Problem 53 - Calculus: One and Several Variables

Question

Estimate within 0.001 by series expansion and check your result by carrying out the integration directly. $$\int_{0}^{1 / 2} x \ln (1-x) d x$$

Step-by-Step Solution

Answer

The integral \(\int_{0}^{1/2} x \ln (1-x) dx\) can be estimated within 0.001 using series expansion as follows: 1. Expand \(\ln(1-x)\) using the power series \(\ln(1-x) = - \sum_{n=1}^{\infty} \frac{x^n}{n}\). 2. Substitute the expanded \(\ln(1-x)\) into the integral and integrate term by term. 3. Compute the partial sum until the error term is less than or equal to 0.001, denoting the required term as n=k. 4. Perform direct integration of the given expression and compare the results obtained from the partial sum to the direct integration result to ensure the estimation is within the desired accuracy of 0.001. By following these steps, we can estimate the integral \(\int_{0}^{1/2} x \ln (1-x) dx\) within the desired accuracy.

Step 1: Expand ln(1-x) using power series

Recall that the power series of \(\ln(1-x)\) is given by: \[\ln(1-x) = - \sum_{n=1}^{\infty} \frac{x^n}{n}\]

Step 2: Replace ln(1-x) in the integral and expand

Now, replacing the ln(1-x) in the integral, we get: \[\int_{0}^{1/2} x \ln(1-x) dx = -\int_{0}^{1/2} x \sum_{n=1}^{\infty} \frac{x^n}{n} dx\] We can now expand the sum: \[-\int_{0}^{1/2} x\left(\frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{3} + \cdots\right) dx\]

Step 3: Integrate term by term

Now we can integrate each term in the series: \[-\left(\int_{0}^{1/2} \frac{x^2}{1} dx + \int_{0}^{1/2} \frac{x^3}{2} dx + \int_{0}^{1/2} \frac{x^4}{3}dx + \cdots\right)\] The general term is \( -\int_{0}^{1/2} \frac{x^{n+1}}{n} dx\). Calculating this integral gives: \[-\frac{1}{n} \cdot \frac{x^{n+2}}{n+2}\Bigg|_0^{1/2} = -\frac{1}{n(n+2)} (2^{-(n+2)}-0)\] Since we need to estimate within 0.001, we consider a partial sum of this series starting from n=1, and compute the value until the error term is less than or equal to 0.001.

Step 4: Estimate partial sums and determine the required n for desired accuracy

Compute partial sums until the error term is less than or equal to 0.001: \[-\left(-\frac{1}{1 (1+2)} (2^{-3}) -\frac{1}{2(2+2)} (2^{-4}) -\frac{1}{3(3+2)} (2^{-5}) -\cdots\right) = S \] Keep calculating the terms and adding them until the absolute value of the last term is less than 0.001. Let's denote the required term as n= k.

Step 5: Perform the direct integration and compare results

Now, we directly integrate the given expression, \[\int_{0}^{1/2} x \ln(1-x) dx\] Use integration by parts, letting \(u = x\), \(dv = \ln(1-x) dx\) \(du = dx\), \(v = -\int \ln(1-x) dx = -(x\ln(1-x)+x-x\ln(x))\Bigg|_0^{\frac{1}{2}}\) We can now find the integral: \[\int u dv = uv - \int v du = xu\ln(1-x)\Bigg|_0^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} \left[ - x - x \ln(1-x) + x \ln(x) \right] dx\] Next, calculate the integral, which results in I. Finally, compare the result obtained from the partial sum S, calculated in Step 4, to the direct integration result, I, to ensure that the estimation is within the desired accuracy of 0.001.

Key Concepts

Power Series

A power series is an infinite sum of the form \(\sum_{n=0}^{\text{infinity}} a_n (x - c)^n\), where \( a_n \) are the coefficients of the series, \( x \) is the variable, and \( c \) is the center of the series. Power series can represent a wide range of functions, and they're especially handy for functions that might not have simpler algebraic expressions.

When applying power series to problems like estimating integrals, it's important to ensure the series converges within the interval of integration. For example, the series expansion of the natural logarithm \(\ln(1-x)\) converges for \( -1 < x < 1 \). One of the strengths of power series is that once you've verified convergence, you can manipulate the series much like a polynomial, making complex integrations more manageable.

Term-by-Term Integration

Term-by-term integration allows us to integrate a power series by integrating each term individually, as if it were a finite sum. We can do this under the condition that the series converges to the function within the region of integration.

For instance, integrating the power series for \(\ln(1-x)\) term by term within the interval [0,1/2] makes the problem simpler. We integrate each term \(\frac{x^n}{n}\) to get \(\frac{x^{n+1}}{n(n+1)}\), which can then be evaluated at the endpoints of the interval. Notably, this process assumes that the function represented by the power series is continuous and differentiable on the interval, which is the case for \(\ln(1-x)\) on [0,1/2].

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation and is given by the formula \(\int u(x)dv(x) = uv(x) - \int v(x)du(x)\). It is particularly useful when dealing with products of functions, where each function has an easier antiderivative.

In our case, to integrate \(x\ln(1-x)\), we choose \(u=x\) and \(dv=\ln(1-x)dx\) cleverly so that \(du\) is simpler (just \(dx\)) and \(v\) is an antiderivative of \(\ln(1-x)\). This allows us to break down the problem into more elementary parts, simplifying our computation while ensuring the accuracy of the integral, provided the bounds of integration align with the conditions for the integration by parts to be valid.

Convergence of Series

Convergence is a crucial concept when dealing with infinite series, as it tells us whether the series approaches a finite value as the number of terms increases. A series converges if the partial sums tend to a limit, which in the context of power series means that the function is well-defined.

For the proper application of series to problems in calculus, not only must we establish convergence, but we also need to determine the rate of convergence, which can impact the accuracy of our results. When estimating an integral to a certain accuracy, such as 0.001, we must find a partial sum where the remainder (or error) of the series beyond that partial sum is less than the desired accuracy. This often involves estimating or bounding the remainder term and is essential for ensuring that our term-by-term integration yields valid and precise results.