# Approximate the integral $\int_0^1 \sin(x^2) dx$

I'd like to ask if someone can please give me a little push with this assignment:

Approximate the value of the integral $\int_0^1 \sin(x^2) dx$ using only $\mathbb{N}$ numbers and basic operations $(+,-,*,/)$. For example use Simpson's rule or the Trapezoidal rule to calculate the integral, then use Taylor's series to determine the value of sinus.

I have used Simpson's rule to approximate the integral and got:

$\int_0^1 \sin(x^2) dx \approx \frac{2}{3}\sin(\frac{1}{4}) + \frac{1}{6}\sin(1)$

But I don't know what to do with the Taylor series. Should I compute it for $\frac{2}{3}\sin(x^2) + \frac{1}{6}\sin(x^2)$ and then express it for $x = \frac{1}{4}$ and $x = 1$ or is my thinking bad?

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But I don't know what to do with the Taylor series.

What you do here is to figure out the Maclaurin expansion for the function $\int_0^x \sin(u^2)\,\mathrm du$. You can figure that by starting with the Maclaurin series for $\sin\,x$ and then replacing the $x$ with $x^2$. Use that series when you're called to evaluate $\sin(x^2)$ for trapezoidal or Simpson's purposes.

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Ok. So you're saying that I should use Maclaurin expansion instead of Taylor's? Excuse me if I sound a bit silly, but I have never heard about the Maclaurin expansion before. If that's what you're saying than I'm afraid that I couldn't do that, since the assignment states that I should use Taylor's series. – Alexandar Živkovič Oct 29 '11 at 20:14
Sorry for the confusion; a Maclaurin series is a Taylor series corresponding to the expansion point 0: $f(x)=f(0)+f'(0)x+\dots$ – J. M. Oct 29 '11 at 20:17
Because the teacher asks them not to use integrals, he should use the Taylor series to approximate $\frac{2}{3}\sin(\frac{1}{4}) + \frac{1}{6}\sin(1)$. – N. S. Oct 29 '11 at 20:21
Thanks, now I understand :) So I should compute it just for $sin(x)$ at point $0$, then replace $x$ with $x^2$ and then substitute the result for the $sin(x^2)$ functions in the expression I got from applying the Simpson's rule and then assign $\frac{1}{4}$ to the first one and $1$ to the second one, right? – Alexandar Živkovič Oct 29 '11 at 20:24
@user9176 Well I thought that was the case. Since I'm getting a feeling that J.M. is telling me to approximate the integral itself with Taylor's series. – Alexandar Živkovič Oct 29 '11 at 20:26