Question

Use Maclaurin series to evaluate: ${\int }_0^{0.1}\frac{dx}{\sqrt{1+x^3}}$

Short answer

0.0999875

Step-by-step solution

Write the function in the integrand

The given function in the integrand is $\frac{1}{\sqrt{1+x^3}}$.

Expand using Maclaurin series

Recall the Maclaurin series expansion for $(1+u{)}^{-1/2}$, which is $1-\frac{u}{2}+\frac{3u^2}{8}-\frac{5u^3}{16}+\dots$. Here, $u=x^3$, so substituting $u=x^3$ we get: $$\frac{1}{\sqrt{1+x^3}}=1-\frac{x^3}{2}+\frac{3x^6}{8}-\frac{5x^9}{16}+\dots$$

Integrate term by term

Now integrate each term in the series: $${\int }_0^{0.1}1dx-{\int }_0^{0.1}\frac{x^3}{2}dx+{\int }_0^{0.1}\frac{3x^6}{8}dx-{\int }_0^{0.1}\frac{5x^9}{16}dx+\dots$$

Evaluate the integrals

Compute each integral separately: $$-{\int }_0^{0.1}1dx=x{|}_0^{0.1}=(0.1)-(0)=0.1$$ $${\int }_0^{0.1}\frac{x^3}{2}dx=\frac{x^4}{8}{|}_0^{0.1}=\frac{(0.1{)}^4}{8}-\frac{(0{)}^4}{8}=\frac{0.0001}{8}=0.0000125$$ $${\int }_0^{0.1}\frac{3x^6}{8}dx=\frac{3x^7}{56}{|}_0^{0.1}=\frac{3(0.1{)}^7}{56}-\frac{3(0{)}^7}{56}=\frac{0.0000001}{56}\approx 1.7857142857\times {10}^{-9}$$ $${\int }_0^{0.1}\frac{5x^9}{16}dx=\frac{5x^{10}}{160}{|}_0^{0.1}=\frac{5(0.1{)}^{10}}{160}-\frac{5(0{)}^{1}0}{160}=\frac{0.0000000001}{160}\approx 3.125\times {10}^{-13}$$

Sum the evaluated integrals

Combine the evaluated integrals: $$0.1-0.0000125+1.7857142857\times {10}^{-9}-3.125\times {10}^{-13}+\cdots \approx 0.0999875$$

Key concepts

Integration Techniques

When we talk about integration techniques, we are referring to various methods to solve integrals, especially those that are not straightforward. For example, standard techniques include substitution, integration by parts, partial fraction decomposition, and special series like the Maclaurin series.
The Maclaurin series is a powerful tool for integration, especially when dealing with complex or non-standard integrands. It expresses a function as an infinite sum of its derivatives at a single point, usually x = 0.
This is useful for approximating the value of integrals that are difficult or impossible to evaluate analytically.
In our problem, integrating $\text{Could not find of for root}$ directly would be challenging, but using the Maclaurin series simplifies it significantly.

Series Expansion

Series expansion is the process of expressing a function as the sum of a sequence of terms. The Maclaurin series is a specific case where the expansion is around x = 0. This method involves using the derivatives of the function at this point to approximate the function. For example, the Maclaurin series for $1/(1+u{)}^{1/2}$ is \1 - u/2 + 3u^2/8 - 5u^3/16 + \dots\ . By substituting \ u = x^3 \, we can rewrite \ \frac{1}{\root1+x^3} \ as an infinite series.

Using this series expansion, we get:

• Integration becomes easier as we deal with simple power series.
• The approximation can be as accurate as desired by including more terms in the series.

It's a versatile technique, especially helpful for approximating solutions to integrals and differential equations.

Approximate Integration

Approximate integration is essential when exact integration is not feasible. Here, we use the Maclaurin series to break down a complex integrand into simpler terms, which can then be integrated individually.
After expanding $\text{Could not find of for root}$\ into a series, we integrate each term from 0 to 0.1. This step-by-step integration helps in getting an approximate value for the integral.
For instance:

• The integral of 1 from 0 to 0.1 is 0.1.
• The integral of \ (x^3/2) \ from 0 to 0.1 is \ (0.1^4/8) = 0.0000125 \ .
• The integral of higher-order terms quickly becomes negligible, making it easier to sum the contributions and get the final value: 0.0999875.
  Approximate integration is useful and practical. It gives good results and is easier than finding the exact integral, especially when used with series expansions like the Maclaurin series.