# Solving an integral using first two terms of the Mclaurin series for $f(x)$

I have a math problem that I am struggling with.

Assume that $\sin(x)$ equals its Maclaurin series for all x. Use the Maclaurin series for $\sin(4 x^2)$ to evaluate the integral $\int_0^{0.73} \sin(4 x^2) \ dx$ . Your answer will be an infinite series. Use the first two terms to estimate its value.

I have used the Mclaurin series to find that the first two terms are $4x^2$ and $-\frac{32}{3}x^6$. I then plug these terms into my integral and solve:

$$\int _0^{0.73} (4x^2-\tfrac{32}{3}x^6)dx = 0.350349$$

The answer above is incorrect according to my online homework. Can anybody help point me in the right direction? I must be making a silly mistake. Thanks!

• Why do you say it is incorrect? Since $\int_0^{0.73}\sin(4x^2)\,\mathrm dx=0.372\,848\ldots$, ot looks that you got the right result. Commented Aug 4, 2017 at 14:55
• My online homework say it is the wrong answer.
– user283028
Commented Aug 4, 2017 at 14:58
• Well, then I say it is the right answer. Commented Aug 4, 2017 at 15:04
• what is the "correct" answer in your online homework? Commented Aug 4, 2017 at 15:53
• Seems your calculator rounded too much. The exact value is $$\int_0^{0.73} 4x^2 - \frac{32}{3}x^6\,dx = \frac{22991588980903}{65625000000000} \approx 0.35034802256614095.$$ Commented Aug 4, 2017 at 16:01

When you take only two terms, the approximation of $\sin(4x^2)$ by only these two terms in not good enough in a left neighborhood of $0.73$, so you need to take three terms. Calculating the integral with three terms gives: $$\int_0^{0.73}4x^2-\frac{32 x^6}{3}+\frac{128 x^{10}}{15}\,dx=0.3746$$