Integration by parts and roots

[jhayes25]

Ok. I have a definite integral of e^sqroot(x) dx from x=0 to x=1.

I would use u=sqroot(x) and du=1/2*sqroot(x), but I'm confused what I would set v=?

 

[rocomath]

\int_0^1 e^{\sqrt x}dx

 

[jhayes25]

\int_0^1 e^{\sqrt x}dx

yes, haha thanks. I knew I would screw it up if i tried.

 

[rocomath]

t=e^{\sqrt x} \rightarrow \ln t=\sqrt x "evaluate for new limits" |_0^1=|_1^e

dt=\frac{e^{\sqrt x}}{2\sqrt x}dx \rightarrow 2\ln tdt=e^{\sqrt x}dx

2\int_1^e\ln tdt

Take it from here.

 

[jambaugh]

Remember that the "dx" is not just notation. In integrating by parts or in substitution you will need to account for it. However if you're trying to do this integral by parts you're not going to want to set u=\sqrt{x} as this is not a factor of your integrand. For integration by parts you need the form:
\int u dv

I think you're first going to want to execute the variable substitution of u=\sqrt{x} and see what integral you get. Then possibly it will be integrable by parts.

Remember that in executing substitution you either need to change your limits to those of the new variable (actually the same in this case) or you should rewrite your final integral back in terms of the original variable before you evaluate the limits.

 

[Count Iblis]

As explained by jambaugh substitute x = t^2, then exp(sqrt(x)) = exp(t) and dx = 2t dt The limits x = 0 and x = 1 correspond to t = 1 and t = 1 respectively.

Try to do this integral using partial integration. Also try to do it by first computing the integral of exp(p t) where p is an arbitrary parameter and then differentiate both sides w.r.t. p.

 

[sutupidmath]

\int_0^1 e^{\sqrt x}dx
another way of doing it would be like this
\int_0^1 e^{\sqrt x}dx=\int_0^1 \sqrt x \frac{e^{\sqrt x}}{\sqrt x } And then take the sub
t=\sqrt x, so, <br /> <br /> dt=\frac{dx}{2 \sqrt x}=&gt;2dt=\frac{dx}{\sqrt x}, x=0 =&gt; t=0, x=1=&gt; t=1 then
\int_0^1 2t e^{t} and now you could apply integration by parts, by letting u=t and v=\int e^{t}dt