# 9.3average Valueap Calculus

Posted By admin On 29/12/21The calculator will find the average value of the function on the given interval, with steps shown.

- 9 3 Average Valueap Calculus 14th Edition
- 9 3 Average Valueap Calculus Calculator
- 9 3 Average Valueap Calculus Equation
- 9 3 Average Valueap Calculus Formula

Eta-Mu modeled multipath propagation of electromagnetic waves. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Let's say that we have the function F of X is equal to X squared plus one and what we want to do is we want to figure out the average value of our function F on the interval, on the closed interval between zero and let's say between zero and three. Adjusted PQI value (AP) and linear regression difference (LD) were calculated as follows: AP = Y2 + P2 LD = AP C1 Where, APAdjusted PQI value LDlinear regression calibration difference The results are attached in the appendix, and Figure 2.14 shows the TD and LD value.

- In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`.
- In general, you can skip parentheses, but be very careful: e^3x is `e^3x`, and e^(3x) is `e^(3x)`.
- Also, be careful when you write fractions: 1/x^2 ln(x) is `1/x^2 ln(x)`, and 1/(x^2 ln(x)) is `1/(x^2 ln(x))`.
- If you skip parentheses or a multiplication sign, type at least a whitespace, i.e. write sin x (or even better sin(x)) instead of sinx.
- Sometimes I see expressions like tan^2xsec^3x: this will be parsed as `tan^(2*3)(x sec(x))`. To get `tan^2(x)sec^3(x)`, use parentheses: tan^2(x)sec^3(x).
- Similarly, tanxsec^3x will be parsed as `tan(xsec^3(x))`. To get `tan(x)sec^3(x)`, use parentheses: tan(x)sec^3(x).
- From the table below, you can notice that sech is not supported, but you can still enter it using the identity `sech(x)=1/cosh(x)`.
- If you get an error, double-check your expression, add parentheses and multiplication signs where needed, and consult the table below.
- All suggestions and improvements are welcome. Please leave them in comments.

Type | Get |

Constants | |

e | e |

pi | `pi` |

i | i (imaginary unit) |

Operations | |

a+b | a+b |

a-b | a-b |

a*b | `a*b` |

a^b, a**b | `a^b` |

sqrt(x), x^(1/2) | `sqrt(x)` |

cbrt(x), x^(1/3) | `root(3)(x)` |

root(x,n), x^(1/n) | `root(n)(x)` |

x^(a/b) | `x^(a/b)` |

x^a^b | `x^(a^b)` |

abs(x) | ` x ` |

Functions | |

e^x | `e^x` |

ln(x), log(x) | ln(x) |

ln(x)/ln(a) | `log_a(x)` |

Trigonometric Functions | |

sin(x) | sin(x) |

cos(x) | cos(x) |

tan(x) | tan(x), tg(x) |

cot(x) | cot(x), ctg(x) |

sec(x) | sec(x) |

csc(x) | csc(x), cosec(x) |

Inverse Trigonometric Functions | |

asin(x), arcsin(x), sin^-1(x) | asin(x) |

acos(x), arccos(x), cos^-1(x) | acos(x) |

atan(x), arctan(x), tan^-1(x) | atan(x) |

acot(x), arccot(x), cot^-1(x) | acot(x) |

asec(x), arcsec(x), sec^-1(x) | asec(x) |

acsc(x), arccsc(x), csc^-1(x) | acsc(x) |

Hyperbolic Functions | |

sinh(x) | sinh(x) |

cosh(x) | cosh(x) |

tanh(x) | tanh(x) |

coth(x) | coth(x) |

1/cosh(x) | sech(x) |

1/sinh(x) | csch(x) |

Inverse Hyperbolic Functions | |

asinh(x), arcsinh(x), sinh^-1(x) | asinh(x) |

acosh(x), arccosh(x), cosh^-1(x) | acosh(x) |

atanh(x), arctanh(x), tanh^-1(x) | atanh(x) |

acoth(x), arccoth(x), cot^-1(x) | acoth(x) |

acosh(1/x) | asech(x) |

asinh(1/x) | acsch(x) |

The average of a set of data is typically defined as the sum of the values divided by the number of data points. But what if you have infinitely many data points? What is the *average value* of a function? Read on to find out!

Defining the average value of a continuous function is not as easy as finding the average of a finite set of data.

## Average Value of Functions

Suppose *f* is a continuous function defined over an interval [*a*, *b*]. In particular *f*(*x*) exists at every one of the infinitely-many points *x* between and including *a* and *b*. So, if you’re looking for the average value of *f* on that interval, it won’t do any good to try adding up those infinitely-many data points.

Instead, the way to tame the infinity is to use calculus. Specifically, we define the average value of a function *f* as the following definite integral.

### The Theory Behind the Formula

But where does the integral formula for average come from?

The key is **sampling**. If you included enough of the function values, say a thousand, a million, or even more, then that should approximate the average of all infinitely-many points!

Let’s illustrate with the following example.

