What is the average value of y=x2 sqrtx3 1

For an explanation of why, read the digression above. The concepts in there really are fundamental to understanding a lot of graphing. Any horizontal translation will affect the domain and leave the range unchanged. Any vertical translation will affect the range and the leave the domain unchanged. Apply the same translation to the domain or range that you apply to the x-coordinates or the y-coordinates.

This works because the domain can be written in interval notation as the interval between two x-coordinates. Likewise for the range as the interval between two y-coordinates. In the following table, remember that domain and range are given in interval notation. If you're not familiar with interval notation, then please check the prerequisite chapter.

The first line is the definition statement and should be used to determine the rest of the answers. Notice on the last two that the order in the range has changed. This is because in interval notation, the smaller number always comes first. Understanding the translations can also help when finding the domain and range of a function.

Begin with what you know. You know this because you know those six common functions on the front cover of your text which are going to be used as building blocks for other functions. And the best part of it is that you understood it! Not only did you understand it, but you were able to do it without graphing it on the calculator. There is nothing wrong with making a graph to see what's going on, but you should be able to understand what's going on without the graph because we have learned that the graphing calculator doesn't always show exactly what's going on.

It is a tool to assist your understanding and comprehension, not a tool to replace it. It is this cohesiveness of math that I want all of you to "get". It all fits together so beautifully. Constant Function. Linear Function. Quadratic Function.

Cubic function. Absolute Value function. Once again, you shouldn't memorize this formula because it actually kind of falls out out of what it actually means. So the average of our function is going to be equal to the definite integral over this interval. So, essentially the area under this curve. So, it's going to be the definite intergral from zero to three of F of X which is X squared plus one DX.

Then we're going to take this area. We're going to take this area right over here and we're going to divide it by the width of our interval to essentially come up with the average height, or the average value of our function.

So, we're going to divide it by B minus A, or three minus zero, which is just going to be three. And so now we just have to evaluate this. So, this is going to be equal to one third times -- Let's see the antiderivative of X squared is X to the third over three.

Antiderivative of one is X, and we're going to evaluate it from zero to three. So, this is going to be equal to one third times when we evaluate it at three. Let me use another color here. When we evaluate it at three it's going to be three to the third divided by three. Well, that's just going to be 27 divided by three. That's nine plus three and then when we evaluated zero, minus zero minus zero.

So, it's just minus. When you evaluated zero it's just going to be zero. And so, we are left with -- I'm going to make the brackets that same color. This is going to be one third times One third times 12, which is equal to four. Which is equal to four. So this is the average value of our function. The average value of our function over this interval is equal to four. Notice, our function actually hits that value at some point in the interval.

At some point in the interval, something lower than two but greater than one. We can maybe call that C. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Asked 7 years, 6 months ago. Active 7 years, 6 months ago.

Viewed 8k times. Add a comment. Active Oldest Votes. I'm just missing some big concept here I think. It makes it easier to calculate, especially on the part of the AP test which forbids the use of a calculator.

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