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Given the function $${F(x)=\int_0^x\sin(5t^2)dt}$$

I must find the MacLaurin polynomial of degree $7$ for $F(x)$. Given as a function of $x$.

Since we already have the first derivative given by the integral, I continued to integrate until I reached a $7$th degree within the function and plugged in. This seems to be clearly wrong and very time consuming. How would I proceed to find the MacLaurin polynomial.

After, I'm asked to find the the value of the integral given below using the polynomial found above. $${\int_0^{0.68}\sin(5x^2)dx}$$

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    $\begingroup$ For this function it is trivial to find the whole Maclaurin series. Please expand $\sin(5t^2)$ first. Use the standard sine Maclaurin expansion and a trivial substitution. Next integrate this series. $\endgroup$ – szw1710 Sep 21 '18 at 21:38
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$$\sin(5t^2)\approx 5t^2-\frac{(5t^2)^3}{3!}+\frac{(5t^2)^5}{5!}-\frac{(5t^2)^7}{7!}+\cdots$$

so that

$$\int_0^x\sin(5t^2)dt\approx\frac{5x^3}3-\frac{5^3x^7}{7\cdot3!}+\frac{5^5x^{11}}{11\cdot5!}-\frac{5^7x^{15}}{15\cdot7!}+\cdots$$

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    $\begingroup$ I'm not sure it's such a good idea to spoon-feed the answer to an OP who's posted so many low-effort questions $\endgroup$ – Jam Sep 21 '18 at 21:45
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    $\begingroup$ @Jam, I agree with you totally, but my own natural reaction in a case like this is to rush in to solve the problem for the OP, and only afterwards to reflect that I probably did to much of the work for them. $\endgroup$ – Lubin Sep 21 '18 at 21:49
  • $\begingroup$ @Jam: " who's posted so many low-effort questions": how do I know ? $\endgroup$ – Yves Daoust Sep 22 '18 at 8:16
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Hint: I'm guessing the Maclaurin polynomial of degree $7$ is the Taylor series at $0$ (up to degree $7$)... So, using Taylor's theorem: $f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$ we plug in the first seven derivatives of $f(x)$ at $0$.

The fundamental theorem of calculus (FTC) gives us the first: $f'(0)=\sin(5\cdot0^2)=\sin0=0$.

Now differentiate $6$ more times (and evaluate at $0$). Start with $f'(x)=\sin(5x^2)$...

To do the second part plug $x=0.68$ into the polynomial.

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$$F(x)=\int_0^x\sin(5t^2)dt=\int\limits_0^x\sum\limits_{k=0}^\infty\frac{(-1)^k5^{2k+1}}{(2k+1)!}t^{4k+2}dt$$$$ =\sum_{k=0}^\infty\frac{(-1)^k5^{2k+1}}{(4k+3)(2k+1)!}x^{4k+3}$$ and since it is up to polynomial of order $7$ we will only need the first two terms, so: $$F(x)\approx\frac{(-1)^05^1}{(3)*1!}x^3+\frac{(-1)^15^3}{(7)*3!}x^7$$$$ =\frac{5}{3}x^3+\frac{125}{42}x^7$$

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