integral with simpson rule

[alla]

hello im student at univ and we have model math numerical analysis the problem is **** calculate arctan(3) for simpson rule of interal and n=4 number of division****
thanks

 

[LCKurtz]

Hint: What definite integral gives [MATH]\text{arctan}(3)[/MATH] for an answer? You need to show some effort.

 

[alla]

thanks for hint i know is integral of 1/1+x^2 = arctang(x) and i know the rule of simpson but the interval Not defined

 

[MarkFL]

You should then know, by the FTOC, that:

[MATH]\int_a^b \frac{1}{x^2+1}\,dx=\arctan(b)-\arctan(a)[/MATH]
What choices for \(a\) and \(b\) do you suppose you could use?

 

[ksdhart2]

thanks for hint i know is integral of 1/1+x^2 = arctang(x) and i know the rule of simpson but the interval Not defined

First, you've made a fairly big error here. What you wrote literally means \(\displaystyle \frac{1}{1} + x^2\) but what you meant was 1/(1 + x^2), or \(\displaystyle \frac{1}{1 + x^2}\). Make sure you understand why those grouping symbols are very important and cannot be omitted. Second, it is true that:

\(\displaystyle \int \frac{1}{1 + x^2} = \arctan(x)\)

We also know, by the Fundamental Theorem of Calculus that:

\(\displaystyle \int\limits_{a}^{b} f(x) = F(b) - F(a)\)

where F(x) is the indefinite integral of f(x). Hence, we want to find an \(a\) and \(b\) such that:

\(\displaystyle \int\limits_{a}^{b} \frac{1}{1 + x^2} = \arctan(b) - \arctan(a) = \arctan(3)\)

What values do you think would work?

 

[alla]

thanks a lot i chose 0 to 3 or -1/2 to 1

 

[ksdhart2]

thanks a lot i chose 0 to 3 or -1/2 to 1

Both of those intervals will work, in that the integral evaluates to \(\arctan(3)\) in both cases, but using [0, 3] is probably the better choice, since the approximation created by Simpson's rule will be slightly more accurate using that interval.