1. ## graph under intervals

I don't even know what to call this - and I am bit baffled by the question:
Q) explain why the graph of the function f(x)=(4-x)In(1/2*x) lies above the x- aixs for (2<x<4) and below the x-axis for 4<x<6.

does this mean x= 3 and 5 and you substitute this in equation to see what you get?

Now I know this- so sorry for posting it. But I have to use this fact that graph is positive 2<x<4 and negative for 4<x<6 and find the area enclosed by the graph anf the x- axis between x=2 and x=6

2. ## Re: graph under intervals

Originally Posted by vidhi96
I don't even know what to call this - and I am bit baffled by the question:
Q) explain why the graph of the function f(x)=(4-x)In(1/2*x) lies above the x- aixs for (2<x<4) and below the x-axis for 4<x<6.
To be clear the function is $\displaystyle f(x) = (4 - x)\ln \left( {\frac{x}{2}} \right)$.

If $\displaystyle 2<x<4$ then $\displaystyle (4-x)>0~\&~\ln \left( {\frac{x}{2}} \right)>0$.

But if $\displaystyle x>4$ then $\displaystyle (4-x)<0.$

3. ## Re: graph under intervals

For f(x) to be positive you know that $\displaystyle 4-x \text{ and } \ln \left(\dfrac{x}{2}\right)$ must have the same sign (and from the domain $\displaystyle x > 0$)

For the graph to be above the x axis then both "parts" of f(x) must have the same sign.

For $\displaystyle 0 < x < 2$ then $\displaystyle 4-x > 0 \text{ and } \ln \left(\dfrac{x}{2}\right) < 0$

For $\displaystyle 2 < x < 4$ then $\displaystyle 4-x > 0 \text{ and } \ln \left(\dfrac{x}{2}\right) > 0$

For $\displaystyle 4 < x$ then $\displaystyle 4-x < 0 \text{ and } \ln \left(\dfrac{x}{2}\right) > 0$

4. ## Re: graph under intervals

thanks but I am not sure how to calculate area of this function between x=2 and x=6 uisng what we just proved

5. ## Re: graph under intervals

hey thanks and I am glad I did the same thing- but I don't know how to calculate the area of the function / graph between x=2 and x=6 - any help will do

6. ## Re: graph under intervals

You posted this in the Calculus section so you should understand that the area under the graph of y= f(x), above the x-axis, for a< x< b is given by $\displaystyle \int_a^b f(x)dx$.

To integrate (4- x)ln(x/2), let u= x/2 so that x= 2u and you have (4- 2u)ln(u)= 4ln(u)- 2uln(u).
You should be able to integrate each of those by parts.

7. ## Re: graph under intervals

I got the part - as I will use integration by substitution hence 2u=x but I am confused over limit of integration so b=4 and a=2 is that correct?

Originally Posted by HallsofIvy
You posted this in the Calculus section so you should understand that the area under the graph of y= f(x), above the x-axis, for a< x< b is given by $\displaystyle \int_a^b f(x)dx$.

To integrate (4- x)ln(x/2), let u= x/2 so that x= 2u and you have (4- 2u)ln(u)= 4ln(u)- 2uln(u).
You should be able to integrate each of those by parts.

8. ## Re: graph under intervals

Originally Posted by HallsofIvy
You posted this in the Calculus section so you should understand that the area under the graph of y= f(x), above the x-axis, for a< x< b is given by $\displaystyle \int_a^b f(x)dx$.

To integrate (4- x)ln(x/2), let u= x/2 so that x= 2u and you have (4- 2u)ln(u)= 4ln(u)- 2uln(u).
You should be able to integrate each of those by parts.
Hi ya I can't figure it out and it seems to complicated can you please show me thanks

9. ## Re: graph under intervals

Let $\displaystyle u = \ln \left(\dfrac{x}{2}\right) \text{ and } v' = 4-x$

Integration by parts it $\displaystyle \displaystyle \int u\,dv = uv - \int v\,du$

Spoiler:
$\displaystyle u' = \dfrac{1}{x}$ and $\displaystyle v = 4x-\dfrac{x^2}{2}$

$\displaystyle \displaystyle \ln \left(\dfrac{x}{2}\right) \cdot \left(4x-\dfrac{x^2}{2}\right) - \int \left[\left(4x-\dfrac{x^2}{2}\right) \cdot \dfrac{1}{x} \right]\ dx$

Many thanks

11. ## Re: graph under intervals

hey do you mind if I complete the whole thing and send it to you for a check.

Just as a started Integral of (4x-x^2/2)*1/x dx = (7x^2/4)+c is this correct?

12. ## Re: graph under intervals

You can cancel an x in the integral though because we know it cannot be 0 and we're not integrating from 0.

$\displaystyle \int \left[\left(4x-\dfrac{x^2}{2}\right) \cdot \dfrac{1}{x} \right]\ dx = \int \left(\dfrac{4x}{x} -\dfrac{x^2}{2x}\right) \ dx$

$\displaystyle = \int \left(4 - \dfrac{x}{2}\right) \dx = \int 4\, dx - \dfrac{1}{2} \int x\ dx$

I'm not sure where 7/4 comes from and you will need to take care with your signs when you subtract it with the first part of the expression

13. ## Re: graph under intervals

that was my error- i worked it all out as this integration was part of big question. I iwll upload entire question once I finish typing it - thnaks again.