Difficult anti differentiation question

**Bryanna** · Mar 6th 2013, 10:28 PM

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

**BobP** · Mar 7th 2013, 03:45 AM

First thing is to check on the wording of the question.
The two areas are the same aren't they ?

**MINOANMAN** · Mar 7th 2013, 04:11 AM

I have sent you a private msg

**hollywood** · Mar 7th 2013, 07:45 AM

So the area above the line and below the curve is $\int_0^{7 - \sqrt{m-10}} (x^3 - 14x^2 + 59x -70)-(mx-70)\,dx$ and the area below the line and above the curve is $\int_{7 - \sqrt{m-10}}^{7 + \sqrt{m-10}} (mx-70)-(x^3 - 14x^2 + 59x -70)\,dx$ . 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

**HallsofIvy** · Mar 7th 2013, 04:43 PM

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"?

x=0
x= 7 - square root of m-10
x= 7 + square root of m-10

WHAT NOW

**Bryanna** · Mar 7th 2013, 09:00 PM

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

**Bryanna** · Mar 7th 2013, 09:00 PM

yeah I mean that

**BobP** · Mar 8th 2013, 12:29 AM

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 ?

**hollywood** · Mar 8th 2013, 12:33 AM

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

$\int_0^{7 - \sqrt{m-10}} (x^3 - 14x^2 + 59x -70)-(mx-70)\,dx =$

$\int_0^{7 - \sqrt{m-10}} x^3 - 14x^2 + (59-m)x \,dx =$

$\left[\frac{x^4}{4} - \frac{14x^3}{3} + \frac{(59-m)x^2}{2} \right]_0^{7 - \sqrt{m-10}}$

Which will be messy if you try to calculate $(7 - \sqrt{m-10})^4$ . Maybe you should do the other integral and hope that something cancels.

- Hollywood

**Bryanna** · Mar 8th 2013, 10:04 PM

Above the curve enclosed by the line,
area below the curve enclosed by the line

**jomex42** · Mar 9th 2013, 12:19 AM

Difficult anti differentiation question-untitled.jpg

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can you solve this and you will get the result of
arctan(α/s)

**Bryanna** · Mar 9th 2013, 01:24 AM

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

**hollywood** · Mar 9th 2013, 09:27 AM

Here's what you should get:

$\int_0^{7 - \sqrt{m-10}} x^3 - 14x^2 + (59-m)x \,dx$
$= -\frac{m^2}{4}+\frac{14m}{3}\sqrt{m-10}-\frac{39m}{2}-\frac{140}{3}\sqrt{m-10}+\frac{5041}{12}$

$\int_{7 - \sqrt{m-10}}^{7 + \sqrt{m-10}} (mx-70)-(x^3 - 14x^2 + 59x -70)\,dx$

$=\frac{28m}{3}\sqrt{m-10}-\frac{280}{3}\sqrt{m-10}$

so setting the second to half of the first gives

$-\frac{m^2}{8}-7m\sqrt{m-10}-\frac{39m}{4}+70\sqrt{m-10}+\frac{5041}{24}=0$

and the solution is

$m = 745+\frac{1274}{\sqrt{3}}-98\sqrt{112+\frac{194}{\sqrt{3}}}$

- 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.

**Bryanna** · Mar 9th 2013, 06:50 PM

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

**hollywood** · Mar 10th 2013, 12:15 AM

We need powers of $7 - \sqrt{m-10}$ :

$(7 - \sqrt{m-10})^2 = 49 - 14\sqrt{m-10} + (m-10) = 39 + m - 14\sqrt{m-10}$

$(7 - \sqrt{m-10})^3 =$
$((39 + m) - 14\sqrt{m-10} )*(7-\sqrt{m-10})=$
$(273+7m)-(39+m+98)\sqrt{m-10}+14(m-10) =$
$133+21m-(137+m)\sqrt{m-10}$

$(7 - \sqrt{m-10})^4 =$
$(133+21m-(137+m)\sqrt{m-10})*(7 - \sqrt{m-10})=$
$(931+147m)-(959+7m)\sqrt{m-10})-$
$(133+21m)(-\sqrt{m-10})+(137+m)(m-10)=$
$(931+147m)-(1092+28m)\sqrt{m-10})+(m^2+127m-1370)=$
$m^2+274m-439-(1092+28m)\sqrt{m-10})$

So the integral is:

$\int_0^{7 - \sqrt{m-10}} x^3 - 14x^2 + (59-m)x \,dx =$

$\left[ \frac{1}{4}x^4 - \frac{14}{3}x^3 + \frac{59-m}{2}x^2 \right]_0^{7 - \sqrt{m-10}}=$

$\frac{1}{4}(7 - \sqrt{m-10})^4 - \frac{14}{3}(7 - \sqrt{m-10})^3 + \frac{59-m}{2}(7 - \sqrt{m-10})^2 =$

$\frac{1}{4}(m^2+274m-439-(1092+28m)\sqrt{m-10})) -$

$\frac{14}{3}(133+21m-(137+m)\sqrt{m-10}) + \frac{59-m}{2}(39 + m - 14\sqrt{m-10}) =$

$\frac{1}{4}(m^2+274m-439) -\frac{14}{3}(133+21m) + \frac{59-m}{2}(39 + m) -$

$\left(\frac{1}{4}(1092+28m) - \frac{14}{3}(137+m) + \frac{59-m}{2}(14)\right)\sqrt{m-10} =$

$-\frac{m^2}{4}+\frac{14m}{3}\sqrt{m-10}-\frac{39m}{2}-\frac{140}{3}\sqrt{m-10}+\frac{5041}{12}$

- Hollywood

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