Question

Calc Question. Will Medal! Evaluate the problem below.

I know for a fact the answer is ex + c. But what are the steps?

Answer 1

\[\int\limits_{?}^{?}\frac{ e \sqrt{x} }{ sqrt{x} }\]

Answer 2

Sorry for the format

Answer 3

\[\large\rm \int\limits \frac{e^{\sqrt x}}{\sqrt x}dx\quad=\quad \int\limits e^{\sqrt x}\left(\frac{1}{\sqrt x}dx\right)\]

Answer 4

Making the substitutiion, \(\large\rm u=\sqrt x\)

Answer 5

Ah ok. So then du = \[\frac{ 2 }{ 3 }x ^{\frac{ 3 }{ 2 }}\]

Answer 6

Oh you integrated by mistake.

Answer 7

You want to `differentiate` to find your du.

Answer 8

*facepalm* Sorry!

Answer 9

du = \[\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\]

Answer 10

Bahhh you can do your 1/2 power if you want.
But square root comes up so often that you should just put this derivative to memory,\[\large\rm \frac{d}{dx}\sqrt{x}\quad=\quad \frac{1}{2\sqrt x}\]Derivative of square root is one over two square roots.

Answer 11

Good to know! Will do!

Answer 12

\[\large\rm du=\frac{1}{2\sqrt x}dx\]Hmm this is ALMOST exactly what we bracketed in our integral.
Is there any way we can get rid of the 2?

Answer 13

Put it outside the integral, making it look like: \[2\int\limits_{?}^{?}\frac{ 1 }{ \sqrt{x} }\]

Answer 14

Well, yes but I meant in our substitution, not in the integral.
You can't just make a 2 magically appear like that :)\[\large\rm du=\frac{1}{2\sqrt x}dx\]Multiplying both sides by 2 gives us,\[\large\rm 2du=\left(\frac{1}{2\sqrt x}dx\right)\]

Answer 15

\[\large\rm \int\limits\limits e^{\sqrt x}\left(\frac{1}{\sqrt x}dx\right)\quad=\quad \int\limits e^u\left(2du\right)\]

Answer 16

Ahh I made a typo :(

Answer 17

I see what your saying!

Answer 18

Multiplying by 2 gave us,\[\large\rm 2du=\left(\frac{1}{\sqrt x}dx\right)\]

Answer 19

Ok! So knowing this, we can now make it so that the 2 is upfront and the function left (e^sqrt(x)). So now would we just have to integrate?

Answer 20

\[\large\rm =2\int\limits e^u du\]Yes, we can integrate from here.

Answer 21

Awesome! So the final answer would be \[2e ^{u}\] and substituting in the property of u, making it \[2e ^{\sqrt{x}} + C\]

Answer 22

Yes, good job!

Answer 23

That was similar than I thought! Haha thank you!

Answer 24

*simpler

Answer 25

heh :) np