Re: graph under intervals

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Re: graph under intervals

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

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

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

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 .

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.

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

.

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.

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

.

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

Re: graph under intervals

Let

Integration by parts it

Re: graph under intervals

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?

Re: graph under intervals

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

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.