Use the Simpson method to estimate \(\displaystyle \displaystyle\int_{0}^{1}\cos(x^2)dx\) 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 \(\displaystyle \displaystyle\int_{0}^{1}\cos(x^2)dx\) with an error less than 0.001.

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

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

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:

\(\displaystyle f^5(x)=120x\sin(x^2)+160x^3\cos(x^2)-32x^5\sin(x^2)\)

From here I did was ask:

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

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.

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

The problem is that when evaluated

\(\displaystyle f^6(0)=120\sin(0^2)=0\)

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.

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

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

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

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:

\(\displaystyle f^5(x)=120x\sin(x^2)+160x^3\cos(x^2)-32x^5\sin(x^2)\)

From here I did was ask:

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

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.

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

The problem is that when evaluated

\(\displaystyle f^6(0)=120\sin(0^2)=0\)

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.

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