Question

How do you evaluate $\int{\left[ \dfrac{x}{\sqrt{2x-1}} \right]}dx$ for [1, 5]?

Answer 1

In this question, we have to find the value of definite integral. Thus, we will use the integration and the basic mathematical rules to get the solution. First, we will apply the substitution method in the denominator that is let $2x-1={{t}^{2}}$ . Then, we will differentiate both the sides with respect to dx and dt. After that, we will substitute the value in the integral and thus make the necessary calculations. Then, we will substitute the value of t in the equation and after that we will apply the integral formula $\int\limits_{a}^{b}{f(x)dx=f(b)-f(a)}$ and make the necessary mathematical calculations, to get the required solution for the problem.

Complete step by step solution:
According to the question, we have to find the value of definite integral.
Thus, we will use the integration and the basic mathematical rules to get the solution.
The integral to be solved is $\int{\left[ \dfrac{x}{\sqrt{2x-1}} \right]}dx$ for [1, 5] -------------- (1)
First, we will use the substitution method, that is let $2x-1={{t}^{2}}$ -------- (2)
Now, we will differentiate equation (2), which is
$\Rightarrow {{\left( 2x-1 \right)}^{\prime }}={{\left( {{t}^{2}} \right)}^{\prime }}$
On solving the above equation, we get
$\Rightarrow 2dx=2tdt$
Now, we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}dx=\dfrac{2t}{2}dt$
Therefore, we get
$\Rightarrow dx=tdt$ ---------- (3)
Now, we will find equation (2) in terms of x that is we will add 1 on both sides in the equation (2), we get
$\Rightarrow 2x-1+1={{t}^{2}}+1$
As we now, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 2x={{t}^{2}}+1$
Now, we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}x=\dfrac{{{t}^{2}}+1}{2}$
On furthers solving, we get
$\Rightarrow x=\dfrac{{{t}^{2}}}{2}+\dfrac{1}{2}$ ------------ (4)
Also, we will find the value of t from equation (2) that is we will take the square root on both sides in the above equation, we get
$\Rightarrow \sqrt{2x-1}=\sqrt{{{t}^{2}}}$
Thus, we get
$\Rightarrow \sqrt{2x-1}=t$ ------- (5)
Now, the integral we need to evaluate is $\int\limits_{1}^{5}{\left[ \dfrac{x}{\sqrt{2x-1}} \right]}dx$
So, now we will put the value of equation (2), (3), and (4) in the above equation, we get
$\Rightarrow \int\limits_{{}}^{{}}{\left[ \dfrac{\dfrac{{{t}^{2}}}{2}+\dfrac{1}{2}}{\sqrt{{{t}^{2}}}} \right]}t.dt$
On further simplification, we get
$\Rightarrow \int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}}{2}+\dfrac{1}{2}}dt$
Now, we will split the terms with respect to addition, we get
$\Rightarrow \int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}}{2}}dt+\int\limits_{{}}^{{}}{\dfrac{1}{2}}dt$
So, we will apply the integration formula $\int{{{x}^{n}}dx=}\dfrac{{{x}^{n+1}}}{n+1}$ in the above equation, we get
$\Rightarrow \dfrac{{{t}^{2+1}}}{2.(3)}+\dfrac{1}{2}t$
Therefore, we get
$\Rightarrow \dfrac{{{t}^{3}}}{6}+\dfrac{1}{2}t$
Now, we will put the value of equation (5) in the above equation, we get
$\Rightarrow \left[ \dfrac{{{\left( \sqrt{2x-1} \right)}^{3}}}{6}+\dfrac{1}{2}\sqrt{2x-1} \right]_{1}^{5}$
So, now we will apply the definite integral formula $\int\limits_{a}^{b}{f(x)dx=f(b)-f(a)}$ in the above equation, we get
$\Rightarrow \left( \dfrac{{{\left( \sqrt{2\left( 5 \right)-1} \right)}^{3}}}{6}+\dfrac{1}{2}\sqrt{2\left( 5 \right)-1} \right)-\left( \dfrac{{{\left( \sqrt{2\left( 1 \right)-1} \right)}^{3}}}{6}+\dfrac{1}{2}\sqrt{2\left( 1 \right)-1} \right)$
On further simplification, we get
$\Rightarrow \left( \dfrac{{{\left( \sqrt{10-1} \right)}^{3}}}{6}+\dfrac{1}{2}\sqrt{10-1} \right)-\left( \dfrac{{{\left( \sqrt{2-1} \right)}^{3}}}{6}+\dfrac{1}{2}\sqrt{2-1} \right)$
$\Rightarrow \left( \dfrac{{{\left( \sqrt{9} \right)}^{3}}}{6}+\dfrac{1}{2}\sqrt{9} \right)-\left( \dfrac{{{\left( \sqrt{1} \right)}^{3}}}{6}+\dfrac{1}{2}\sqrt{1} \right)$
$\Rightarrow \left( \dfrac{9\sqrt{9}}{6}+\dfrac{3}{2} \right)-\left( \dfrac{1}{6}+\dfrac{1}{2} \right)$
On solving the above equation, we get
$\Rightarrow \left( \dfrac{3\sqrt{9}}{2}+\dfrac{3}{2} \right)-\left( \dfrac{1}{6}+\dfrac{1}{2} \right)$
Now, we will take the LCM of denominator in the above equation, we get
$\Rightarrow \left( \dfrac{3\sqrt{9}+3}{2} \right)-\left( \dfrac{1+3}{6} \right)$
$\Rightarrow \left( \dfrac{3\left( \sqrt{9}+1 \right)}{2} \right)-\left( \dfrac{4}{6} \right)$
Thus, we get
$\Rightarrow \left( \dfrac{3\left( \sqrt{9}+1 \right)}{2} \right)-\dfrac{2}{3}$
Now, we will again take the LCM of denominator in the above equation, we get
$\Rightarrow \dfrac{9\left( \sqrt{9}+1 \right)-4}{6}$ which is the required answer.
Therefore, for the integral $\int{\left[ \dfrac{x}{\sqrt{2x-1}} \right]}dx$ for [1, 5] , the value is equal to $\dfrac{9\left( \sqrt{9}+1 \right)-4}{6}$

Note: While solving this problem, do mention all the steps properly to avoid confusion and mathematical error. One of the alternative methods to solve this problem is find the limits of t from the substitution equation and then put it in $\dfrac{{{t}^{3}}}{6}+\dfrac{1}{2}t$ step, to get the required solution for the problem.