Evaluate ∫0^0.4 x ln(1+x) d x using Maclaurin's theorem,...

Evaluate $\int_{0}^{0.4} x \ln (1+x) \mathrm{d} x$ using Maclaurin's theorem, correct to 3 decimal places. From Problem 6 , $$ \ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots $$ Hence $\int_{0}^{0.4} x \ln (1+x) \mathrm{d} x$ $$ \begin{aligned} &=\int_{0}^{0.4} x\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots\right) \mathrm{d} x \\ &=\int_{0}^{0.4}\left(x^{2}-\frac{x^{3}}{2}+\frac{x^{4}}{3}-\frac{x^{5}}{4}+\frac{x^{6}}{5}-\cdots\right) \mathrm{d} x \end{aligned} $$ $$ \begin{aligned} =&\left[\frac{x^{3}}{3}-\frac{x^{4}}{8}+\frac{x^{5}}{15}-\frac{x^{6}}{24}+\frac{x^{7}}{35}-\cdots\right]_{0}^{0.4} \\ =&\left(\frac{(0.4)^{3}}{3}-\frac{(0.4)^{4}}{8}+\frac{(0.4)^{5}}{15}-\frac{(0.4)^{6}}{24}\right.\\ &\left.\quad+\frac{(0.4)^{7}}{35}-\cdots\right)-(0) \end{aligned} $$ $$ =0.02133-0.0032+0.0006827-\cdots $$ $=\mathbf{0 . 0 1 9}$, correct to 3 decimal places.

Evaluate $\int_{0}^{0.4} x \ln (1+x) \mathrm{d} x$ using Maclaurin's theorem, correct to 3 decimal places.
From Problem 6 ,
$$
\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots
$$
Hence $\int_{0}^{0.4} x \ln (1+x) \mathrm{d} x$
$$
\begin{aligned}
&=\int_{0}^{0.4} x\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots\right) \mathrm{d} x \\
&=\int_{0}^{0.4}\left(x^{2}-\frac{x^{3}}{2}+\frac{x^{4}}{3}-\frac{x^{5}}{4}+\frac{x^{6}}{5}-\cdots\right) \mathrm{d} x
\end{aligned}
$$
$$
\begin{aligned}
=&\left[\frac{x^{3}}{3}-\frac{x^{4}}{8}+\frac{x^{5}}{15}-\frac{x^{6}}{24}+\frac{x^{7}}{35}-\cdots\right]_{0}^{0.4} \\
=&\left(\frac{(0.4)^{3}}{3}-\frac{(0.4)^{4}}{8}+\frac{(0.4)^{5}}{15}-\frac{(0.4)^{6}}{24}\right.\\
&\left.\quad+\frac{(0.4)^{7}}{35}-\cdots\right)-(0)
\end{aligned}
$$
$$
=0.02133-0.0032+0.0006827-\cdots
$$
$=\mathbf{0 . 0 1 9}$, correct to 3 decimal places.

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Key Concepts

Maclaurin Series

A Maclaurin series is a special case of a Taylor series expanded about zero. It expresses a function as an infinite sum of terms calculated from the derivatives of the function evaluated at zero. This series is particularly useful for approximating functions near the origin and for simplifying complex functions into an easily manipulable polynomial form.

Power Series Integration

Integrating a function represented as a power series can be performed term-by-term within the series' radius of convergence. This method transforms the problem of integrating a potentially difficult function into the simpler task of integrating polynomials, thus facilitating numerical approximations and analytic solutions where direct integration may be challenging.

Series Approximation and Error Control

When using a truncated power series to approximate a function, it is important to consider the error introduced by neglecting higher-order terms. By retaining enough terms in the series expansion, one can achieve the desired accuracy, such as three decimal places, while ensuring that the approximation error remains within acceptable limits.

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