Estimate the average value of the function *f*(*x*) = √(*x*) + 1 over the interval [1, 3].

Since we’re just estimating, let’s pick four sample points. (The more sample points you pick, the better your estimated average will be.)

Divide the interval [1, 3] into four equal subintervals, and let’s agree to choose the midpoint of each subinterval. Then plug those midpoints into *f* to find the sample values.

Midpoints | Height: f(x) = √(x) + 1 |
---|---|

1.25 | √(1.25) + 1 = 2.12 |

1.75 | √(1.75) + 1 = 2.32 |

2.25 | √(2.25) + 1 = 2.50 |

2.75 | √(2.75) + 1 = 2.66 |

Finally, use the familiar old averaging formula. Add up the data and divide by the number of data points:

(2.12 + 2.32 + 2.50 + 2.66)/4 = 2.4

So the (approximate) average of the function is 2.4.

### Taking it to the Limit

However, what we’ve just done will not give us an exact answer because we’ve essentially ignored most of the function! What about *f*(1.3) or *f*(2.95234)? No matter how many sample points we include, there will always be some missing… Unless we can use the magic of calculus to catch them all.

The sampling process should remind you of a **Riemann Sum**. For a quick reminder, feel free to check out AP Calculus Review: Riemann Sums.

Midpoint Riemann sum. In the limit as the number of sample points goes to ∞, the Riemann sum becomes a definite integral.

Let *n* be any whole number and *x _{k}*

^{*}stand for the various sample

*x*-values. Then the estimated average is the sum:

Next, allow *n* → ∞ using a limit. We also need to get Δ*x* into the act somehow. The trick is to multiply and divide by (*b* – *a*). Remember, Δ*x* = (*b* – *a*)/*n*.

## Examples

Ok, now that you’ve seen the theory, let’s use the formula in practice!

### Problem 1

Find the exact average value of *f*(*x*) = √(*x*) + 1 over the interval [1, 3].

#### Solution

Above, we only estimated the average to be 2.4. Now we’ll use the integral formula to determine the average value precisely.

(It’s interesting to compare our estimate with the exact value above. √(3) + 2/3 ≈ 2.39871747, which means that our estimate of 2.4 was actually pretty good!)

## 9 3 Average Valueap Calculus 14th Edition

### Problem 2

The amount of energy associated with a certain chemical reaction is given by *E* = *x* ln *x*, where 1 ≤ *x* ≤ *e*, and *x* represents the amount of one of the reactants. Find the average energy of the reaction over the range of possible levels of reactant.

#### Solution

This problem seems more like chemistry than math!

Chemistry can be fun too! But what does this have to do with calculus??

However, the keyword *average* tells us that mathematics plays a major role in this problem. In fact, they are simply asking for the *average value* of *f*(*x*) = *x* ln *x*, over the interval [1, *e*].

First set up the integral formula with *a* = 1 and *b* = *e*. Then work out the integration, which involves Integration By Parts in this case.

Thus the average energy of the reaction is (*e*^{2} + 1)/[ 4(*e* – 1) ], or roughly 1.22.

## Mean Value Theorem for Integrals

Averages are also called **means**. So you may use the same formula to find the **mean value** of a function.

There is also an important result in calculus that relates the mean value to a particular function value on the given interval.

**The Mean Value Theorem for Integrals (MVTI).** If *f* is continuous on a closed interval [*a*, *b*], then there is at least one point *x* = *c* in that interval such that the mean value of the function is equal to *f*(*c*). That is,

(* Caution:* There is also a Mean Value Theorem for Derivatives. It’s important not to confuse the two.)

### Problem 3

Let *f*(*x*) = 6*x*^{2} – 8*x* + 1. Determine the value of *c* at which the mean value of *f* on [-1, 1] is the same as *f*(*c*).

#### Solution

According to the Mean Value Theorem for Integrals, there must be at least one such value *c*. Let’s set up the formula and find it!

At this point, we will need to solve a quadratic equation. Don’t forget your Quadratic Formula!

6*c*^{2} – 8*c* – 2 = 0

*c* = (2 ± √(7))/3

It seems as though there may be two answers. However, only one lies within the given interval [-1, 1].

*c* = (2 + √(7))/3 ≈ 1.549, not in the interval.

*c* = (2 – √(7))/3 ≈ -0.215, in the interval.

Therefore, the only value that satisfies the MVTI is *c* = (2 – √(7))/3.

## 9 3 Average Valueap Calculus Calculator

## Summary

Although average value and the Mean Value Theorem for Integrals are specialized topics and only show up in a few problems on any given AP Calculus test, they are important concepts to master. For one thing, they illustrate how integral calculus can be used in applications.

Moreover, working out the average value of a function is no more difficult than computing a definite integral. So now when you see these kinds of problems on the AP Calculus Exam, you can rise to the challenge!

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##### About Shaun Ault

## 9 3 Average Valueap Calculus Equation

Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!

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## 9 3 Average Valueap Calculus Formula

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