# Homework Help: Simpson approximation method

1. Aug 8, 2010

### Telemachus

1. The problem statement, all variables and given/known data
Use the Simpson method to estimate $$\displaystyle\int_{0}^{1}\cos(x^2)dx$$ with an approximation error less than 0.001.

Well, I have a problem. Actually I'm looking for a bound for the error of approximate method integration by using Simpson's method.

I have to bring $$\displaystyle\int_{0}^{1}\cos(x^2)dx$$ with an error less than 0.001.

I started looking for the fourth order derivative, and got:

$$f^4(x)=48x^2\sin(x^2)-12\cos(x^2)+16x^4\cos(x^2)$$

Now, I have to find a bound K for this derivative in the interval [0,1]. What I did do was watch for if they had maximum and minimum on the interval, then calculate the derivative of order five:
$$f^5(x)\in{[0,1]}\Rightarrow{f^5(x)=0\Longleftrightarrow{x=0}}$$

From here I did was ask:

If $$f^5(x)\in{[0,1]}\Rightarrow{f^5(x)=0\Longleftrightarrow{x=0}}$$

So here I know is that zero is a maximum or minimum. Then I wanted to look for on the concavity of the curve, and I thought the easiest thing would be to look at the sixth derivative, to know how it would behave the fourth derivative of the original function.

$$f^6(x)=720x^2\cos(x^2)-480x^4\sen(x^2)-64x^6\cos(x^2)+120\sen(x^2)$$

The problem is that when evaluated

$$f^6(0)=120\sen(0^2)=0$$

So, I get zero the derivative sixth, and I do not say whether it is concave upwards or downwards is concave.

What should I do? there may be a less cumbersome to work this, if so I would know.

Greetings.

Last edited: Aug 8, 2010
2. Aug 8, 2010

### vela

Staff Emeritus
You could just plot the function and see it hits a maximum on that interval at x=1. Alternately, you can use arguments like x2sin x2≤x2 to find an upper bound for the function. In these types of problems, it's usually not terribly important to find the actual maximum; an upper bound for the maximum is good enough.

3. Aug 8, 2010

### Telemachus

I complicated it too much. Just by evaluating the 6th derivative in any other point of the interval I could get the concavity of the curve. An other way was just seeing for the value of the 5th derivative, as the concavity changes only in zero then I would know if it creases or decreases.

Bye, and thanks.

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