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# How do you evaluate the integral ${{e}^{\sqrt{x}}}dx$ when $a=1$ and $b=9$?

Last updated date: 10th Aug 2024
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Hint: In this problem we have to calculate the integration of the given function. We can observe that we have limits in the problem so we need to calculate the definite integral.so we will calculate the indefinite integral and after that we will apply the limits. First, we will take the substitution $\sqrt{x}=t$ and calculate the value of $dx$ by differentiating the value $\sqrt{x}=t$. Now we will substitute these values in the integration of the given function. Now we will simplify the equation and use integration by parts rule to get the integration value. Now we will apply the given limits to get the definite integral value.

Given function is ${{e}^{\sqrt{x}}}dx$, limits are $a=1$ and $b=9$.
We are going to calculate the indefinite integral of the function ${{e}^{\sqrt{x}}}dx$. For this we are going to use the substitution method.
Let $\sqrt{x}=t$
Squaring on both sides of the above equation, then we will get
$\Rightarrow x={{t}^{2}}$
Differentiating the above equation with respect to $x$, then we will get
$\Rightarrow dx=2tdt$
Now the indefinite integral of the function ${{e}^{\sqrt{x}}}dx$ is
\begin{align} & \Rightarrow \int{{{e}^{\sqrt{x}}}dx}=\int{{{e}^{t}}\left( 2tdt \right)} \\ & \Rightarrow \int{{{e}^{\sqrt{x}}}dx}=2\int{t{{e}^{t}}dt} \\ \end{align}
The above equation is in the form of integration by parts. From the ILATE rule the first function $u=t$, second function is $v={{e}^{t}}$. Substituting these values in the integration by parts formula which is $\int{uvdx}=u\int{vdx}-\int{\left[ {{\left( u \right)}^{'}}\int{vdx} \right]dx}$, then we will get
$\Rightarrow \int{{{e}^{\sqrt{x}}}dx}=2\left[ t\int{{{e}^{t}}dt}-\int{\left[ {{\left( t \right)}^{'}}\int{{{e}^{t}}dt} \right]dt} \right]$
We know that $\int{{{e}^{x}}dx}={{e}^{x}}+C$, $\dfrac{d}{dx}\left( x \right)=1$. Applying these formulas in the above equation, then we will get
\begin{align} & \Rightarrow \int{{{e}^{\sqrt{x}}}dx}=2\left[ t{{e}^{t}}-\int{{{e}^{t}}dt} \right] \\ & \Rightarrow \int{{{e}^{\sqrt{x}}}dx}=2\left[ t{{e}^{t}}-{{e}^{t}} \right]+C \\ \end{align}
Resubstituting $\sqrt{x}=t$ in the above equation, then we will get
$\Rightarrow \int{{{e}^{\sqrt{x}}}dx}=2\left[ \sqrt{x}{{e}^{\sqrt{x}}}-{{e}^{\sqrt{x}}} \right]+C$
The above calculated value is the indefinite integral value. In the problem we have given the limits $a=1$ and $b=9$. Applying these limits to the above integration
\begin{align} & \Rightarrow \int\limits_{a=1}^{b=9}{{{e}^{\sqrt{x}}}dx}=2\left[ \sqrt{x}{{e}^{\sqrt{x}}}-{{e}^{\sqrt{x}}} \right]_{1}^{9} \\ & \Rightarrow \int\limits_{a=1}^{b=9}{{{e}^{\sqrt{x}}}dx}=2\left[ \left[ \sqrt{9}{{e}^{\sqrt{9}}}-{{e}^{\sqrt{9}}} \right]-\left[ \sqrt{1}{{e}^{\sqrt{1}}}-{{e}^{\sqrt{1}}} \right] \right] \\ \end{align}
We have the value $\sqrt{9}=3$, $\sqrt{1}=1$, then the above equation is modified as
\begin{align} & \Rightarrow \int\limits_{a=1}^{b=9}{{{e}^{\sqrt{x}}}dx}=2\left[ \left[ 3{{e}^{3}}-{{e}^{3}} \right]-\left[ {{e}^{1}}-{{e}^{1}} \right] \right] \\ & \Rightarrow \int\limits_{a=1}^{b=9}{{{e}^{\sqrt{x}}}dx}=2\left[ 2{{e}^{3}} \right] \\ & \Rightarrow \int\limits_{a=1}^{b=9}{{{e}^{\sqrt{x}}}dx}=4{{e}^{3}} \\ \end{align}

Hence the integration of the given function ${{e}^{\sqrt{x}}}dx$ $a=1$ and $b=9$ is $4{{e}^{3}}$.

Note: In this problem we have used the terms definite integral and indefinite integral. If you observe any limits in the given problem then it belongs to the definite integral and if there are no limits in the problem then it belongs to an indefinite integral. We can directly find the definite integral without calculating the indefinite integral but it is not advisable to do. Because the process includes several steps in a single step.