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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!

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  • $\begingroup$ 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. $\endgroup$
    – José Carlos Santos
    Commented Aug 4, 2017 at 14:55
  • $\begingroup$ My online homework say it is the wrong answer. $\endgroup$
    – user283028
    Commented Aug 4, 2017 at 14:58
  • $\begingroup$ Well, then I say it is the right answer. $\endgroup$
    – José Carlos Santos
    Commented Aug 4, 2017 at 15:04
  • $\begingroup$ what is the "correct" answer in your online homework? $\endgroup$
    – uniquesolution
    Commented Aug 4, 2017 at 15:53
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    $\begingroup$ Seems your calculator rounded too much. The exact value is $$\int_0^{0.73} 4x^2 - \frac{32}{3}x^6\,dx = \frac{22991588980903}{65625000000000} \approx 0.35034802256614095.$$ $\endgroup$
    – Daniel Fischer
    Commented Aug 4, 2017 at 16:01

1 Answer 1

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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: $$\int_0^{0.73}4x^2-\frac{32 x^6}{3}+\frac{128 x^{10}}{15}\,dx=0.3746$$ The orange curve - two terms approximation; the blue line - the original function.

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  • $\begingroup$ Still incorrect, according to the online homework.. $\endgroup$
    – user283028
    Commented Aug 4, 2017 at 15:47
  • $\begingroup$ The fact that the number is not equal to the answer in your online homework does not necessarily imply that any of those answer are "incorrect". Since we are dealing with approximations, there is no "correct" answer, but only a more accurate answer, and clearly approximating with the green curve gives a more accurate answer than that which is obtained from the orange line. $\endgroup$
    – uniquesolution
    Commented Aug 4, 2017 at 15:50
  • $\begingroup$ I understand what you're saying but I am still unsure of how to get a value that the online homework will accept as "correct". $\endgroup$
    – user283028
    Commented Aug 4, 2017 at 16:37

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