calculate integral approximations

**abilitiesz** · November 1st 2009, 10:20 PM

Hi, I need help on solving this problem.

I tried to find the antiderivative of this equation first and got:

-ln / 4x^4

I then tried to find the interbals of the width so I did.. 8-2/5 = 6/5 = 1.2

So, thus I added 1.2 to the endpoints and got.. 2, 3.2, 4.4, 5.6, 6.8, 8.

I then plugged it into the equation I solved for to try to find T5, but I keep getting the wrong answer. Is my anti derivative even correct?

Thank you.

**CaptainBlack** · November 1st 2009, 10:34 PM

Originally Posted by abilitiesz

Hi, I need help on solving this problem.

I tried to find the antiderivative of this equation first and got:

-ln / 4x^4

I then tried to find the interbals of the width so I did.. 8-2/5 = 6/5 = 1.2

So, thus I added 1.2 to the endpoints and got.. 2, 3.2, 4.4, 5.6, 6.8, 8.

I then plugged it into the equation I solved for to try to find T5, but I keep getting the wrong answer. Is my anti derivative even correct?

Thank you.

Someone may be along in a short while who can make a good guess at what you are asking, but in the mean time post the question as asked and explain where you are having difficulties.

CB

**abilitiesz** · November 1st 2009, 10:45 PM

Oh ok, I edited my first post and put the question along with the equations. Basically, all I'm asking for is if my anti derivative is correct. I know how to solve it after that.

**mr fantastic** · November 1st 2009, 11:29 PM

Originally Posted by abilitiesz

Hi, I need help on solving this problem.

I tried to find the antiderivative of this equation first and got:

-ln / 4x^4

I then tried to find the interbals of the width so I did.. 8-2/5 = 6/5 = 1.2

So, thus I added 1.2 to the endpoints and got.. 2, 3.2, 4.4, 5.6, 6.8, 8.

I then plugged it into the equation I solved for to try to find T5, but I keep getting the wrong answer. Is my anti derivative even correct?

Thank you.

I will guess from the notation that you're expected to get the approximate value of the integral using the Trapeziodal rule (with n = 5) and the Midpoint rule (with n = 5). There are formulas in which you substitute the stuff you've been given to get the answer. It involves nothing more than simple but tedious arithmetic.

I don't know why you would be trying to get an anti-derivative when the question is clearly not asking you to.

**CaptainBlack** · November 2nd 2009, 12:06 AM

Originally Posted by abilitiesz

Hi, I need help on solving this problem.

I tried to find the antiderivative of this equation first and got:

-ln / 4x^4

The anti-derivative of $[\ln(x)]^5$ is not what you have here. Also you are unlikely be able to find the anti-derivative, which is one of the points of asking you to integrate the function numerically.

WolframAlpha gives:

$\int \log^5(x) dx = -120 x+x \log^5(x)-5 x \log^4(x)+$ $20 x \log^3(x)-60 x \log^2(x)+120 x \log(x)+C$

which can be obtained by repeated integration by parts.

CB

**mr fantastic** · November 2nd 2009, 01:09 AM

Originally Posted by CaptainBlack

The anti-derivative of $[\ln(x)]^5$ is not what you have here. Also you are unlikely be able to find the anti-derivative, which is one of the points of asking you to integrate the function numerically.

WolframAlpha gives:

$\int \log^5(x) dx = -120 x+x \log^5(x)-5 x \log^4(x)+$ $20 x \log^3(x)-60 x \log^2(x)+120 x \log(x)+C$

which can be obtained by repeated integration by parts.

CB

Since the integrand appears to be $\frac{1}{(\log x)^5}$ the business of attempting to get an anti-derivative is even more dire (and even more clearly not intended) ....

**CaptainBlack** · November 2nd 2009, 01:43 AM

Originally Posted by mr fantastic

Since the integrand appears to be $\frac{1}{(\log x)^5}$ the business of attempting to get an anti-derivative is even more dire (and even more clearly not intended) ....

Not really dire-er, the main difference is that the chain of integrations by parts terminates with $\text{li}(x)$ rather than something more elementary

CB

**mr fantastic** · November 2nd 2009, 02:04 AM

Originally Posted by CaptainBlack

Not really dire-er, the main difference is that the chain of integrations by parts terminates with $\text{li}(x)$ rather than something more elementary

CB

Well, it looks like this question has been $\text{li}x\text{ed}$ into shape ....

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