Question

9. Evaluate the following integral with Gauss quadrature formula: \[ I=\int_{0}^{\infty} e^{-x} d x \]

Answer 1

To evaluate the integral using the Gauss quadrature formula, we first need to express the integral as a definite integral over a finite interval. We can do this by making a substitution:
`\sf u = e^(-x)`. The limits of integration will also change accordingly.

When
`\sf x = 0`,
`\sf u = e^(-0) = 1`.

When
`\sf x = \infty`,
`\sf u = e^(-\infty) = 0`.

So the integral can be rewritten as:

`\sf I = \int_(0)^(\infty) e^(-x) dx = \int_(1)^(0) -(du)/(u)`

Now, we can apply the Gauss quadrature formula, which states that for the integral of a function
`\sf f(x)` over an interval
`\sf [a, b]`, we can approximate it using the weighted sum:

`\sf I \approx \sum_(i=1)^(n) w_i f(x_i)`

where
`\sf w_i` are the weights and
`\sf x_i` are the nodes.

For our specific integral, we have
`\sf f(u) = -(1)/(u)`. We can use the Gauss-Laguerre quadrature formula, which is specifically designed for integrating functions of the form
`\sf f(u) = e^(-u) g(u)`.

Using the Gauss-Laguerre weights and nodes, we have:

`\sf I \approx (1)/(2) \left( f(x_1) + f(x_2) \right)`

where
`\sf x_1 = 0.5858` and
`\sf x_2 = 3.4142`.

Plugging in the function values and evaluating the expression, we get:

`\sf I \approx (1)/(2) \left( -(1)/(x_1) - (1)/(x_2) \right) \approx (1)/(2) \left( -(1)/(0.5858) - (1)/(3.4142) \right) \approx 0.5`

Therefore, the approximate value of the integral using the Gauss quadrature formula is
`\sf I \approx 0.5`.

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