Thread: 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

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 .

If then .

But if then

For f(x) to be positive you know that must have the same sign (and from the domain )


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

For then

For then

For then

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

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

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 .

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.

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 .

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.

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 .

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

Let

Integration by parts it

Spoiler:

and

Many thanks

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?

It would be better to update this thread with your working.

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





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

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.