{2}} \right) + 2\end{array}$
Substituting $f\left( {\dfrac{\pi }{2}} \right) = a$ and $f\left( {\dfrac{{3\pi }}{2}} \right) = b$, we get
$\Rightarrow \int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx = \dfrac{{3\pi }}{2}b - \dfrac{\pi }{2}a + 2$
Taking $\dfrac{\pi }{2}$ common, we get
$\therefore \int\limits_{\pi /2}^{3\pi /2}{f\left( x \right)}dx=\dfrac{\pi }{2}\left( 3b-a \right)+2$

Therefore, the value of the definite integral $\int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx$ is $\dfrac{\pi }{2}\left( {3b - a} \right) + 2$.

Note:
We have used integration by parts to simplify the integral $\int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx$. A common mistake we can make is to first integrate $f'\left( x \right) = \dfrac{{\cos x}}{x}$ to find $f\left( x \right)$, and then substitute it into the integral $\int\limits_{\pi /2}^{3\pi /2} {f\left( x \right)} dx$. We will avoid doing this because $f'\left( x \right) = \dfrac{{\cos x}}{x}$ cannot be integrated using any elementary functions (it can be integrated in terms of infinite series, using Taylor series expansion of $\cos x$).