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and the proof is complete.
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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
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
The average value is found by multiplying the area by $1\text{/}\left(4-0\right).$ Thus, the average value of the function is
Set the average value equal to $f\left(c\right)$ and solve for c .
At $c=2,f\left(2\right)=4.$
Find the average value of the function $f\left(x\right)=\frac{x}{2}$ over the interval $\left[0,6\right]$ and find c such that $f\left(c\right)$ equals the average value of the function over $[0,6].$
$\text{Average value}=1.5;c=3$
Given ${\int}_{0}^{3}{x}^{2}dx}=9,$ 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
Replacing $f\left(c\right)$ with c ^{2} , we have
Since $\text{\u2212}\sqrt{3}$ is outside the interval, take only the positive value. Thus, $c=\sqrt{3}$ ( [link] ).
Given ${\int}_{0}^{3}\left(2{x}^{2}-1\right)dx}=15,$ find c such that $f\left(c\right)$ equals the average value of $f(x)=2{x}^{2}-1$ over $[0,3].$
$c=\sqrt{3}$
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.
If $f\left(x\right)$ is continuous over an interval $\left[a,b\right],$ and the function $F\left(x\right)$ is defined by
then ${F}^{\prime}\text{(}x)=f\left(x\right)$ over $\left[a,b\right].$
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.
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