# Definite Integral Confusion

Hi,

So this might be overwhelmingly stupid... But the fundamental theorem of calculus states:
$$\int_{a}^{b}f(x)dx=F(b)-F(a)$$
Where F is any antiderivative of f.

So I have this very simple integral that I'm trying to solve...:

$$2\pi\int_{0}^{2}x^3\sqrt{1+9x^4}dx\rightarrow \ u=1+9x^4$$
$$=\frac{\pi}{18}*\frac{2}{3}u^{3/2}\mid_{1}^{145}=\frac{\pi}{27}(1+9x^4)^{3/2}\mid_{1}^{145}$$

For the answer, I'm given the following:
$$\frac{\pi}{27}(145\sqrt{145}-1)$$

However, when I use the fundamental theorem of calculus, I get something nasty:
$$\frac{\pi}{27}[(1+9(145)^4)^{3/2}-(1+9(1)^4)^{3/2}]$$
$$\mbox{calculator gives}=\frac{2}{27}[(1989227813\sqrt{1989227813}-5\sqrt{5})*\pi*\sqrt{2}]}$$

I don't know why I keep getting the wrong thing. I'm obviously making some sort of stupid error. Any suggestions?

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0rthodontist
Plug the two expressions (yours and the calculator's) into your calculator to evaluate them as numbers.

The long one is 2.91x10^13
The shorter one with the sqrt(145) comes to 203.044...

dav2008
Gold Member
Why are you evaluating from 1 to 145? Those are the limits for u, not x.

Either keep it in terms of u and evaluate from 1 to 145 or change it back to x (extra work) and evaluate from 0 to 2:

$$=\frac{\pi}{27}u^{3/2}\mid_{1}^{145}=\frac{\pi}{27}(1+9x^4)^{3/2}\mid_{0}^{2}$$

In that line it's pointless to change back to x since it's a definite integral. Just evaluate u from 1 to 145 and you're done.

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Ah.

There's the confusion. ...For some reason I guess I thought I had to replace the u by x as with indefinite integrals.

Thanks much for helping me realize my mistake.

Thanks,
Heather

dav2008
Gold Member
There's nothing stopping you from rewriting it in terms of x, but like I said you have to be careful to write the limits in terms of x. Plus, it's just easier to keep things in u.

Here's another problem where I'm screwing up the substitution... I've got the limits straight, (thanks to you), but I don't know why I'm getting this wrong.

$$2\pi\int_{4}^{9}\sqrt{x+\frac{1}{4}dx}\rightarrow \ u=x+\frac{1}{4} \\ = 2\pi\int_{17/4}^{37/4}\sqrt{u}du \\ =2\pi*\frac{2}{3}u^{3/2}\mid_{17/4}^{37/4}$$

Somewhere along the line, my integral is supposed to come out to this:
$$2\pi*\frac{1}{6}(4x+1)^{3/2}$$

...Incorrect substitution...??
u=x+1/4
du=dx

Either way, my answer is off by 1/2 when I get down to it with what I've done.

$$\frac{\pi}{3}(37\sqrt{37}-17\sqrt{17})$$
$$\frac{\pi}{6}(37\sqrt{37}-17\sqrt{17})$$

Nevermind. I figured this out. Quick multiplication happened as a result of me not moving the denominators out of the expression with the limits plugged in.

With the correction, I get the correct pi/6 multiplier.

I would however like to know how the book and how maple is getting the answer with the 1/6 multiplier... Can someone explain that?
...It's different from what I got.

Is
$$2\pi*\frac{1}{6}(4x+1)^{3/2}$$
what you got, or what your book/maple got?

what the book/maple got for the integral w/out limits