Math Problem Statement

Approximate ∫0to6 1/1+x^2 𝑑π‘₯ using (i). Trapezoidal rule

Solution

To approximate the integral ∫0611+x2 dx\int_{0}^{6} \frac{1}{1+x^2} \, dx using the Trapezoidal rule, we will follow these steps:

  1. Divide the interval [0,6][0, 6] into nn subintervals of equal width Ξ”x\Delta x.
  2. Apply the Trapezoidal rule formula to approximate the integral.

The Trapezoidal rule formula is given by: ∫abf(x) dxβ‰ˆΞ”x2[f(x0)+2βˆ‘i=1nβˆ’1f(xi)+f(xn)]\int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{2} \left[ f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right]

Here, a=0a = 0, b=6b = 6, and Ξ”x=bβˆ’an=6βˆ’0n=6n\Delta x = \frac{b-a}{n} = \frac{6-0}{n} = \frac{6}{n}.

Let's choose n=6n = 6 for simplicity, so Ξ”x=1\Delta x = 1.

The points are x0=0x_0 = 0, x1=1x_1 = 1, x2=2x_2 = 2, x3=3x_3 = 3, x4=4x_4 = 4, x5=5x_5 = 5, and x6=6x_6 = 6.

The function f(x)=11+x2f(x) = \frac{1}{1+x^2} evaluated at these points:

  • f(x0)=f(0)=11+02=1f(x_0) = f(0) = \frac{1}{1+0^2} = 1
  • f(x1)=f(1)=11+12=12f(x_1) = f(1) = \frac{1}{1+1^2} = \frac{1}{2}
  • f(x2)=f(2)=11+22=15f(x_2) = f(2) = \frac{1}{1+2^2} = \frac{1}{5}
  • f(x3)=f(3)=11+32=110f(x_3) = f(3) = \frac{1}{1+3^2} = \frac{1}{10}
  • f(x4)=f(4)=11+42=117f(x_4) = f(4) = \frac{1}{1+4^2} = \frac{1}{17}
  • f(x5)=f(5)=11+52=126f(x_5) = f(5) = \frac{1}{1+5^2} = \frac{1}{26}
  • f(x6)=f(6)=11+62=137f(x_6) = f(6) = \frac{1}{1+6^2} = \frac{1}{37}

Now, applying the Trapezoidal rule formula: ∫0611+x2 dxβ‰ˆ12[1+2(12+15+110+117+126)+137]\int_{0}^{6} \frac{1}{1+x^2} \, dx \approx \frac{1}{2} \left[ 1 + 2 \left( \frac{1}{2} + \frac{1}{5} + \frac{1}{10} + \frac{1}{17} + \frac{1}{26} \right) + \frac{1}{37} \right]

Calculating inside the brackets: 2(12+15+110+117+126)=2(0.5+0.2+0.1+0.0588+0.0385)=2(0.8973)=1.79462 \left( \frac{1}{2} + \frac{1}{5} + \frac{1}{10} + \frac{1}{17} + \frac{1}{26} \right) = 2 \left( 0.5 + 0.2 + 0.1 + 0.0588 + 0.0385 \right) = 2 \left( 0.8973 \right) = 1.7946

So the approximation becomes: [ \frac{1}{2} \left[ 1 + 1.7946 + \frac{1}{37} \right] = \frac{1}{2} \left[ 1 + 1.7946 + 0.0270 \right] = \frac{1}{2} \left[ 2.8216 \right] = 1.4108 ]

Thus, the approximation of the integral ∫0611+x2 dx\int_{0}^{6} \frac{1}{1+x^2} \, dx using the Trapezoidal rule with n=6n=6 is approximately 1.41081.4108.

Would you like more details on any step or have any questions?

Here are five related questions to explore further:

  1. How does the accuracy of the Trapezoidal rule compare to other numerical integration methods like Simpson's rule?
  2. What happens to the approximation as the number of subintervals nn increases?
  3. Can you derive the exact value of ∫0611+x2 dx\int_{0}^{6} \frac{1}{1+x^2} \, dx and compare it to the approximation?
  4. How would you implement the Trapezoidal rule for this integral in a programming language like Python?
  5. What are the error bounds for the Trapezoidal rule, and how can they be estimated for this integral?

Tip: Increasing the number of subintervals nn generally improves the accuracy of the Trapezoidal rule approximation.

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Math Problem Analysis

Mathematical Concepts

Numerical Integration
Trapezoidal Rule
Definite Integrals

Formulas

Trapezoidal rule formula: ∫a to b f(x) dx β‰ˆ Ξ”x/2 [f(x0) + 2Ξ£f(xi) + f(xn)]
f(x) = 1/(1+x^2)

Theorems

Trapezoidal Rule

Suitable Grade Level

Undergraduate or Advanced High School (Grades 11-12)