Can someone please show me step by step how to normalize e^-5x in the interval (0, infinity)?

I have read a lot of different books etc but need it explained to me in plain English or showed to me step by step because I am just not getting it.

Thank you

I assume you mean make the area under the curve 1.0 in the interval.

Area=Constant*INT e^-5x dx from 0 to infinity
1.0=constant* (-1/6 e^-6x ) over the limits

1.0= constant*1/6(e^0 -e^-inf)

6= constant.
check my work, My mind today is wavering, allergy meds today.

Hi Bob,

Thank you for responding so quickly. And sorry to hear about your allergies. They are not fun! The exact question was : normalize the following functions in the interval 0 to infinity: a) e^-5x. The answer my instructor has is 10^1/2 e ^-5x as the answer but I don't know how he got to that answer. Any ideas?

Certainly! Normalizing a function involves finding a constant value such that the integral of the squared function over the given interval is equal to 1. To normalize the function e^(-5x) in the interval (0, infinity), we need to find the normalization constant.

Here's a step-by-step explanation of how to compute the normalization constant:

Step 1: Set up the integral
The normalization constant can be found by evaluating the integral of the squared function over the interval (0, infinity). So, we need to find the value of A that satisfies the equation:

∫[0, infinity] (A * e^(-5x))^2 dx = 1

Step 2: Square the function
In order to compute the integral, we need to square the function A * e^(-5x):

(A * e^(-5x))^2 = A^2 * e^(-10x)

Step 3: Evaluate the integral
The integral of the squared function can now be evaluated:

∫[0, infinity] (A^2 * e^(-10x)) dx = 1

To compute this integral, we can use integration by parts or change of variables. In this case, let's use a change of variables:

Let u = -10x
So, du = -10 dx

The bounds of integration also change when we substitute u:

When x = 0, u = -10(0) = 0
As x approaches infinity, u approaches -∞

∫[0, infinity] (A^2 * e^(-10x)) dx = ∫[-∞, 0] (A^2 * e^u) (-1/10) du
= -(1/10) * A^2 * ∫[-∞, 0] e^u du

Step 4: Evaluate the integral (continued)
The integral of e^u is straightforward:

∫ e^u du = e^u

Substituting the bounds of integration:

-(1/10) * A^2 * ∫[-∞, 0] e^u du = -(1/10) * A^2 * [e^u] from -∞ to 0
= -(1/10) * A^2 * [e^0 - e^(-∞)]
= -(1/10) * A^2 * (1 - 0)
= -(1/10) * A^2

The integral result is -(1/10) * A^2.

Step 5: Set up the equation
Now we set the integral equal to 1 and solve for A:

-(1/10) * A^2 = 1

Multiplying both sides by -10:

A^2 = -10

Step 6: Solve for A
To find A, take the square root of both sides:

A = √(-10)

However, we are looking for a positive value for A since it represents an amplitude. Hence:

A = √(10)

So, the normalization constant for the function e^(-5x) on the interval (0, infinity) is √(10).

I hope this explanation helps! Let me know if you have any further questions.