/sci/ - Science & Math

[Anonymous No.5818877]

Does anyone know how to solve this. I have been on it for the last hr.

[Anonymous No.5818882]

yep, i know how to solve it.

[Anonymous No.5818887]

That makes one of us. Any help will be appreciated.

[metric !1tensorX8Q No.5818889]

Hint: fundamental theorem of calculus.

[Anonymous No.5818899]

hint: read your text book
if you take an hour to solve this, you dont know what you're doing. there's a theorem for dealing specifically with problems like this.

[Anonymous No.5818907]

Sub some value in for u.

[Anonymous No.5818911]

>>5818907

children are weeping

[Anonymous No.5818924]

>>5818911
Whoops, it says find, not prove.

[Anonymous No.5818931]

so would you break the integral into two so you can have the two y limits of integration on the top then solve?

[Anonymous No.5818936]

>>5818931
answer is e^(y^2) / (y^2)

[Anonymous No.5818942]

thats the answer if there was only one y limit of integration.

[Anonymous No.5818957]

http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

The last term is 0 because the e^t/t is independent of y.

e^(y^2)/y^2 * 2y - e^sqrt(y)/sqrt(y) * 1/(2sqrt(y))

[Anonymous No.5818973]

>>5818957
Thats what I have but trying to simplify more.

[SciDropout No.5819026]

I don't know what it means.

[SciDropout No.5819061]

looks like 15*exp(y)/(4*y) to me.

[Anonymous No.5819074]

>>5819061
thank you! I have the answer and now can see how to simplify to reach it. df/dy= 1/2y(4e^2-e^(y^1/2))

[SciDropout No.5819089]

>>5819074 u r welcome.

[Anonymous No.5819103]

Let F(y) be the antiderivative of <span class="math">\frac{e^t}{t}[/spoiler].

Then, the <span class="math">\int\limits_{\sqrt{y}}^{y^2} \frac{e^t}{t} =F(y^2)-F(\sqrt{y})=f(y)[/spoiler]

Then, <span class="math">\frac{df}{dy}=\frac{d}{dy} (F(y^2)-F(\sqrt{y})) [/spoiler]

Fortunately, we know <span class="math">F'(u)=\frac{e^u}{u}[/spoiler], the only trick is the chain rule.