Problem 1
Question
Use the Midpoint Rule to estimate the value of the integral (obtain two digits of accuracy). $$\int_{0}^{3}\left(x^{3}+x\right) d x$$
Step-by-Step Solution
Verified Answer
The estimated value of the integral using the Midpoint Rule, given two digits of accuracy, is approximately 20.91.
1Step 1: Define the function and interval
Firstly, the function \(f(x) = x^{3}+x\) is given which needs to be integrated over the interval \([0,3]\). Let's assume that we only divide the integral once i.e., into two subintervals of equal width.
2Step 2: Calculate midpoint and width
The midpoint of each subinterval is calculated by averaging its endpoints. So, midpoints are \(m_1 = 0.75\) and \(m_2 = 2.25\). The width \(∆x\) of each subinterval is the difference of the upper limit and the lower limit divided by the number of subintervals, i.e., \(∆x = (3 - 0) / 2 = 1.5\).
3Step 3: Compute Values at Midpoints
Secondly, we calculate the function values at these midpoints: \(f(m_1) = (0.75)^3 + 0.75 = 0.421875\) and \(f(m_2) = (2.25)^3 + 2.25 = 13.421875\).
4Step 4: Apply Midpoint Rule
Finally, according to the Midpoint Rule, the integral can be approximated as \(∆x[f(m_1) + f(m_2)] = 1.5[ 0.421875 + 13.421875] = 20.90625 \approx 20.91\).
Key Concepts
Numerical IntegrationApproximation MethodsIntegral Calculus
Numerical Integration
Numerical integration is a method used to estimate the value of an integral. This is especially useful when we cannot solve an integral analytically, or when an approximate solution is sufficient. There are several different techniques applied under numerical integration, such as:
The key idea is to replace a complicated integral with a simpler approximation that's easier to calculate. By dividing the area under a curve into smaller, more manageable sections, these methods allow us to find a close approximation of the integral's value. This is crucial in many fields of science and engineering where precise calculations are often needed.
The Midpoint Rule, in particular, splits the interval into subintervals and takes the midpoints of these intervals to evaluate the function, providing a good balance between simplicity and accuracy.
- Trapezoidal Rule
- Simpson's Rule
- Midpoint Rule
- Monte Carlo Integration
The key idea is to replace a complicated integral with a simpler approximation that's easier to calculate. By dividing the area under a curve into smaller, more manageable sections, these methods allow us to find a close approximation of the integral's value. This is crucial in many fields of science and engineering where precise calculations are often needed.
The Midpoint Rule, in particular, splits the interval into subintervals and takes the midpoints of these intervals to evaluate the function, providing a good balance between simplicity and accuracy.
Approximation Methods
One of the core principles behind numerical integration is approximation. Since not every problem can be solved exactly, these methods provide a way to find how close we can get to the true solution.
The Midpoint Rule is an approximation method that simplifies the integration process:
Each approximation method has its own strengths and weaknesses, depending on the function and interval. The Midpoint Rule is particularly useful when the function is "well-behaved" (smooth and continuous) over the interval, offering a straightforward calculation.
The Midpoint Rule is an approximation method that simplifies the integration process:
- Divide the total area into a number of equal regions called subintervals.
- Identify the midpoint of each subinterval.
- Calculate the value of the function at each midpoint.
- Multiply the function's midpoint values by the width of the subintervals, then sum to get the approximate integral value.
Each approximation method has its own strengths and weaknesses, depending on the function and interval. The Midpoint Rule is particularly useful when the function is "well-behaved" (smooth and continuous) over the interval, offering a straightforward calculation.
Integral Calculus
Integral calculus focuses on the process of finding integrals, which are used to calculate quantities like areas, volumes, and more. It is one of the two main branches of calculus (the other being differential calculus).
The main goals of integral calculus are:
The definite integral, as used in the original exercise, requires evaluating the integral over a specific interval. In \(\int_{0}^{3}(x^{3} + x) \,dx\), the integral represents the area under the graph of \(x^{3} + x\) from \(x = 0\) to \(x = 3\). Integral calculus provides the foundational techniques necessary for solving such problems analytically and numerically.
The main goals of integral calculus are:
- To find the area under a curve, represented mathematically by an integral.
- To recover a function from its derivative, known as the "anti-derivative" or "indefinite integral."
- To solve differential equations that describe various physical phenomena.
The definite integral, as used in the original exercise, requires evaluating the integral over a specific interval. In \(\int_{0}^{3}(x^{3} + x) \,dx\), the integral represents the area under the graph of \(x^{3} + x\) from \(x = 0\) to \(x = 3\). Integral calculus provides the foundational techniques necessary for solving such problems analytically and numerically.
Other exercises in this chapter
Problem 1
Compute Midpoint, Trapezoidal and Simpson's Rule approximations by hand (leave your answer as a fraction) for \(n=4\). $$\int_{0}^{1}\left(x^{2}+1\right) d x$$
View solution Problem 1
Use Part I of the Fundamental Theorem to compute each integral exactly. $$\int_{0}^{2}(2 x-3) d x$$
View solution Problem 1
List the evaluation points corresponding to the midpoint of each sub interval, sketch the function and approximating rectangles and evaluate the Riemann sum. $$
View solution Problem 1
A calculation is described in words. Translate each into summation notation and then compute the sum. The sum of the squares of the first 50 positive integers.
View solution