
Solve the following integral and obtain the answer after performing the integration
$\int\limits_{0}^{2}{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}$
a) $\dfrac{2}{\ln 3}\left( {{3}^{\sqrt{2}}}-1 \right)$
b) 0
c) $\dfrac{2\sqrt{2}}{\ln 3}$
d) $\dfrac{{{3}^{\sqrt{2}}}}{\sqrt{2}}$
Answer
587.1k+ views
Hint: In this question , we note that all the expressions are in terms of $\sqrt{x}$, therefore, we can change the variables from x to $\sqrt{x}$ and then perform the integral to obtain the required answer. We should also change the limits of the integration according to the change in variables.
Complete step-by-step answer:
The integral to be evaluated is
$\int\limits_{0}^{2}{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}$
We note that if we change the variable from x to $\sqrt{x}$, then if $\sqrt{x}=t$
$\begin{align}
& dt=d\left( \sqrt{x} \right)=\dfrac{1}{2\sqrt{x}}dx \\
& \Rightarrow dx=2\sqrt{x}d\left( \sqrt{x} \right)=2tdt.........(1.1) \\
\end{align}$
Also,
When $x\to 0\Rightarrow t=\sqrt{x}\to 0.............(1.2)$
When $x\to 2\Rightarrow t=\sqrt{x}\to \sqrt{2}...............(1.3)$
Therefore, we can rewrite the integral as
$\int\limits_{0}^{2}{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}=\int\limits_{0}^{\sqrt{2}}{\dfrac{{{3}^{t}}}{t}2tdt}=2\int\limits_{0}^{\sqrt{2}}{{{3}^{t}}dt}.........(1.4)$
Also, we know the following form of the integral
\[\int{{{a}^{x}}dx=\dfrac{{{a}^{x}}}{\ln a}}\]
Thus, using this form of the integral in equation (1.4) with a=3 and x=t, we get
$\int\limits_{0}^{2}{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}=2\int\limits_{0}^{\sqrt{2}}{{{3}^{t}}dt}=2\times \left[ \dfrac{{{3}^{t}}}{\ln 3} \right]_{0}^{\sqrt{2}}=\dfrac{2}{\ln 3}\left( {{3}^{\sqrt{2}}}-{{3}^{0}} \right)=\dfrac{2}{\ln 3}\left( {{3}^{\sqrt{2}}}-1 \right)$
Which matches option (a) given in the question. Thus, we obtain our answer to the given question as
$\int\limits_{0}^{2}{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}=\dfrac{2}{\ln 3}\left( {{3}^{\sqrt{2}}}-1 \right)$
Note: While solving this question, we should be careful to change the limits of the integration when we change the variables. However, in another way, we can perform the indefinite integral with the new variable and then rewrite it in terms of the old variable in which case we can keep the limits of integration unchanged. For example, in this case, we could also have done
\[\int{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}=\int{\dfrac{{{3}^{t}}}{t}2tdt}=2\dfrac{{{3}^{t}}}{\ln 3}=2\dfrac{{{3}^{\sqrt{x}}}}{\ln 3}\] and then use the limits of the variable x from 0 to 2. However, the answer will remain the same in both the methods.
Complete step-by-step answer:
The integral to be evaluated is
$\int\limits_{0}^{2}{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}$
We note that if we change the variable from x to $\sqrt{x}$, then if $\sqrt{x}=t$
$\begin{align}
& dt=d\left( \sqrt{x} \right)=\dfrac{1}{2\sqrt{x}}dx \\
& \Rightarrow dx=2\sqrt{x}d\left( \sqrt{x} \right)=2tdt.........(1.1) \\
\end{align}$
Also,
When $x\to 0\Rightarrow t=\sqrt{x}\to 0.............(1.2)$
When $x\to 2\Rightarrow t=\sqrt{x}\to \sqrt{2}...............(1.3)$
Therefore, we can rewrite the integral as
$\int\limits_{0}^{2}{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}=\int\limits_{0}^{\sqrt{2}}{\dfrac{{{3}^{t}}}{t}2tdt}=2\int\limits_{0}^{\sqrt{2}}{{{3}^{t}}dt}.........(1.4)$
Also, we know the following form of the integral
\[\int{{{a}^{x}}dx=\dfrac{{{a}^{x}}}{\ln a}}\]
Thus, using this form of the integral in equation (1.4) with a=3 and x=t, we get
$\int\limits_{0}^{2}{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}=2\int\limits_{0}^{\sqrt{2}}{{{3}^{t}}dt}=2\times \left[ \dfrac{{{3}^{t}}}{\ln 3} \right]_{0}^{\sqrt{2}}=\dfrac{2}{\ln 3}\left( {{3}^{\sqrt{2}}}-{{3}^{0}} \right)=\dfrac{2}{\ln 3}\left( {{3}^{\sqrt{2}}}-1 \right)$
Which matches option (a) given in the question. Thus, we obtain our answer to the given question as
$\int\limits_{0}^{2}{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}=\dfrac{2}{\ln 3}\left( {{3}^{\sqrt{2}}}-1 \right)$
Note: While solving this question, we should be careful to change the limits of the integration when we change the variables. However, in another way, we can perform the indefinite integral with the new variable and then rewrite it in terms of the old variable in which case we can keep the limits of integration unchanged. For example, in this case, we could also have done
\[\int{\dfrac{{{3}^{\sqrt{x}}}}{\sqrt{x}}dx}=\int{\dfrac{{{3}^{t}}}{t}2tdt}=2\dfrac{{{3}^{t}}}{\ln 3}=2\dfrac{{{3}^{\sqrt{x}}}}{\ln 3}\] and then use the limits of the variable x from 0 to 2. However, the answer will remain the same in both the methods.
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