How do you determine If $\int\limits_{0}^{1}{\dfrac{3}{4x-1}dx}$ converges or diverges?
Answer
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Hint: To determine If $\int\limits_{0}^{1}{\dfrac{3}{4x-1}dx}$ converges or diverges, we have to split the limits. We can see that the integral gets an indefinite solution when $x=\dfrac{1}{4}$ . Hence, we will split the given integral as $\int\limits_{0}^{1}{\dfrac{3}{4x-1}dx}=\int\limits_{0}^{\dfrac{1}{4}}{\dfrac{3}{4x-1}dx}+\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{3}{4x-1}dx}$ . Then we have to find $\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{b}{\dfrac{3}{4x-1}dx} \right]$ and $\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \int\limits_{b}^{1}{\dfrac{3}{4x-1}dx} \right]$ . If the limits in each case exist or are finite, then the given integral converges. If the limits are undefined or plus or minus infinite, then the given integral diverges.
Complete step by step solution:
We have to determine If $\int\limits_{0}^{1}{\dfrac{3}{4x-1}dx}$ converges or diverges. We can see that the integral gets an indefinite solution when $x=\dfrac{1}{4}$ . Let us split the limits. We know that $\int\limits_{a}^{c}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{b}^{c}{f\left( x \right)dx}$ . Hence, we can write the given integral as
$\int\limits_{0}^{1}{\dfrac{3}{4x-1}dx}=\int\limits_{0}^{\dfrac{1}{4}}{\dfrac{3}{4x-1}dx}+\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{3}{4x-1}dx}...\left( i \right)$
Let us consider $\int\limits_{0}^{\dfrac{1}{4}}{\dfrac{3}{4x-1}dx}$ . We can write this integral in terms of limits as
$\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{b}{\dfrac{3}{4x-1}dx} \right]$
First let us integrate the given integral. Let us substitute $u=4x-1$ . Let us differentiate this equation on both the sides. We will get $du=4dx$ . We have to multiply and divide the given integral by 4.
$\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{n}{\dfrac{3}{4x-1}dx} \right]=\left( \dfrac{1}{4}\times 3 \right)\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{b}{\dfrac{4}{4x-1}dx} \right]$
We can now substitute the values of u and du in the above equation.
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{b}{\dfrac{du}{u}} \right]$
We know that $\int{\dfrac{1}{x}dx}=\log \left| x \right|+C$ . Thus the above equation becomes
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| u \right| \right]_{0}^{b}$
Let us now substitute the value of u. We will get
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4x-1 \right| \right]_{0}^{b}$
Now, we have to apply the intervals.
$\begin{align}
& \Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right|-\log \left| 0-1 \right| \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right|-\log \left| -1 \right| \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right|-\log 1 \right] \\
\end{align}$
We know that $\log 1=0$ . Hence, we can write the above equation as
$\begin{align}
& \Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right|-0 \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right| \right] \\
\end{align}$
Now, we have to apply the limits. We will get
\[\begin{align}
& \Rightarrow \dfrac{3}{4}\left[ \log \left| 4\times \dfrac{1}{4}-1 \right| \right] \\
& =\dfrac{3}{4}\left[ \log \left| 1-1 \right| \right] \\
& =\dfrac{3}{4}\left[ \log \left| 0 \right| \right] \\
& =\infty \\
& \Rightarrow \displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{b}{\dfrac{3}{4x-1}dx} \right]=\infty ...(ii) \\
\end{align}\]
Now let us evaluate $\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{3}{4x-1}dx}$ .
We can write this integral in terms of limits as
$\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \int\limits_{b}^{1}{\dfrac{3}{4x-1}dx} \right]$
On integrating similar to the previous term, we will get
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4x-1 \right| \right]_{b}^{1}$
Now, we have to apply the intervals.
