Improper integral question

[Sheikh al-Majaneen]

The problem:

\int_{-2}^{14} \! \frac{14}{\sqrt[4]{x+2}}\, \mathrm{d}x

What I get, unless I am wrong, is

\lim_{t\to-2}14\ln{\sqrt[4]{x+2}}\Biggr]_t^{14}

...Which, as far as I know, should be divergent, since ln 0 is undefined, approaching negative infinity. However, (since this is online and I can attempt this problem multiple times) I know that it converges, but I don't understand how (unless I am right and webassign is wrong).

I actually do know the answer, which is 448/3 (which I arrived at with a calculator and the ability to try a problem almost exactly the same, whose answer I could see after not getting it right--not the most enlightening process), but I don't understand how it is 448/3. So, if anybody understand why it is that...please help me see why.

[Dason]

Your integral is wrong. Note that you can write your expression as 14(x+2)^(-1/4)

[Sheikh al-Majaneen]

Oh...right.

/Double Facepalm

Thank you.

[skullturf]

I teach calculus. Both you and my students know the formula

\int \frac1x dx = \ln x

which, casually paraphrased, could be said as "integral of 1 over something is natural log of something."

I often tell my students to be careful with that casual paraphrase. Notice that, for example,

\int \frac1{x^2+1} dx = \arctan x

which has absolutely nothing to do with logarithms.

It is true that

\int \frac1{x^2+1} 2x dx = \ln(x^2+1)

but that's a totally different integral.

Perhaps the above casual paraphrase would be better expressed as "integral of 1 over something times d something is natural log of something".

[Talith]

I find it much easier to remember that the integral of f'(x)/f(x) is log(f(x)).

[antonfire]

I often tell my students to be careful with that casual paraphrase. Notice that, for example,

\int \frac1{x^2+1} dx = \arctan x

which has absolutely nothing to do with logarithms.

Well, not nothing. 1/(x²+1) = i/2 (1/(i+x) + 1/(i-x)). So its integral is i/2 (log(i+x) - log(i-x)) = i/2 log((i+x)/(i-x)). You can see that this is the same thing as arctan x, since when x is real it is the imaginary part of log(1+ix).

[Sagekilla]

I teach calculus. Both you and my students know the formula

\int \frac1x dx = \ln x

which, casually paraphrased, could be said as "integral of 1 over something is natural log of something."

This is why I beat people over the head for saying those kinds of things, even if it's a casual paraphrase.
I find it better to say "integral of 1 over x plus something is is natural log of x plus something".

It's not much harder to say, but you won't get tripped up on something like 1 / (x^2 + 1).

[Annihilist]

Substitution, people!

The problem is:

\int_{-2}^{14} \! \frac{14}{\sqrt[4]{x+2}}\, \mathrm{d}x

Now, let:

u=x+2

\frac{du}{dx}\ = 1

du=dx

When:

x=14, u = 16

x=-2, u = 0

So the new integral problem is:

=\int_{0}^{16} \! \frac{14}{\sqrt[4]{u}}\, \mathrm{d}u

=\int_{0}^{16} \! 14u^\frac{-1}{4}\, \mathrm{d}u

=\frac{4}{3}*14u^\frac{3}{4}\Biggr]_0^{16}

=(\frac{56}{3}16^\frac{3}{4})-(0)

=\frac{448}{3}\

Can I ask where you got this problem from? Homework or otherwise - where did you find it? What are you doing it for?

[Annihilist]

I teach calculus. Both you and my students know the formula

\int \frac1x dx = \ln x

which, casually paraphrased, could be said as "integral of 1 over something is natural log of something."

I often tell my students to be careful with that casual paraphrase. Notice that, for example,

\int \frac1{x^2+1} dx = \arctan x

which has absolutely nothing to do with logarithms.

It is true that

\int \frac1{x^2+1} 2x dx = \ln(x^2+1)

but that's a totally different integral.

Perhaps the above casual paraphrase would be better expressed as "integral of 1 over something times d something is natural log of something".

\int \frac{f'(x)}{f(x)} dx = \ln f(x)+c

(Edited the +c in)

[jestingrabbit]

Substitution, people!

Meh. Anytime that you get du/dx = 1, you should just write out the antiderivative imo.

[Annihilist]

Substitution, people!

Meh. Anytime that you get du/dx = 1, you should just write out the antiderivative imo.

Yeah thats what I did. Just slower.

[Proginoskes]

\int \frac{f'(x)}{f(x)} dx = \ln f(x)

Sorry, you lose a point. The correct answer is

\int \frac{f'(x)}{f(x)} dx = \ln f(x) + C

[Annihilist]

\int \frac{f'(x)}{f(x)} dx = \ln f(x)

Sorry, you lose a point. The correct answer is

\int \frac{f'(x)}{f(x)} dx = \ln f(x) + C

Ah shit. Thanks.

No one else was using 'c's so I forgot to add it.

EDIT: "c = 0" is a valid answer though, unless you have been given a specific point which the function passes through.

[gfauxpas]

\int \frac{f'(x)}{f(x)} dx = \ln f(x)

Sorry, you lose a point. The correct answer is

\int \frac{f'(x)}{f(x)} dx = \ln f(x) + C

Well if you want to play that game:

\displaystyle \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C

where f(x) ≠ 0

[gorcee]

\int \frac{f'(x)}{f(x)} dx = \ln f(x)

Sorry, you lose a point. The correct answer is

\int \frac{f'(x)}{f(x)} dx = \ln f(x) + C

Ah shit. Thanks.

No one else was using 'c's so I forgot to add it.

EDIT: "c = 0" is a valid answer though, unless you have been given a specific point which the function passes through.

Sure, we'll give you partial credit for getting one point.

+1/infinity points.