Question

Show that the exact value of \(\int_{0}^{1} x e^{-x} d x=1-\frac{2}{e} .\) Find the absolute error if you approximate the integral using the midpoint rule with 16 subdivisions.

Short answer

Approximate with the midpoint rule and calculate the error as the absolute difference from \(1 - \frac{2}{e}\).

Step-by-step solution

Exact Value of the Integral

First, we need to solve the integral \(\int_{0}^{1} x e^{-x} \: dx\). We can use integration by parts where we let \( u = x \) and \( dv = e^{-x} \, dx \). Hence, \( du = dx \) and \( v = -e^{-x} \).Using the integration by parts formula:\[ \int u \, dv = uv - \int v \, du \]we get:\[ -xe^{-x} \bigg|_{0}^{1} + \int_{0}^{1} e^{-x} \, dx \]Evaluating, we have:\[ -1 \cdot e^{-1} - (0 \cdot e^0) + \left[ -e^{-x} \right]_{0}^{1} = -\frac{1}{e} + \left[ -e^{-1} + 1 \right] = 1 - \frac{2}{e} \]The exact value is \(1 - \frac{2}{e}\).

Midpoint Rule Approximation

To approximate the integral using the midpoint rule with 16 subdivisions, we calculate the width of each subinterval as \( \Delta x = \frac{1-0}{16} = \frac{1}{16} \).Determine the midpoints \(x_i\) \( (i = 0, 1, ..., 15) \):\[ x_i = (i + 0.5)\Delta x = \frac{i+0.5}{16} \]The midpoint approximation is given by:\[ M_n = \Delta x \sum_{i=0}^{15} \left( x_i e^{-x_i} \right) \]Calculate \( M_n \) by computing each midpoint's value in the function and then summing them up.

Calculate Absolute Error

The absolute error is the difference between the exact value and the approximation. If \( M_n \) is the midpoint rule approximation:\[ \text{Absolute Error} = \left| 1 - \frac{2}{e} - M_n \right| \]Compute this value once the midpoint approximation \( M_n \) is calculated.

Key concepts

Midpoint Rule

The midpoint rule is a numerical method used to approximate the value of definite integrals. It is particularly useful when finding an exact integral is difficult or impossible.
This rule approximates the area under a curve by dividing it into smaller rectangles and summing their areas. Each rectangle's height is determined by the value of the function at the midpoint of the base.
Here's how the midpoint rule works:

• Divide the interval \[a, b\] into \(n\) equal parts, creating subintervals each with width \(\Delta x = \frac{b-a}{n}\).
• Determine the midpoints \(x_i\) of each subinterval: \[ x_i = a + \left(i + \frac{1}{2}\right) \Delta x \]
• Approximate the integral by computing \[ M_n = \Delta x \sum_{i=0}^{n-1} f(x_i) \]

In practice, this method provides a good balance between computational simplicity and precision, especially with a higher number of subdivisions.

Absolute Error

Absolute error is a way to measure the accuracy of an approximation. It quantifies the difference between the approximate value and the exact value of an expression. This measurement is crucial when using methods like the midpoint rule to approximate integrals.
When computing absolute error, the formula used is:

• \[ \text{Absolute Error} = \left| \text{Exact Value} - \text{Approximate Value} \right| \]

This provides a non-negative number representing the error size.
If the absolute error is small, it indicates that the approximation is close to the exact value, verifying the approximation's reliability. Frequent use of the midpoint rule or any other numerical technique will include calculating the absolute error to ensure that the approximations are satisfactorily close to the truth.

Definite Integral

A definite integral is used to calculate the area under a curve within a specified interval \[a, b\]. Instead of finding the area for the indefinite integral (an antiderivative with a general form), the definite integral provides a specific numeric result.

• The definite integral of a function \(f(x)\) from \(a\) to \(b\) is represented as \(\int_{a}^{b} f(x) \, dx\). This provides the net area, considering negative values under the x-axis as subtracting from the area.

The method to solve a definite integral involves finding the antiderivative \(F(x)\) and evaluating at the upper and lower limits:

• \[ F(b) - F(a) \]

Understanding definite integrals is key to various fields, such as physics for calculating distances over time when velocity changes. It also forms the foundational concept for advanced calculus operations like integration by parts.