Definite Integral of e to a function
- Thread starter azadder
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According to The Integrator, the indefinite integral is:
. . . . .\(\displaystyle \L \int\, e^{\sqrt{x}}\, dx\, =\, 2\, e^{\sqrt{x}}\, \left(\sqrt{x}\, -\, 1\right)\)
But, off the top of my head, I don't see how to get that....
Fortunately, there are many people surfing by here who are much smarter than me, so I'm sure we'll see a good hint soon.
Eliz.
. . . . .\(\displaystyle \L \int\, e^{\sqrt{x}}\, dx\, =\, 2\, e^{\sqrt{x}}\, \left(\sqrt{x}\, -\, 1\right)\)
But, off the top of my head, I don't see how to get that....
Fortunately, there are many people surfing by here who are much smarter than me, so I'm sure we'll see a good hint soon.
Eliz.
You can use substituion by letting:
\(\displaystyle \L\\\int\frac{\sqrt{x}e^{\sqrt{x}}}{\sqrt{x}}dx\)
Let \(\displaystyle \L\\u=\sqrt{x} \;\ and \;\ du=\frac{1}{2\sqrt{x}}dx \;\ and \;\2du=\frac{1}{\sqrt{x}}dx\)
Then,
\(\displaystyle \L\\2\int{ue^{u}}du\)
\(\displaystyle \L\\2(u-1)e^{u}\)
Resub:
\(\displaystyle \L\\2(\sqrt{x}-1)e^{\sqrt{x}}\)
\(\displaystyle \L\\\int\frac{\sqrt{x}e^{\sqrt{x}}}{\sqrt{x}}dx\)
Let \(\displaystyle \L\\u=\sqrt{x} \;\ and \;\ du=\frac{1}{2\sqrt{x}}dx \;\ and \;\2du=\frac{1}{\sqrt{x}}dx\)
Then,
\(\displaystyle \L\\2\int{ue^{u}}du\)
\(\displaystyle \L\\2(u-1)e^{u}\)
Resub:
\(\displaystyle \L\\2(\sqrt{x}-1)e^{\sqrt{x}}\)