1. ## Integral.

Explain why the graph of the function f(x) = (x2 + 1) ln(1/2x) lies
below the x-axis for 1 < x < 2 and above the x-axis for 2 < x < 3.

The graph of the function f(x)=(x^2+1) ln⁡(1/2x) is positive above the x-axis and negative below the x-axis.

Use this fact to find the area enclosed by the graph and the x-axis
between x = 1 and x = 3, giving your answer to five decimal
places.

Can anyone help me.

2. ## Re: Integral.

I can't really read what your question is asking. From the context, I am assuming this is f(x):

$\displaystyle f(x) = (x^2+1)\ln\left(\dfrac{x}{2}\right)$

Is that correct?

Assuming it is, then $\displaystyle \ln 1 = 0$ since $\displaystyle e^0 = 1$. So, for all $\displaystyle \dfrac{1}{2} < \dfrac{x}{2} < 1$, $\displaystyle \ln\left(\dfrac{x}{2}\right)<0$ and for all $\displaystyle 1<\dfrac{x}{2}<\dfrac{3}{2}$, $\displaystyle \ln\left(\dfrac{x}{2}\right)>0$.

Also, $\displaystyle x^2+1>0$ for all real values of $\displaystyle x$. So, you have a positive times a negative for all values of x where $\displaystyle 1 < x < 2$ and you have a positive times a positive for all $\displaystyle 2<x<3$.

Do you know how to use integration by parts? What method are you using to find the area enclosed by the graph and the x-axis? Are you treating area below the x-axis as "negative" area?

3. ## Re: Integral.

Originally Posted by thomasthetankengine
Explain why the graph of the function f(x) = (x2 + 1) ln(1/2x) lies
below the x-axis for 1 < x < 2 and above the x-axis for 2 < x < 3.

The graph of the function f(x)=(x^2+1) ln⁡(1/2x) is positive above the x-axis and negative below the x-axis.

Use this fact to find the area enclosed by the graph and the x-axis
between x = 1 and x = 3, giving your answer to five decimal
places.

Can anyone help me.
First of all, since you are asked for the AREA, which is a QUANTITY, you need to split the region up into two parts (one where you are above the x axis and one where you are below), and from there set up two integrals. The area is the sum of the MAGNITUDES of these amounts.

4. ## Re: Integral.

I understand what i'm being asked now. I'm struggling, however, integrating ln(x/2).

5. ## Re: Integral.

Originally Posted by thomasthetankengine
I understand what i'm being asked now. I'm struggling, however, integrating ln(x/2).
Let u = x/2. Then $\displaystyle \int ln(x/2)~dx = \int ln(u) \cdot (2 du)$

-Dan

6. ## Re: Integral.

can anybody help with this?
I acnt even find any similar examples online. I cant find the area of the curve if there are 2 'functions ' multiplying.

ln(1/2 x)(x^2+1). I have no idea how to go about this. I can do this type of problem if it was only (x^2 +1) on its own.

7. ## Re: Integral.

Do you not have a text book? Every Calculus text I know shows that $\displaystyle \int ln(x)dx= (x- 1)ln(x)+ C$ as a clever example of "integration by parts". And integrating $\displaystyle \int ln(x/2)(x^2+ 1)dx$ can also be done using integration by parts.

8. ## Re: Integral.

\int ln(x/2)(x^2+1) = ln(1/2x) * 2x - \int 1/x * 2x

9. ## Re: Integral.

my text book is rubbish.

10. ## Re: Integral.

An example of integration by parts.

The underlying idea here is that we may write $\displaystyle \int u~dv = uv - \int v~du$.

So let's look at $\displaystyle \int (x^2)~ln(x)~dx$

Letting $\displaystyle u = ln(x)$ and $\displaystyle dv = x^2~dx$ (the LHS of the template in u, v above), we may then say that $\displaystyle du = \frac{1}{x}~dx$ and $\displaystyle v = \int x^2~dx = \frac{1}{3}x^3$

$\displaystyle \int (x^2)~ln(x)~dx = \frac{1}{3} x^3~ln(x) - \int \frac{1}{3}~x^3 \cdot \frac{1}{x}~dx$

The last integral should be easy.

Sorry about all the reposts. I finally fixed it!

-Dan

11. ## Re: Integral.

Thanks for thAt again. Is the last integral x^3/9 ???

12. ## Re: Integral.

Originally Posted by NED1
Thanks for thAt again. Is the last integral x^3/9 ???
Yup.

If you want some practice, I'd recommend $\displaystyle \int x~sin(x)~dx$ and $\displaystyle \int x^2~e^x~dx$. (But not in this thread.)

-Dan

13. ## Re: Integral.

I'm really stuck. I have done what you said and this is the result but can't go from here. ∫▒〖(x^2+1)(ln⁡(x/2) )=〗 〖(x〗^2+1)(x-1) ln⁡(x)-∫▒2x xln(x)=〖(x〗^2+1)(x-1) ln⁡(x)-2∫▒〖x(x-1) ln⁡(x) 〗I don't see where i'm going wrong.

14. ## Re: Integral.

Originally Posted by thomasthetankengine
I'm really stuck. I have done what you said and this is the result but can't go from here. ∫▒〖(x^2+1)(ln⁡(x/2) )=〗 〖(x〗^2+1)(x-1) ln⁡(x)-∫▒2x xln(x)=〖(x〗^2+1)(x-1) ln⁡(x)-2∫▒〖x(x-1) ln⁡(x) 〗I don't see where i'm going wrong.
$\displaystyle \int (x^2 - 1)~ln \left ( \frac{x}{2} \right )~dx$

$\displaystyle = \int x^2~ln \left ( \frac{x}{2} \right )~dx - \int ln \left ( \frac{x}{2} \right )~dx$

For both integrals, let $\displaystyle u = x/2 \implies dx = 2 du$. Thus
$\displaystyle = \int (2u)^2~ln(u)~(2~du) - \int ln(u)~(2~du)$

$\displaystyle = 8 \int u^2~ln(u)~du - 2 \int ln(u)~du$

Can you take it from here?

-Dan

15. ## Re: Integral.

Originally Posted by topsquark
$\displaystyle \int (x^2 - 1)~ln \left ( \frac{x}{2} \right )~dx$

$\displaystyle = \int x^2~ln \left ( \frac{x}{2} \right )~dx - \int ln \left ( \frac{x}{2} \right )~dx$

For both integrals, let $\displaystyle u = x/2 \implies dx = 2 du$. Thus
$\displaystyle = \int (2u)^2~ln(u)~(2~du) - \int ln(u)~(2~du)$

$\displaystyle = 8 \int u^2~ln(u)~du - 2 \int ln(u)~du$

Can you take it from here?

-Dan
The integral is that I need to solve is (x^2+1)(ln⁡(x/2)). I have tried to solve it using integration by parts. Let f(x)=x^2+1 and g' (x)=ln⁡(x/2); then f'(x)=2x and g(x) can be solved by the method of integration by substitution. Let u=x/2 then ∫ln⁡(x/2)dx=∫ln⁡(u)(2du)=(x-1) ln⁡(x). I then used the integration by parts and got ∫(x^2+1)(ln⁡(x/2) )=(x^2+1)(x-1) ln⁡(x)-∫2x xln(x)=(x^2+1)(x-1) ln⁡(x)-2∫x(x-1)ln⁡(x)=
I don't get where the two integrals come in. Is that because you used integration by parts twice? I'm really sorry, I'm getting really confused.

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