# graph under intervals

• July 2nd 2011, 08:51 AM
vidhi96
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
• July 2nd 2011, 09:02 AM
Plato
Re: graph under intervals
Quote:

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 $f(x) = (4 - x)\ln \left( {\frac{x}{2}} \right)$.

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

But if $x>4$ then $(4-x)<0.$
• July 2nd 2011, 09:02 AM
e^(i*pi)
Re: graph under intervals
For f(x) to be positive you know that $4-x \text{ and } \ln \left(\dfrac{x}{2}\right)$ must have the same sign (and from the domain $x > 0$)

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

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

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

For $4 < x$ then $4-x < 0 \text{ and } \ln \left(\dfrac{x}{2}\right) > 0$
• July 2nd 2011, 09:19 AM
vidhi96
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
• July 2nd 2011, 09:20 AM
vidhi96
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
• July 2nd 2011, 09:49 AM
HallsofIvy
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 $\int_a^b f(x)dx$.

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.
• July 2nd 2011, 10:19 AM
vidhi96
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?

Quote:

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 $\int_a^b f(x)dx$.

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.

• July 2nd 2011, 02:18 PM
vidhi96
Re: graph under intervals
Quote:

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 $\int_a^b f(x)dx$.

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
• July 2nd 2011, 02:33 PM
e^(i*pi)
Re: graph under intervals
Let $u = \ln \left(\dfrac{x}{2}\right) \text{ and } v' = 4-x$

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

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

$\displaystyle \ln \left(\dfrac{x}{2}\right) \cdot \left(4x-\dfrac{x^2}{2}\right) - \int \left[\left(4x-\dfrac{x^2}{2}\right) \cdot \dfrac{1}{x} \right]\ dx$
• July 2nd 2011, 02:56 PM
vidhi96
Re: graph under intervals
Many thanks
• July 2nd 2011, 03:07 PM
vidhi96
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?
• July 2nd 2011, 03:21 PM
e^(i*pi)
Re: graph under intervals
$\int \left[\left(4x-\dfrac{x^2}{2}\right) \cdot \dfrac{1}{x} \right]\ dx = \int \left(\dfrac{4x}{x} -\dfrac{x^2}{2x}\right) \ dx$
$= \int \left(4 - \dfrac{x}{2}\right) \dx = \int 4\, dx - \dfrac{1}{2} \int x\ dx$