Question

For h=0.5,. 25,.125 solve integration from 0 to 1, dx/1+x2 and improve the accuracy by romberg integration. Compare your value with the real value.

Answer 1

Answer on Question #74242 – Math – Quantitative Methods

Question

For $h=0.5, 25, 125$ solve integration from $\int_0^1\frac{1}{1 + x^2} dx$ and improve the accuracy by Romberg integration. Compare your value with the real value.

Solution

Evaluate an integral $\int_0^1\frac{1}{1 + x^2} dx$ and improve the accuracy by Romberg integration. We begin by using the Trapezoidal Rule, or, equivalently, the Composite Trapezoidal Rule

$$\int_{a}^{b} f(x) dx \approx \frac{h}{2} \left[ f_{0} + 2 \sum_{i=1}^{N-1} f_{i} + f_{N} \right], \quad h = \frac{b - a}{N}, \quad x_{i} = x_{0} + ih, \quad x_{0} = a, \quad x_{N} = b$$

We get

$$N = 1, \quad h = 1, \quad I_{T} = \frac{h}{2} \left[ f_{0} + f_{1} \right] = 0.750000.$$ $$N = 2, \quad h = \frac{1}{2}, \quad I_{T} = \frac{h}{2} \left[ f_{0} + 2 f_{1} + f_{2} \right] = 0.775000$$ $$N = 4, \quad h = \frac{1}{4}, \quad I_{T} = \frac{h}{2} \left[ f_{0} + 2 f_{1} + 2 f_{2} + 2 f_{3} + f_{4} \right] = 0.782794$$ $$N = 8, \quad h = \frac{1}{8}, \quad I_{T} = \frac{h}{2} \left[ f_{0} + 2 f_{1} + 2 f_{2} + 2 f_{3} + 2 f_{4} + 2 f_{5} + 2 f_{6} + 2 f_{7} + f_{8} \right] = 0.784752$$ $$N = 8, \quad h = \frac{1}{8},$$ $$I_{T} = \frac{h}{2} \left[ \begin{array}{l} f_{0} + 2 f_{1} + 2 f_{2} + 2 f_{3} + 2 f_{4} + 2 f_{5} + 2 f_{6} + 2 f_{7} + 2 f_{8} + 2 f_{9} + 2 f_{10} \\ + 2 f_{11} + 2 f_{12} + 2 f_{13} + 2 f_{14} + 2 f_{15} + f_{16} \end{array} \right] = 0.785452$$

Exact solution of the given integral is as follows:

$$\int_{0}^{1} \frac{1}{1 + x^{2}} dx = \left[ \tan^{-1} x \right]_{0}^{1} = \tan^{-1} 1 - \tan^{-1} 0 = 0.785398 \text{ in radians}.$$

Compare. Approximate value is 0.785452 when $h=1/16$ and the real value is 0.785398.

Using trapezoidal rule with Romberg integration to achieve the accuracy of $10^{-6}$

$$R_{1} = \frac{h^{2}}{12} f^{-}(\xi), \quad 0 < \xi < 1$$

Since $f(x) = \frac{1}{1 + x^2}$ for $0 \leq x \leq 1$, for achieving accuracy of $10^{-6}$, we require $h = 0.007$. Hence at least $(1-0)/0.007 = 144$ function evaluations to achieve this accuracy if trapezoidal rule is used directly.

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