$\begin{align}
& \Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4-1 \right|-\log \left| 4b-1 \right| \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 3 \right|-\log \left| 4b-1 \right| \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log 3-\log \left| 4b-1 \right| \right] \\
\end{align}$
We know that $\log 3=0.477$ . Hence, we can write the above equation as
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ 0.477-\log \left| 4b-1 \right| \right]$
We know that $\displaystyle \lim_{x \to \infty }\left( a-b \right)=\displaystyle \lim_{x \to \infty }a-\displaystyle \lim_{x \to \infty }b$ and limit of a constant is constant itself. Thus, we can write the above equation as
\[\begin{align}
& \Rightarrow \dfrac{3}{4}\left\{ \displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}0.477-\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4b-1 \right| \right] \right\} \\
& =\dfrac{3}{4}\left\{ 0.477-\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4b-1 \right| \right] \right\} \\
\end{align}\]
Let us apply distributive property in the above equation.
\[\begin{align}
& \Rightarrow \dfrac{3}{4}\times 0.477-\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4b-1 \right| \right] \\
& =0.357-\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4b-1 \right| \right] \\
\end{align}\]
Now, we have applied the limits. We will get
\[\begin{align}
& \Rightarrow 0.357-\dfrac{3}{4}\left[ \log \left| 4\times \dfrac{1}{4}-1 \right| \right] \\
& =0.357-\dfrac{3}{4}\left[ \log \left| 1-1 \right| \right] \\
& =0.357-\dfrac{3}{4}\left[ \log \left| 0 \right| \right] \\
& =0.357-\infty \\
& =-\infty \\
& \Rightarrow \displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \int\limits_{b}^{1}{\dfrac{3}{4x-1}dx} \right]=-\infty ...\left( iii \right) \\
\end{align}\]
We can see that the limit of the integrals from (ii) and (iii) are plus and minus infinite respectively. Hence, the given integral diverges.
Note: Students must know when an integral converges and diverges. If the limits exist or are finite, then the given integral converges. If the limits are undefined or plus or minus infinite, then the given integral diverges. We can find when the given integral gets indefinite solution by equating the denominator of the given integral to 0.
\[\begin{align}
& \Rightarrow 4x-1=0 \\
& \Rightarrow 4x=1 \\
& \Rightarrow x=\dfrac{1}{4} \\
\end{align}\]
Complete step by step solution:
We have to determine If $\int\limits_{0}^{1}{\dfrac{3}{4x-1}dx}$ converges or diverges. We can see that the integral gets an indefinite solution when $x=\dfrac{1}{4}$ . Let us split the limits. We know that $\int\limits_{a}^{c}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{b}^{c}{f\left( x \right)dx}$ . Hence, we can write the given integral as
$\int\limits_{0}^{1}{\dfrac{3}{4x-1}dx}=\int\limits_{0}^{\dfrac{1}{4}}{\dfrac{3}{4x-1}dx}+\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{3}{4x-1}dx}...\left( i \right)$
Let us consider $\int\limits_{0}^{\dfrac{1}{4}}{\dfrac{3}{4x-1}dx}$ . We can write this integral in terms of limits as
$\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{b}{\dfrac{3}{4x-1}dx} \right]$
First let us integrate the given integral. Let us substitute $u=4x-1$ . Let us differentiate this equation on both the sides. We will get $du=4dx$ . We have to multiply and divide the given integral by 4.
$\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{n}{\dfrac{3}{4x-1}dx} \right]=\left( \dfrac{1}{4}\times 3 \right)\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{b}{\dfrac{4}{4x-1}dx} \right]$
We can now substitute the values of u and du in the above equation.
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{b}{\dfrac{du}{u}} \right]$
We know that $\int{\dfrac{1}{x}dx}=\log \left| x \right|+C$ . Thus the above equation becomes
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| u \right| \right]_{0}^{b}$
Let us now substitute the value of u. We will get
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4x-1 \right| \right]_{0}^{b}$
Now, we have to apply the intervals.
