Mean value theorem for integrals
What is the Mean Value Theorem for integrals?
The mean value theorem for integrals tells us that, for a continuous function ???f(x)???, there’s at least one point ???c??? inside the interval ???[a,b]??? at which the value of the function will be equal to the average value of the function over that interval.
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This means we can equate the average value of the function over the interval to the value of the function at the single point.
In other words,
???\frac{1}{b-a}\int^b_a{f(x)}\ dx=f(c)???
The equation above sets the average value of the function over the interval ???[a,b]??? (on the left), equal to the value of the function at the point ???c??? (on the right). If we multiply both sides by ???(b-a)???, we get the mean value theorem for integrals:
???\int^b_a{f(x)}\ dx=f(c)(b-a)???
Solving integration problems using the Mean Value Theorem for integrals
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Finding the point that satisfies the Mean Value Theorem in a particular interval
Example
Find the point ???c??? that satisfies the mean value theorem for integrals on the interval ???[1,4]???.
???f(x)=3x^2-2x???
Looking at the equation we can see that it is a polynomial and is therefore continuous. This means that we can go ahead and use the mean value theorem for integrals.
???\int^b_a{f(x)}\ dx=f(c)(b-a)???
???\int^4_1{3x^2-2x}\ dx=\left(3c^2-2c\right)(4-1)???
???\int^4_1{3x^2-2x}\ dx=9c^2-6c???
Now we can break up the integral to make it easier to solve.
???\int^4_13x^2\ dx+\int^4_1-2x\ dx=9c^2-6c???
???3\int^4_1x^2\ dx-2\int^4_1x\ dx=9c^2-6c???
This means we can equate the average value of the function over the interval to the value of the function at the single point.
Integrate.
???\left[3\left(\frac{x^3}{3}\right)-2\left(\frac{x^2}{2}\right)\right]\bigg|^4_1=9c^2-6c???
???\left(x^3-x^2\right)\Big|^4_1=9c^2-6c???
Now we can evaluate over the interval.
???\left[(4)^3-(4)^2\right]-\left[(1)^3-(1)^2\right]=9c^2-6c???
???48=9c^2-6c???
???0=9c^2-6c-48???
Now we need to solve for ???c???.
???0=3\left(3c^2-2c-16\right)???
???0=3c^2-2c-16???
???0=(3c-8)(c+2)???
Setting each of the factors equal to ???0??? individually to solve for ???c???, we get
???3c-8=0???
???c=\frac83???
and
???c+2=0???
???c=-2???
Only one of these values, ???c=8/3???, falls in the interval ???[1,4]???, which means it’s the only solution.
It’s possible to find more than one valid answer for ???c???, but in the last example, there’s only one point, ???c=8/3???, at which the value of the function is equal to the average value of the function over the interval.
