Difficult anti differentiation question

I have a curve, x^3 - 14x^2 + 59x -70 , and I need to run this line through it, mx - 70 , to make the area above the curve enclosed by the line equal to HALF of the area below the line enclosed by the curve..... So, I need to find the gradient (m) of the line to run through it for this to be possible.

So far I have derived that the x intersects are :

x=0

x= 7 - square root of m-10

x= 7 + square root of m-10

WHAT NOW (Worried)

Re: Difficult anti differentiation question

First thing is to check on the wording of the question.

The two areas are the same aren't they ?

Re: Difficult anti differentiation question

I have sent you a private msg

Re: Difficult anti differentiation question

So the area above the line and below the curve is and the area below the line and above the curve is . You need to calculate these two integrals (treating m as a constant), then set the second equal to half of the first and solve for m.

The process is straightforward, but kind of messy.

- Hollywood

Re: Difficult anti differentiation question

Quote:

Originally Posted by

**Bryanna** I have a curve, x^3 - 14x^2 + 59x -70 , and I need to run this line through it, mx - 70 , to make the area above the curve enclosed by the line equal to HALF of the area below the line enclosed by the curve..... So, I need to find the gradient (m) of the line to run through it for this to be possible.

So far I have derived that the x intersects are :

The x-intercepts of **what**? There are two graphs refered to, the graph of y= x^3- 14x^2+ 59x- 70 and y= mx- 70 and **neither** has what you give as x- intercepts. Do you mean where the **graphs** intersect, not "x-interepts"?

Quote:

x=0

x= 7 - square root of m-10

x= 7 + square root of m-10

WHAT NOW (Worried)

Re: Difficult anti differentiation question

So how would I calculate it treating m as a constant?

Re: Difficult anti differentiation question

Re: Difficult anti differentiation question

Am I misreading this question or is everyone else ?

'... the area above the curve enclosed by the line ...'

'... the area below the line enclosed by the curve ...'

They describe the same area don't they ?

Re: Difficult anti differentiation question

It just means pretend that m is a constant like 3 or 7. So for example,

Which will be messy if you try to calculate . Maybe you should do the other integral and hope that something cancels.

- Hollywood

Re: Difficult anti differentiation question

Above the curve enclosed by the line,

area below the curve enclosed by the line

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Re: Difficult anti differentiation question

Re: Difficult anti differentiation question

I tried it but I dont get how it would cancel out on the other side

Re: Difficult anti differentiation question

Here's what you should get:

so setting the second to half of the first gives

and the solution is

- Hollywood

P.S. jomex2: You should start a new thread with your problem. Few (if any) people besides Bryanna and me are reading this thread now.

Re: Difficult anti differentiation question

ok so I am starting to see where you are coming from... But... you know how you did the first bit = - m^2/4 + 14m/3 (square root) m-10..... That answer, how did you get it? I am expanding it and I dont get how you got that

Re: Difficult anti differentiation question

We need powers of :

So the integral is:

- Hollywood