$\begin{align}
& \Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right|-\log \left| 0-1 \right| \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right|-\log \left| -1 \right| \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right|-\log 1 \right] \\
\end{align}$
We know that $\log 1=0$ . Hence, we can write the above equation as
$\begin{align}
& \Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right|-0 \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \log \left| 4b-1 \right| \right] \\
\end{align}$
Now, we have to apply the limits. We will get
\[\begin{align}
& \Rightarrow \dfrac{3}{4}\left[ \log \left| 4\times \dfrac{1}{4}-1 \right| \right] \\
& =\dfrac{3}{4}\left[ \log \left| 1-1 \right| \right] \\
& =\dfrac{3}{4}\left[ \log \left| 0 \right| \right] \\
& =\infty \\
& \Rightarrow \displaystyle \lim_{b \to {{\dfrac{1}{4}}^{+}}}\left[ \int\limits_{0}^{b}{\dfrac{3}{4x-1}dx} \right]=\infty ...(ii) \\
\end{align}\]
Now let us evaluate $\int\limits_{\dfrac{1}{4}}^{1}{\dfrac{3}{4x-1}dx}$ .
We can write this integral in terms of limits as
$\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \int\limits_{b}^{1}{\dfrac{3}{4x-1}dx} \right]$
On integrating similar to the previous term, we will get
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4x-1 \right| \right]_{b}^{1}$
Now, we have to apply the intervals.
$\begin{align}
& \Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4-1 \right|-\log \left| 4b-1 \right| \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 3 \right|-\log \left| 4b-1 \right| \right] \\
& =\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log 3-\log \left| 4b-1 \right| \right] \\
\end{align}$
We know that $\log 3=0.477$ . Hence, we can write the above equation as
$\Rightarrow \dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ 0.477-\log \left| 4b-1 \right| \right]$
We know that $\displaystyle \lim_{x \to \infty }\left( a-b \right)=\displaystyle \lim_{x \to \infty }a-\displaystyle \lim_{x \to \infty }b$ and limit of a constant is constant itself. Thus, we can write the above equation as
\[\begin{align}
& \Rightarrow \dfrac{3}{4}\left\{ \displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}0.477-\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4b-1 \right| \right] \right\} \\
& =\dfrac{3}{4}\left\{ 0.477-\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4b-1 \right| \right] \right\} \\
\end{align}\]
Let us apply distributive property in the above equation.
\[\begin{align}
& \Rightarrow \dfrac{3}{4}\times 0.477-\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4b-1 \right| \right] \\
& =0.357-\dfrac{3}{4}\displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \log \left| 4b-1 \right| \right] \\
\end{align}\]
Now, we have applied the limits. We will get
\[\begin{align}
& \Rightarrow 0.357-\dfrac{3}{4}\left[ \log \left| 4\times \dfrac{1}{4}-1 \right| \right] \\
& =0.357-\dfrac{3}{4}\left[ \log \left| 1-1 \right| \right] \\
& =0.357-\dfrac{3}{4}\left[ \log \left| 0 \right| \right] \\
& =0.357-\infty \\
& =-\infty \\
& \Rightarrow \displaystyle \lim_{b \to {{\dfrac{1}{4}}^{-}}}\left[ \int\limits_{b}^{1}{\dfrac{3}{4x-1}dx} \right]=-\infty ...\left( iii \right) \\
\end{align}\]
We can see that the limit of the integrals from (ii) and (iii) are plus and minus infinite respectively. Hence, the given integral diverges.
Note: Students must know when an integral converges and diverges. If the limits exist or are finite, then the given integral converges. If the limits are undefined or plus or minus infinite, then the given integral diverges. We can find when the given integral gets indefinite solution by equating the denominator of the given integral to 0.
\[\begin{align}
& \Rightarrow 4x-1=0 \\
& \Rightarrow 4x=1 \\
& \Rightarrow x=\dfrac{1}{4} \\
\end{align}\]
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