# Thread: Integration question (pretty fun xD)

1. ## Integration question (pretty fun xD)

Just these 2 questions if anyone can lend a hand

1.Find the minimum area of the part bounded by the parabola $y=a^3x^2-a^4x\ (a>0)$ and the line $y = x$.

2. Evaluate $\,\int_0^1\;\frac{x-1}{\ln\,x}\;dx\,$

First time on this forum, any help will be greatly appreciated!!! XD

2. Originally Posted by usagi_killer
Just these 2 questions if anyone can lend a hand

1.Find the minimum area of the part bounded by the parabola $y=a^3x^2-a^4x\ (a>0)$ and the line $y = x$.
you want to minimize the area, so this is an optimization problem. those are all done the same way, so you should have an idea of how to proceed.

the area will be given by an integral, first set this up. find the limits and then set up the integral for (top function) - (bottom function) over those limits. you want to minimize the resulting intral. note that you may not have to do the integral, though it is simple enough to do if you choose. you can apply the second fundamental theorem of calculus to find the derivative of the function defined as the integral

2. Evaluate $\,\int_0^1\;\frac{x-1}{\ln\,x}\;dx\,$
substitution works here. first note that you can write the integral as $\int_0^1 \frac {x^2 - x}{x \ln x}~dx$ by multiplying by $\frac xx$. Now, let $u = \ln x$ and continue via integration by substitution.

First time on this forum, any thanks will be greatly appreciated!!! XD
Welcome! ...you meant to say "help" as opposed to "thanks" right?

3. Originally Posted by Jhevon
you want to minimize the area, so this is an optimization problem. those are all done the same way, so you should have an idea of how to proceed.

the area will be given by an integral, first set this up. find the limits and then set up the integral for (top function) - (bottom function) over those limits. you want to minimize the resulting intral. note that you may not have to do the integral, though it is simple enough to do if you choose. you can apply the second fundamental theorem of calculus to find the derivative of the function defined as the integral

substitution works here. first note that you can write the integral as $\int_0^1 \frac {x^2 - x}{x \ln x}~dx$ by multiplying by $\frac xx$. Now, let $u = \ln x$ and continue via integration by substitution.

Welcome! ...you meant to say "help" as opposed to "thanks" right?
Thanks for the help, for the 2nd question i got let u = ln(x)

$\frac{du}{dx} = \frac{1}{x}$

$du = \frac{1}{x}dx$

$\int_0^1\frac{1}{u}(x^2-x)du = ...$

how do you continue from here?

4. Originally Posted by usagi_killer
Thanks for the help, for the 2nd question i got let u = ln(x)

\frac{du}{dx} = \frac{1}{x}

du = \frac{1}{x}dx

\int_0^1\frac{1}{u}(x^2-x)du = ...

how do you continue from here?
Hint: $u = \ln x \implies x = e^u$

by the way, we have an improper integral here, so keep that in mind

5. ## Solution to ist question

See attachment

6. I must admit, the second problem is more complicated than i realized (where's Krizalid when you need him?!) my substitution was a knee-jerk reaction . there is no primitive integral here, but with the limits, things work out (Maple gave an answer of ln(2)), so we have to employ some trick. i am thinking about using double integrals. have you done those?

7. ## Not much help but.....

I am as bothered by this as much as y'all

I did find in an old CRC manual from 1945 in general:

the integral from 0 to 1 of (x^p-x^q)/ln(x) = ln[(p+1)/(q+1)]

Which does yield ln(2) but the question remains why?

8. Originally Posted by Jhevon

I must admit, the second problem is more complicated than i realized (where's Krizalid when you need him?!)
Here I am, sorry for the delay.

$\int_{0}^{1}{\frac{x^{p}-x^{q}}{\ln x}\,dx}=\int_{0}^{1}{\int_{q}^{p}{x^{t}\,dt}\,dx}= \int_{q}^{p}{\frac{dt}{t+1}}=\ln \frac{p+1}{q+1},$ provided that $p,q>-1.$

9. Originally Posted by Krizalid
Here I am, sorry for the delay.

$\int_{0}^{1}{\frac{x^{p}-x^{q}}{\ln x}\,dx}=\int_{0}^{1}{\int_{q}^{p}{x^{t}\,dt}\,dx}= \int_{q}^{p}{\frac{dt}{t+1}}=\ln \frac{p+1}{q+1},$ provided that $p,q>-1.$
well, at least i was barking up the right tree. that's not the double interal i was thinking of though. maybe mine would have gotten me in trouble.

10. ## thanks

I learned something today