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Question: How do you evaluate the integral \(\int{{{e}^{\sqrt{x}}}dx}\)?...

How do you evaluate the integral ∫exdx\int{{{e}^{\sqrt{x}}}dx}?

Explanation

Solution

We will use the method of substitution to solve this integral. We will also use the integration by parts formula. We will be using the differentiation of x\sqrt{x}, which is a standard derivative given by ddx(x)=12x\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{2\sqrt{x}}. We will also use the integration of the exponential function. It is a standard integral given as ∫exdx=ex\int{{{e}^{x}}dx}={{e}^{x}}.

Complete step-by-step solution:
We have to find the value of the following integral,
I=∫exdxI=\int{{{e}^{\sqrt{x}}}dx}
Let us substitute u=xu=\sqrt{x}. The differentiation of x\sqrt{x}, which is a standard derivative given by ddx(x)=12x\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{2\sqrt{x}}. So, we also have
du=12xdx ∴dx=2xdu \begin{aligned} & du=\dfrac{1}{2\sqrt{x}}dx \\\ & \therefore dx=2\sqrt{x}du \\\ \end{aligned}
And since u=xu=\sqrt{x}, we can write dx=2ududx=2udu. Therefore, the integral becomes the following,
I=2∫euuduI=2\int{{{e}^{u}}udu}
Integration by parts for two functions, ff and gg, is given by the following formula,
∫f⋅gdx=f∫gdx−∫(dfdx∫gdx)dx\int{f\cdot gdx}=f\int{gdx}-\int{\left( \dfrac{df}{dx}\int{gdx} \right)dx}
Let f=uf=u and g=eug={{e}^{u}}. Using the above formula, we get the following,
I=2∫ueudu ∴I=2(u∫eudu−∫(ddu(u)∫eudu)du) \begin{aligned} & I=2\int{u{{e}^{u}}du} \\\ & \therefore I=2\left( u\int{{{e}^{u}}du}-\int{\left( \dfrac{d}{du}\left( u \right)\int{{{e}^{u}}du} \right)du} \right) \\\ \end{aligned}
The integration of the exponential function is a standard integral given as ∫exdx=ex\int{{{e}^{x}}dx}={{e}^{x}}. We know that the differentiation is ddu(u)=1\dfrac{d}{du}\left( u \right)=1. Substituting these values in the above equation, we get the following,
I=2(ueu−∫eudu)I=2\left( u{{e}^{u}}-\int{{{e}^{u}}du} \right)
Using the integral of the exponential function again, we get
I=2(ueu−eu) ∴I=2eu(u−1) \begin{aligned} & I=2\left( u{{e}^{u}}-{{e}^{u}} \right) \\\ & \therefore I=2{{e}^{u}}\left( u-1 \right) \\\ \end{aligned}
Now, we will re-substitute the value of the variable u=xu=\sqrt{x}. So, we obtain the following
I=2ex(x−1)I=2{{e}^{\sqrt{x}}}\left( \sqrt{x}-1 \right)
Thus, we have obtained the value of the given integral.

Note: It is important to change the infinitesimal in the integral according to the variable while using the method of substitution. If the given integral is a definite integral, then we have to also change the limits with respect to the variable being substituted. We should be familiar with the integration and differentiation of standard functions since they are useful in such types of questions.