5.3 The fundamental theorem of calculus  (Page 2/11)

and the proof is complete.

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Finding the average value of a function

Find the average value of the function $f\left(x\right)=8-2x$ over the interval $\left[0,4\right]$ and find c such that $f\left(c\right)$ equals the average value of the function over $[0,4].$

The formula states the mean value of $f\left(x\right)$ is given by

$\frac{1}{4-0}{\displaystyle {\int }_0^4\left(8-2x\right)dx}.$

We can see in [link] that the function represents a straight line and forms a right triangle bounded by the x - and y -axes. The area of the triangle is $A=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right).$ We have

$A=\frac{1}{2}\left(4\right)\left(8\right)=16.$

The average value is found by multiplying the area by $1\text{/}\left(4-0\right).$ Thus, the average value of the function is

$\frac{1}{4}\left(16\right)=4.$

Set the average value equal to $f\left(c\right)$ and solve for c .

$\begin{array}{ccc}8-2c\hfill & =\hfill & 4\hfill \\ \hfill c& =\hfill & 2\hfill \end{array}$

At $c=2,f\left(2\right)=4.$

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Finding the point where a function takes on its average value

Given ${\displaystyle {\int }_0^3{x}^{2}dx}=9,$ find c such that $f\left(c\right)$ equals the average value of $f(x)=x^2$ over $[0,3].$

We are looking for the value of c such that

$f\left(c\right)=\frac{1}{3-0}{\displaystyle {\int }_0^3{x}^{2}dx}=\frac{1}{3}\left(9\right)=3.$

Replacing $f\left(c\right)$ with c ² , we have

$\begin{array}{ccc}{c}^2\hfill & =\hfill & 3\hfill \\ c\hfill & =\hfill & \text{±}\sqrt{3}.\hfill \end{array}$

Since $\text{−}\sqrt{3}$ is outside the interval, take only the positive value. Thus, $c=\sqrt{3}$ ( [link] ).

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Fundamental theorem of calculus part 1: integrals and antiderivatives

As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1 , is stated here. Part 1 establishes the relationship between differentiation and integration.

Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, $F\left(x\right),$ as the definite integral of another function, $f\left(t\right),$ from the point a to the point x . At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it’s a function. The key here is to notice that for any particular value of x , the definite integral is a number. So the function $F\left(x\right)$ returns a number (the value of the definite integral) for each value of x .

Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the Fundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative.