How to find the integral of e^(sqrt(x)) as x goes from 0 to 1

[ricekrispie]

Homework Statement

\int_0^1 e ^(sqrt. x) dx

Homework Equations

The Attempt at a Solution

i don´t know what to let u equal to... maybe sqrt.x? but what good would that do when u is differentiated?

 

[rocomath]

hmm, your latek messed up

\int_{0}^{1} e^{\sqrt{x}}dx

yes?

 

[ricekrispie]

yes, thanks... i´m still figuring out how to use it

 

[rocomath]

Let u=\sqrt{x}

then, use another "simple" substitution so that you only have 1 variable.

then, you will need to use integration by parts.

 

[ricekrispie]

i still haven´t figured the problem itself though...

 

[rocomath]

yes, thanks... i´m still figuring out how to use it

shoot, at least you've already almost got it down. there are ppl with like 500+ posts and still don't use latek, hurts my eyes :p

 

[ricekrispie]

that was before learning integration by parts so i guess i´m not supposed to use it

 

[rocomath]

that was before learning integration by parts so i guess i´m not supposed to use it

what technique are you not allowed to use?

 

[ricekrispie]

Let u=\sqrt{x}

then, use another "simple" substitution so that you only have 1 variable.

then, you will need to use integration by parts.

integration by parts... that's why i´m stuck

 

[rocomath]

there's no way around this problem without the use of integration by parts, want to work this problem together by use of parts?

 

[ricekrispie]

do i let u = e^\sqrt{x} ?

 

[rocomath]

the derivative of your substitution would then become

u=e^{\sqrt{x}}

du=\frac{e^{\sqrt{x}}}{2\sqrt{x}}dx

messy. then you will need to take the ln of your u-substitution so you can get your problem in terms of 1 variable.

 

[ricekrispie]

so then u is just \sqrt{x} ?

 

[rocomath]

so then u is just \sqrt{x} ?

yes and i can help you through integration by parts, b/c we can't integrate this problem by simple substitutions.

 

[ricekrispie]

ok... u = \sqrt{x}
and du = du=\frac{1}{2\sqrt{x}}dx

 

[ricekrispie]

what would dv be?

 

[rocomath]

well first, let's make it a "t-substitution" b/c we'll need to use U and V for parts.

so

dx=2\sqrt{x}dt

now we need our integration in terms of 1 variable, so

t=\sqrt{x}

thus

\int e^{\sqrt{x}}dx \rightarrow 2\int te^{t}dt

now we do integration by parts.

 

[rocomath]

what would dv be?

no, the last step was a simple substitution ... this next step we will use parts.

 

[ricekrispie]

wait.. i´m confused... now i have \sqrt{x} as u
and then i have to square it?

 

[rocomath]

2\int te^{t}dt

\begin{array}{cc}u=t & dV=e^{t}dt \\ du=dt & V=e^{t}\end{array}

i'm sure you can take it from here! (judging from your other post)

 

[ricekrispie]

omg i´m sorry hehe
where did u get the t from?

 

[ricekrispie]

oh my god.. nevermind... i got it

 

[ricekrispie]

i didn´t read your other post! thankss hehehe

 

[rocomath]

i didn´t read your other post! thankss hehehe

i just updated it again, idk why i squared it. lol sorry, it's not supposed to be squared.

at times you will need to square it, but not in this situation. (you could, but you'll get the same result which is a waste of time)