Problem 161
Question
\(\int_{0}^{1} \frac{\sin \pi x}{1+x^{2}} d x\) lies between (A) \(\frac{1}{\pi}\) and \(\frac{2}{\pi}\) (B) \(\frac{\pi}{2}\) and \(\pi\) (C) \(\pi\) and \(\frac{3 \pi}{2}\) (D) None of these
Step-by-Step Solution
Verified Answer
The integral lies between \( \frac{1}{\pi} \) and \( \frac{2}{\pi} \) (option A).
1Step 1: Understand the Integral
We are asked to find the integral \( \int_{0}^{1} \frac{\sin \pi x}{1+x^{2}} \, dx \), which is a definite integral with bounds 0 and 1. The function \( \frac{\sin \pi x}{1+x^{2}} \) is continuous on the interval \([0, 1]\), so the integral is defined.
2Step 2: Evaluate the Function at Specific Points
Evaluate the function \( f(x) = \frac{\sin \pi x}{1+x^{2}} \) at specific points on the interval \([0, 1]\). For example:- At \(x = 0\), \(f(0) = 0\).- At \(x = 0.5\), \(f(0.5) = \frac{\sin \frac{\pi}{2}}{1+\frac{1}{4}} = \frac{1}{1.25} = 0.8\).- At \(x = 1\), \(f(1) = \frac{\sin \pi}{2} = 0\).These evaluations help us understand the behavior of the function within the integration limits.
3Step 3: Estimate the Integral Using Numerical Methods
Since an analytic solution for the integral is complex, approximate the integral using numerical methods such as Simpson's Rule or Trapezoidal Rule. For example, using Simpson's Rule gives us an approximation.Calculate the approximation:- Divide the interval \([0, 1]\) into small subintervals.- Apply Simpson's rule formula: \( \frac{b-a}{6}(f(a) + 4f((a+b)/2) + f(b)) \).- For \(b = 1\) and \(a = 0\), this simplifies to \( \frac{1}{6}(f(0) + 4f(0.5) + f(1)) \approx \frac{1}{6}(0 + 4(0.8) + 0) = \frac{3.2}{6} = 0.53\approx \frac{1}{2}\).
4Step 4: Compare Integral Approximation with Given Ranges
Compare the numerical approximation of \( \approx 0.53 \) with the given answer options.- Option (A): \( \frac{1}{\pi} \approx 0.318 \) and \( \frac{2}{\pi} \approx 0.636 \)- Option (B): \( \frac{\pi}{2} \approx 1.57 \) and \( \pi \approx 3.14 \)- Option (C): \( \pi \approx 3.14 \) and \( \frac{3\pi}{2} \approx 4.71 \)From these comparisons, we see that \( 0.53 \) falls within the range of option (A).
5Step 5: Conclude the Correct Option
Given that the approximation \( 0.53 \) lies within the interval specified by option (A) (\( \frac{1}{\pi} \approx 0.318 \) to \( \frac{2}{\pi} \approx 0.636 \)), we can conclude that this is the correct range the integral fits into.
Key Concepts
Numerical MethodsSimpson's RuleTrapezoidal Rule
Numerical Methods
Numerical methods are techniques used to approximately solve mathematical problems when analytic solutions are difficult or impossible to obtain. These methods are essential tools in fields like engineering and applied mathematics.
When dealing with definite integrals, especially those that do not have simple antiderivatives, numerical methods provide a means of estimating the area under curves within specific bounds. Such methods are instrumental when we want accurate solutions without heavy symbolic computation.
Three main benefits of numerical methods include:
When dealing with definite integrals, especially those that do not have simple antiderivatives, numerical methods provide a means of estimating the area under curves within specific bounds. Such methods are instrumental when we want accurate solutions without heavy symbolic computation.
Three main benefits of numerical methods include:
- Flexibility: Applied to various functions regardless of complexity.
- Precision: With finer intervals, they yield highly accurate results.
- Ease of Use: Implementations using algorithms are straightforward in modern computing tools.
Simpson's Rule
Simpson's Rule is a popular numerical method for approximating definite integrals. It is named after Thomas Simpson, who formalized the formula in the 19th century.
The rule provides a robust method for approximation by using quadratic polynomials. These polynomials come from evaluating the integrand function at an even number of intervals on the given domain.
Here’s how Simpson’s Rule works:
The rule provides a robust method for approximation by using quadratic polynomials. These polynomials come from evaluating the integrand function at an even number of intervals on the given domain.
Here’s how Simpson’s Rule works:
- Divide the interval \([a, b]\) into an even number of subintervals of equal width \(h\).
- Calculate function values at the endpoints and midpoints of each subinterval.
- Apply the formula: \[ \frac{b-a}{6}(f(a) + 4f((a+b)/2) + f(b)) \]This weighted sum assigns higher weight to the midpoint, reflecting greater polynomial influence.
Trapezoidal Rule
The Trapezoidal Rule is another fundamental numerical approach for estimating definite integrals. It approximates the region under the curve by segmenting it into trapezoids rather than rectangles.
This method excels due to its simplicity and straightforward implementation, that's why it is often one of the first numerical methods taught in calculus. Here's how it works:
This method excels due to its simplicity and straightforward implementation, that's why it is often one of the first numerical methods taught in calculus. Here's how it works:
- Divide the domain \([a, b]\) of the integral into \(n\) subintervals, which are all equal in width \(h = \frac{b-a}{n}\).
- Calculate the function values at each endpoint and sum them with double weight for all intermediate points.
- Apply the formula: \[ T = \frac{h}{2} \left(f(a) + 2\sum_{i=1}^{n-1} f(x_i) + f(b)\right) \]This formula averages the start and end trapezoid heights, creating a balance based on the curve.
Other exercises in this chapter
Problem 159
If \(f(x)= \begin{cases}\frac{1}{2^{n}}, & \text { when } \frac{1}{2^{n+1}}
View solution Problem 160
If \(f(x)= \begin{cases}\frac{1}{n}, & \frac{1}{n+1}
View solution Problem 162
\(\int_{\pi / 6}^{\pi / 2} \frac{x}{\sin x} d x\) lies between (A) \(\frac{\pi^{2}}{3}\) and \(\frac{2 \pi^{2}}{3}\) (B) \(\frac{\pi^{2}}{9}\) and \(\frac{2 \pi
View solution Problem 163
\(\int_{0}^{1} \frac{x^{2}}{\sqrt{1+x^{2}}}\) lies between (A) \(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{3}}\) (B) \(\frac{1}{\sqrt{2}}\) and 1 (C) \(\frac{1}
View solution