Definite integral involving base e and square root of variable as exponent

[phyzmatix]

Homework Statement

Calculate

$$\int_{0}^{4}\sqrt{x} e^{\sqrt{x})$$

The Attempt at a Solution

At first I tried substitution, but this didn't bring me anywhere since the integral is not of the form

$$\int f(g(x))g'(x)$$

My attempt at integration by parts also leads to an endless loop. I used

$$u = e^{\sqrt{x}}$$

Please give me some pointers here. I know I didn't include a whole attempt at a solution, but it's only to save myself some time with the typing (latex is still slow-going). However, if you really need me to include all I've done, I can do so.

 

[gamesguru]

Use substitution, u=Sqrt[x], du=1/(2Sqrt[x])dx so 2u du=dx.
Hope that helps.

 

[CompuChip]

I didn't work this out so I may be pointing you in the wrong direction. But what if you write the integrand as
$$\sqrt{x} e^{\sqrt{x}} \propto x \frac{\mathrm d}{\mathrm dx} e^{\sqrt{x}}$$

 

[phyzmatix]

Use substitution, u=Sqrt[x], du=1/(2Sqrt[x])dx so 2u du=dx.
Hope that helps.

Thanks gamesguru, but unless I'm doing something wrong, this doesn't seem to bring me any closer to the answer.

I didn't work this out so I may be pointing you in the wrong direction. But what if you write the integrand as
$$\sqrt{x} e^{\sqrt{x}} \propto x \frac{\mathrm d}{\mathrm dx} e^{\sqrt{x}}$$

Hi CompuChip. Sorry, but I have no idea what this is that you've just done here Would you mind explaining?

EDIT: I just noticed that I forgot the "dx" in the OP. Sorry.

 

[CompuChip]

Hi CompuChip. Sorry, but I have no idea what this is that you've just done here Would you mind explaining?

I've just rewritten the thing under the integral sign as a product of a (simple) function of x and a derivative w.r.t. x. If you work out the right hand side you'll see you get the original question back, maybe up to some numeric factors. Then you can try partial integration, which will move the d/dx to the x so maybe it becomes simpler. I don't know if it would help though, I think you'd still have to do $$\int e^{\sqrt{x}} \, \mathrm dx$$ in the end.

 

[gamesguru]

If you use my substitution, $u=\sqrt{x}, 2u du=dx$ then the integral becomes:
$$\int \sqrt{x}e^{\sqrt{x}}dx=\int u e^u 2u du=2\int u^2 e^u du$$.
Use integration by parts now, differentiate $u^2$ and integrate $e^u du$.
If you do the integration by parts correctly you get the final answer,
$$2e^{\sqrt{x}}(x-2\sqrt{x}+2)$$.
Hopefully you get the same thing.

 

[phyzmatix]

I've just rewritten the thing under the integral sign as a product of a (simple) function of x and a derivative w.r.t. x. If you work out the right hand side you'll see you get the original question back, maybe up to some numeric factors. Then you can try partial integration, which will move the d/dx to the x so maybe it becomes simpler. I don't know if it would help though, I think you'd still have to do $$\int e^{\sqrt{x}} \, \mathrm dx$$ in the end.

I'm starting to wonder if there is perhaps a typing error in the question. What makes me wonder even more is that the question only counts 5 marks, so it probably isn't supposed to be a huge calculation. Or is that a dangerous assumption to make and first year Calculus really is supposed to be this tricky?

 

[phyzmatix]

If you use my substitution, $u=\sqrt{x}, 2u du=dx$ then the integral becomes:
$$\int \sqrt{x}e^{\sqrt{x}}dx=\int u e^u 2u du=2\int u^2 e^u du$$.
Use integration by parts now, differentiate $u^2$ and integrate $e^u du$.
If you do the integration by parts correctly you get the final answer,
$$2e^{\sqrt{x}}(x-2\sqrt{x}+2)$$.
Hopefully you get the same thing.

Oh, ok, I see now. I'm going to give that a go. Ridiculous as this sounds, I wasn't sure if I could combine the two approaches and typically, on hindsight, it suddenly seems the logical thing to do

 

[phyzmatix]

gamesguru, using your proposed substitution, I get

$$u = \sqrt{x}, 2\sqrt{x}du = dx$$

as you obviously did, however, I'm struggling to convince myself that it's mathematically correct to substitute into the substitution, i.e. to write

$$2\sqrt{x}du = dx$$

as

$$2udu = dx$$

I'm sorry if I'm being irritating, but I really want to understand this and how it works.

 

[HallsofIvy]

Of course that's correct- it's just algebra.

 

[phyzmatix]

Of course that's correct- it's just algebra.

Thank you for your input. I just wanted confirmation since I didn't find any examples in my textbook or study guide that did the same, which caused some doubt (I'm studying through a distance learning university, so I'm very reliant on examples as a means of explanation and to determine what's allowed and what isn't).

Thanks again!

 

[gamesguru]

gamesguru, using your proposed substitution, I get

$$u = \sqrt{x}, 2\sqrt{x}du = dx$$

as you obviously did, however, I'm struggling to convince myself that it's mathematically correct to substitute into the substitution, i.e. to write

$$2\sqrt{x}du = dx$$

as

$$2udu = dx$$

I'm sorry if I'm being irritating, but I really want to understand this and how it works.

Here's the full reason for that.
We make the substitution $u=\sqrt{x}$.
Differentiate both sides with respect to x,
$$\frac{du}{dx}=\frac{1}{2\sqrt{x}}$$.
With me so far? Bring the dx over,
$$du=\frac{1}{2\sqrt{x}}dx$$.
Now we recognize the original substitution $u=\sqrt{x}$:
$$du=\frac{1}{2u}dx$$.
Finally, bring over the 2u:
$$2u du=dx$$.
That's the reason why, and you can check this by integration, 2u du->u^2 and dx->x so:
$$u^2=x, u=\sqrt{x}$$.
Hope that helps.