Find the average value of f(x)=e^3 x /(1+e^6 x) over the interval [-(ln3) / 6,0].

Name: Problem 30, Chapter 5: Calculus Early Transcendentals
Uploaded: 2023-03-17T12:02:47-08:00
Duration: 5 min 20 s
Channel: William Semus
Description: Find the average value of f(x)=e^3 x /(1+e^6 x) over the interval [-(ln3) / 6,0].

Find the average value of f(x)=e^3 x /(1+e^6 x) over the...

Key Concepts

Integration by Substitution

Integration by substitution is a method used to simplify the process of integrating complex functions by introducing a new variable. This technique transforms an integral into a more manageable form, making it easier to evaluate, especially when dealing with composite functions like those involving exponentials.

Exponential Functions

Exponential functions involve the variable appearing as an exponent, which results in rapid growth or decay. Their unique properties often require specific techniques when performing algebraic manipulations and integration, particularly when combined in complex expressions.

Average Value of a Function

This concept is used to determine a representative value of a function over a specific interval. It is calculated by dividing the definite integral of the function over the interval by the length of that interval, providing insight into the overall behavior of the function within that set range.

Definite Integrals

Definite integrals compute the net area under a curve between two bounds. They are central to applications in calculus, such as calculating areas, accumulated quantities, and average values of functions over a given interval.

Transcript

00:01 So in this question, we want to find the average value of f of x equals e to the 3x over the quantity of 1 plus e to the 6x over the interval from negative ln of 3 over 6 to 0.

00:13 So let's get this average value, and i'm using my formula 1 over b minus a, 1 over 0, minus negative the ln of 3 over 6 times the integral from a.

00:30 To b times the integral from negative ln of 3 over 6 to 0 of my function e to the 3x over 1 plus e to the 6x d x.

00:49 So how am i going to start this? well, i'm going to start by simplifying that thing in front a little bit.

00:55 I'm going to say this is 1 over the ln of 3 over 6 times the integral from negative ln of 3 over 6 to 0 of e to the 3x over 1 plus.

01:15 My trick is i am going to rewrite e to the 6x as e to the 3x quantity square.

01:26 Now i can simplify a little more in front.

01:28 I can say this is 6 over the ln of 3 times the integral from negative ln of 3 over 6 to 0 of e to the 3x over 1 plus e to the 3x quantity being squared dx.

01:49 Now i'm going to do a u substitution.

01:53 Namely, i'm going to let you equal e to the 3x.

02:00 My du is then 3e to the 3x dx or equivalently, du over 3 equals e to the 3x dx.

02:17 Now, i am doing a u substitution within a definite interval.

02:23 So why should change my limits of integration? the hard one is x.

02:28 So if x is the negative ln of 3 over 6, it's that bottom on it's hard.

02:36 If x is negative ln of 3 over 6, what's my u? well, my u is e to the power of 3 times negative ln of 3 over 6.

02:54 Or equivalently, e to the negative 1 1 1⁄2, l .n of 3, which by properties of logs is e to the power of ln of 3 of 3 to the negative 1 1⁄2, my u is 3 to the negative 1 half or 1 over root 3.

03:24 Now what if x is 0? that one's easier.

03:29 Well then my u is just e to the 0...

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