Applied Mathematics for Electrical Engineering (3130908)

Question-4a-OR

BE | Semester-3 Winter-2019 | 26-11-2019

Q4) (a)

Evaluate $\int_{0}^{1} \exp (- x^{2}) d x$ by Gauss integration formula with $n = 3$ .

Three Point Gaussian Quadrature Formula,

\int_{- 1}^{1} f (x) d x = \frac{8}{9} f (0) + \frac{5}{9} [f (- \sqrt{\frac{3}{5}}) + f (\sqrt{\frac{3}{5}})]

First we transform the interval [0,1] to [-1,1] by using,

x = \frac{b + a}{2} + \frac{b - a}{2} \cdot u

Here, a=0,b=1.

x = (\frac{1 + 0}{2}) + (\frac{1 - 0}{2}) \cdot u

x = \frac{1}{2} + \frac{u}{2}

\Rightarrow dx = \frac{du}{2}

Therefore,

\int_{0}^{1} e^{- x^{2}} d x = \int_{- 1}^{1} e^{- {(\frac{u + 1}{2})}^{2}} \frac{d u}{2}

\int_{0}^{1} e^{- x^{2}} d x = \frac{1}{2} \int_{- 1}^{1} e^{- {(\frac{u + 1}{2})}^{2}} d u

\int_{0}^{1} e^{- x^{2}} d x = \frac{1}{2} [\frac{8}{9} e^{- {(\frac{0 + 1}{2})}^{2}} + \frac{5}{9} (e^{- {(\frac{- \sqrt{\frac{3}{5}} + 1}{2})}^{2}} + e^{- {(\frac{\sqrt{\frac{3}{5}} + 1}{2})}^{2}})]

\int_{0}^{1} e^{- x^{2}} d x = \frac{1}{2} [\frac{8}{9} (0.7788) + \frac{5}{9} (0.9874 + 2.1975)]

\int_{0}^{1} e^{- x^{2}} d x = 1.2308

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