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The answer is "numerically". In general it is impossible to find a closed form (elementary) anti-derivative but one certainly can do a numerical integration- that's how the values in a "Gaussian distribution" table are calculated.alba_ei said:how did gauss integrate this function
to obtain the values of the normal distribution?
arildno said:By a sequence of clever tricks:
1. Let us consider the integral
[tex]I=\int_{-\infty}^{\infty}e^{-x^{2}}dx[/tex]
Now, since a dummy variable's name is irrelevant, we may write:
[tex]I^{2}=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)^{2}=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)(\int_{-\infty}^{\infty}e^{-x^{2}}dx)=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)(\int_{-\infty}^{\infty}e^{-y^{2}}dy)[/tex]
2. By Fubini's theorem, we have:
[tex]\int_{-\infty}^{\infty}e^{-x^{2}}dx\int_{-\infty}^{\infty}e^{-y^{2}}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}}e^{-y^{2}}dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy[/tex]
3. Switching to polar coordinates, [itex]x=r\cos\theta,y=r\sin\theta,0\leq{r}\leq\infty,0\leq\theta\leq{2\pi}[/itex], we get:
[tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy=\int_{0}^{\infty}\int_{0}^{2\pi}re^{-r^{2}}d\theta{d}r[/tex]
which is easily integrated to the value [itex]\pi[/tex]
4. Hence, we have [itex]I^{2}=\pi\to{I}=\sqrt{\pi}[/itex]
An integral is a mathematical concept that represents the area under a curve on a graph. It is used to solve problems involving rates of change, such as finding the distance traveled by an object with a given velocity.
Integration is an important tool in mathematics and science because it allows us to find the exact value of a quantity, rather than just approximating it. It is also used to solve a wide range of real-world problems, such as calculating volumes and areas.
A definite integral has specific limits of integration, meaning that it calculates the area under a curve between two given points. An indefinite integral does not have limits of integration and is used to find a general solution to a specific problem.
Integration and differentiation are inverse operations. This means that integration is the reverse process of differentiation. Integration allows us to find the original function when given the derivative, while differentiation allows us to find the rate of change of a function.
Some common techniques for solving integrals include substitution, integration by parts, trigonometric substitution, and partial fractions. These techniques are used to simplify the integrand and make it easier to evaluate the integral.