Q. 35 - Calculus

Question

Find the Maclaurin series for $\sin \left(x^2\right)$, and use it to approximate ${\int }_0^3\sin \left(x^2\right)dx$ to within 0.001 of its value. How many terms would you need to approximate the integral to within ${10}^{-6}$ of its value? 

Step-by-Step Solution

Answer

The Maclaurin series can be given as
$$\begin{gathered} \sin \left(x^2\right)=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+.... \\ {\int }_0^3\sin \left(x^2\right)dx=9-\frac{3^7}{4!}+\frac{3^{11}}{5!·11}-\frac{3^{15}}{8!·15}+... \end{gathered}$$

First six terms are required to get a value 0.0001 of its original values

And First nine terms are required to get a value 0.000001 of its original values

Step 1: Given information

We are given Maclaurin series of sin(x)

Step 2: Find the Maclaurin series of sin ( x 2 )

The Maclaurin series of sin(x) can be given as

$$\begin{gathered} \sin \left(x\right)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+.... \\ Replace x by x^2 \\ \sin \left(x^2\right)=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+.... \end{gathered}$$

Which is the required Maclaurin series

Now integrating we get,

$$\begin{gathered} {\int }_0^3\sin \left(x^2\right)dx={\int }_0^3(x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+....)dx \\ {\int }_0^3\sin \left(x^2\right)dx={{[\frac{x^3}{3}-\frac{x^7}{42}+\frac{x^{11}}{5!·11}-\frac{x^{15}}{8!·15}+....]}^3}_0 \\ {\int }_0^3\sin \left(x^2\right)dx=9-\frac{3^7}{4!}+\frac{3^{11}}{5!·11}-\frac{3^{15}}{8!·15}+... \end{gathered}$$

First six terms are required to get a value 0.0001 of its original values

And First nine terms are required to get a value 0.000001 of its original values