Integral of e^(-ax) from evaluated between neg/pos infinity

Thread starter kkowalko
Start date Apr 27, 2010
Tags

Infinity Integral

In summary, the formula for evaluating the integral of e^(-ax) between negative and positive infinity is ∫e^(-ax)dx = (-1/a)e^(-ax) + C, where C is the constant of integration. To solve this integral, you can use this formula and plug in the limits of integration. The constant "a" in the integral affects the shape and scale of the exponential function, and can also be evaluated using other methods such as integration by parts or substitution. Additionally, the integral of e^(-ax) is closely related to the Laplace transform, as the Laplace transform of e^(-ax) is 1/(s+a) and the inverse Laplace transform can be used to solve

Apr 27, 2010

kkowalko

Hey guys I am working on a Physics problem and can't figure out a little but of math involved. What is this integral? I tried it out and got "1/a + a", assuming "1/a" is the positive half of the integral.

Physics news on Phys.org

Apr 27, 2010

Gigasoft

The integral is infinite.

Apr 27, 2010

zachzach

Graph [tex] f(x) = e^{-x} [/tex] and see if you can tell what the area under it is.

What is the formula for evaluating the integral of e^(-ax) between negative and positive infinity?

The formula for evaluating the integral of e^(-ax) between negative and positive infinity is ∫e^(-ax)dx = (-1/a)e^(-ax) + C, where C is the constant of integration.

How do you solve the integral of e^(-ax) from negative to positive infinity?

To solve the integral of e^(-ax) from negative to positive infinity, you can use the formula ∫e^(-ax)dx = (-1/a)e^(-ax) + C. Plug in the limits of integration (negative and positive infinity) to get the final solution.

What is the significance of the constant "a" in the integral of e^(-ax) from negative to positive infinity?

The constant "a" in the integral of e^(-ax) from negative to positive infinity affects the shape and scale of the exponential function. It determines the rate at which the function decays towards zero as x approaches infinity.

Can the integral of e^(-ax) from negative to positive infinity be evaluated using any other methods?

Yes, the integral of e^(-ax) from negative to positive infinity can also be evaluated using integration by parts or substitution. However, the formula ∫e^(-ax)dx = (-1/a)e^(-ax) + C is the most straightforward and efficient method for solving this integral.

What is the relationship between the integral of e^(-ax) from negative to positive infinity and the Laplace transform?

The integral of e^(-ax) from negative to positive infinity is closely related to the Laplace transform. In fact, the Laplace transform of e^(-ax) is equal to 1/(s+a), where s is the Laplace variable. This means that the integral of e^(-ax) from negative to positive infinity can be solved using the inverse Laplace transform.