Solve the problem.
-Use series to estimate the integral's value to within an error of magnitude less than $10 ^ { - 3 }$.
$$\int _ { 0 } ^ { 0.5 } \ln \left( x ^ { 2 } + 1 \right) d x$$
A) 0.07649
B) 0.2054
C) 0.03225
D) 0.03887

Question

Quizplus · Accepted Answer

To estimate the integral $\int _ { 0 } ^ { 0.5 } \ln \left( x ^ { 2 } + 1 \right) d x$ to within an error of less than $10^{-3}$, we can use the Maclaurin series for $\ln(1+x)$, which is $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$, and adapt it for $\ln(x^2+1)$.Given $\ln(x^2+1)$, let $u = x^2$, so the series becomes $\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$, substituting $u = x^2$ gives $\ln(x^2+1) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \cdots$.To integrate term by term from 0 to 0.5:$$\int_{0}^{0.5} \ln(x^2+1) dx = \int_{0}^{0.5} \left(x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \cdots \right) dx$$$$= \left[\frac{x^3}{3} - \frac{x^5}{10} + \frac{x^7}{21} - \cdots \right]_0^{0.5}$$$$= \frac{(0.5)^3}{3} - \frac{(0.5)^5}{10} + \frac{(0.5)^7}{21} - \cdots$$$$= \frac{1}{24} - \frac{1}{320} + \frac{1}{2688} - \cdots$$Calculating the first few terms gives a sum close to 0.03887, which is within the desired error margin of $10^{-3}$. Therefore, the best estimate for the integral, to within the specified error, is 0.03887.

Solve the Problem 10−310 ^ { - 3 }10−3 ∫00.5ln⁡(x2+1)dx\int _ { 0 } ^ { 0.5 } \ln \left( x ^ { 2 } + 1 \right) d x∫00.5​ln(x2+1)dx

Solve the Problem $10 ^ { - 3 }$ $\int _ { 0 } ^ { 0.5 } \ln \left( x ^ { 2 } + 1 \right) d x$