Mean Value Theorem Made Simple

Let us first discuss what is mean value theorem?The mean value theorem defines that for any given curve between two ending points, there should be a point at which the slope of the tangent to the curve is similar to the slope of the secant through its ending points.

If f(x) is a function, so that f(x) is continuous on the closed interval [p,q] and also differentiable on the open interval (p, q), then there is point r in (p, q) that is, p < r < q such that

f’(r) = f(q) -f(p)/ q-p

Lagrange’s Mean ValueTheorem or first mean value theorem is another name for the mean value theorem. This article discuss about Mean Value Theorem for Integrals, Mean ValueTheorem for Integrals problems and Cauchy Mean Value Theorem

Geometrical Representation of Mean Value Theorem 

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The above mean value theorem graph represents the graph of the function f(x).

• Let us consider the point A as (a,f(a)) and point B as = (b,f(b))]

• The point C in the graph where the tangent passes through the curve is (cf(c)).

• The slope of the tangent line is similar to the secant line i.e.both the tangent line and the secant line are parallel to each other.

What is The Mean Value Theorem for Integrals?

For defining what is mean value theorem for integrals, let us consider

 f (θ) be continuous on [p, q]. 

F(θ) = $${\int }_P^Q$$ f(z) dz

The Fundamental Theorem of Calculus indicates (θ) = f (θ). The Mean Value Theorem indicates  the inclusion of r ϵ (p,q) such that 

F(q)- F(p)/ q-p = F’(r)  or equivalently     F(q)-F(p) - F’(r)(q- p)

Which indicates

$${\int }_p^q$$  f(z)dz = f(r) (q-p)

This theorem is known as the First Mean Value Theorem for Integrals.The point f (r) is determined as  the average value of f (θ) on [p, q].

Along with the "First Mean Value Theorem for integrals", there is also a “Second Mean Value Theorem for Integrals”

Let us learn about the second mean value theorem for integrals.

The number f (r) is known as the g(p)-weighted average of f (θ) on the interval [p, q].

Let f (θ) and g(θ) be continuous on [p,q]. Assume that g(θ) is positive, i.e. g (θ) ≥ 0 for any θ Є [p, q]. Then there includes r Є(p, q) such that

$${\int }_p^q$$ f (z)g(z)dz = f(r) $${\int }_p^q$$ g(z)dz

The number f (r) is called the g(θ)-weighted average of f (θ) on the interval [p, q].

The application of the second mean value theorem may define the Center of Mass of one-dimensional non-homogeneous objects such as a metal rod. If the object is homogeneous and placed on the x-axis from x = p to x = q, then its center of mass will be the midpoint.

 p+q/2

If the object is not homogeneous with  λ (θ) being the density function, then the total mass M is represented as 

M = $${\int }_p^q$$  λ (θ) dθ

The density-weighted average θc is stated through 

$${\int }_p^q$$ θ  λ (θ)dθ = θc  $${\int }_p^q$$  θ  λ (θ)dθ = θc M,

Or equivalently

θc = 1/M $${\int }_p^q$$ θ  λ(θ) dθ  

The point θc is the center of mass of the object.

Cauchy Mean Value Theorem

To define cauchy mean value theorem , we will consider two functions f and k functions  represented on [p,q] such both are continuous in closed interval [p,q] and also both are differentiable on open interval (p,q) k'(x) ≠ 0 for any x ∈ (p,q) then there includes at least one point r ∈ (p,q) such that

f’(r)/k’(r) = f(q) -f(p)/ k(q) -k(p)

If we consider k(x) = x for every x ∈ {p,q} in Cauchy’s mean value theorem, we get

f(q) -f(p)/q-p = f'(r) which is considered as Langrange’s mean value theorem. This is also known as extended mean value theorem.

Mean Value Theorem for Integral Problems

Here, you can see a mean value theorem for integrals problems with solutions.

1. A rod of length Z is placed on the x-axis from x = 0 to x = Z. Suppose that the density (x) of the rod is proportional to the distance from the x = 0 endpoint of the rod. Let us find the total mass M and the center of mass xc of the rod. We have (x) = gx, for some constant g> 0. We have

M $${\int }_z^0$$ λ(x)dx = $${\int }_z^0$$ k x dx = k  Z²/2 =  KZ²/2  

XC 1/M $${\int }_z^0$$ λ(x)dx 1/M $${\int }_z^0$$kx² dx = 1/M k Z³/3 = 2/3Z

If the rod was homogeneous, then the center of mass would be placed at the middle point of the rod. Now it is closer to the ending point x = Z. This is not unexpected as there is more mass at this end.

Solved Example

1. For the function f(x) = ex , p= 0 and q = 1, then find the value of r in the mean value theorem.

Solution: f(q) –f(p)/q-p = f’r

= (eq )- (ep) / q-p = f’r

= e-1/1-0 = er

= c = log(e-1)

2. Assume that f(x) be continuous and increasing on (p,q). Compare,

f(p)(q-p) and $${\int }_p^q$$ f(x)dx

Solution-The First Mean Value Theorem for Integrals implies the existence of r Є (p,q) such that 

f(r)(q-p) = $${\int }_p^q$$ f (x) dx

As f(x) is increasing, then f(p) ≤ f(r).Hence f (p)(q - p) ≤ f(r)(q-p) which implies

f(p)(q-p) ≤ $${\int }_p^q$$ f (x) dx

Quiz Time 

1. The value of c for which mean value theorem f(x) for x in the interval (-1,1) is

1. ½

2. -½

3. 1

4. Non-existent in the interval.

2. Geometrically the mean value theorem assures that there is at least one point on the curve f(x) whose abscissa lies in (a,b) at which tangent is

1. Parallel to the x axis

2. Parallel to the u axis

3. Parallel to the line y = x

4. Parallel to the line joining the endpoints of the curve.

3. For the function f(θ) = [θ] ; θ Є [5,9], the mean value theorem 

1. Is applicable

2. ‘Not applicable as the function is continuous but not differentiable.

3. Not applicable because the function is differentiable but not continuous

4. Not applicable because the function is neither continuous nor differentiable

Facts

• The Mean Value Theorem was initially defined by famous Indian Mathematician and Astronomer Vatasseri Parameshvara Nambudiri. Later the theorem was proved by Augustin Louis Cauchy.

• Augustin Louis Cauchy proved the mean value theorem. The restricted form of this theorem was proved by Augustin Louis Cauchy in 1691. In 1823, Augustin Louis Cauchy stated and proved the modern form of mean value theorem.