Welcome to our community

Be a part of something great, join today!

Maybe add this to the Integral Calculus tutorial...

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Reading the Integral Calculus tutorial, I felt like contributing. Do with it what you wish...

Proposition: You can evaluate areas exactly using Integrals.

Knowing that an integral is an antiderivative, and that derivatives are RATES, it seems odd that going in reverse would give geometric measurements.

To prove this proposition, we need to do some analysis of functions in general.


First, we should note that a function can reach its global maxima and minima only at turning points and endpoints.

Theorem 1: The Extreme Value Theorem (given without proof due to how obvious it seems).

The Extreme Value Theorem states that for any function $\displaystyle f(x) $ defined over an interval $ \displaystyle x \in [\alpha, \beta]$ and continuous over $\displaystyle x \in (\alpha, \beta ) $, it must reach its maximum and minimum at some points in $\displaystyle x \in [\alpha, \beta]$


Theorem 2: Rolle's Theorem (also obvious, but the proof follows nicely from the Extreme Value Theorem).

Rolle's Theorem states that if a function is defined over $\displaystyle x \in [a,b] $ and is continuous over $\displaystyle x \in (a, b)$, if the function has the same value at two points, i.e. if $\displaystyle f(a) = f(b) $, then there must be a turning point in between them, since at some point between them the function has to turn to go back to that function value.

Therefore, if $\displaystyle f(a) = f(b) $, then there exists some $\displaystyle c \in [a, b] $ such that $\displaystyle f'(c) = 0 $.

Proof: If the function's endpoints are $\displaystyle a, b $ and we have $\displaystyle f(a) = f(b) $, then if $\displaystyle f(a) $ is a global minimum, so is $\displaystyle f(b) $, and if $\displaystyle f(a) $ is a global maximum, so is $\displaystyle f(b) $. But by the Extreme Value Theorem, the function must reach its global maximum and minimum at some points in the interval. If $\displaystyle f(a) = f(b) $ is the global maximum, then there must be a turning point as the global minimum. If $\displaystyle f(a) = f(b) $ is the global minimum, then there must be a turning point as the global maximum. If $\displaystyle f(a) = f(b) $ is neither the global maximum or minimum, then there must be two turning points as the global maximum and minimum. Q.E.D.


Theorem 3: The Mean Value Theorem

The Mean Value Theorem states that for any continuous and differentiable function over $\displaystyle x \in [a, b] $, the gradient of the chord between $\displaystyle a $ and $\displaystyle b $ is equal to the gradient of the tangent to the function at some point in between $\displaystyle a $ and $\displaystyle b $.

438px-Lagrange_mean_value_theorem.svg.png

In symbols: $ \displaystyle \frac{f(b) - f(a)}{b - a} = f'(c) $ for some $\displaystyle c \in [a, b] $

Proof (from Wikipedia):

Define $\displaystyle g(x) = f(x) − rx $, where $r$ is a constant. Since $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, the same is true for $g$. We now want to choose $r$ so that $g$ satisfies the conditions of Rolle's theorem. Namely

\[ \displaystyle \begin{align*} g(a) &= g(b) \\ f(a) - r\,a &= f(b) - r\,b \\ r(b - a) &= f(b) - f(a) \\ r &= \frac{f(b) - f(a)}{b - a} \end{align*} \]

By Rolle's theorem, since $g$ is continuous and $g(a) = g(b)$, there is some $c$ in $(a, b)$ for which $\displaystyle g'(c) = 0 $, and it follows from the equality $g(x) = f(x) − r\,x$ that

\[\begin{align*}\displaystyle g'(x) &= f'(x) - r \\ g'(c) &= f'(c) - r \\ 0 &= f'(c) - r \\ r &= f'(c) \\ \frac{f(b) - f(a)}{b - a} &= f'(c) \end{align*}\]

as required. Q.E.D.


Note, we can rearrange the equation from the Mean Value Theorem as $\displaystyle f(b) - f(a) = (b - a)f'(c) $. We will refer back to this later. Now to prove our original proposition...

If we want to evaluate the area underneath a function $\displaystyle f(x) $ between $\displaystyle x = a $ and $\displaystyle x = b $. Start by making $\displaystyle n $ subdivisions on your interval and marking the midpoint of each, so that $\displaystyle a = x_0 < m_1 < x_1 < m_2 < x_2 < \dots < m_n < x_n = b $.

area1.JPG

Then rectangles can be drawn of length $= x_i - x_{i-1}$ and width $= f(m_i)$, so that their area $= (x_i - x_{i-1})\,f(m_i)$.

Then the area bounded by the curve and the $x$ axis between $x = a$ and $x = b$ can be approximated as $ \displaystyle A \approx \sum_{i = 1}^n \,(x_i - x_{i-1})\,f(m_i) $.

By making more subdivisions (making $n \to \infty$), this sum converges upon the true area.

area2.JPGarea3.JPGarea4.JPG

So $ \displaystyle A = \lim_{n \to \infty} \sum_{i = 1}^n \,(x_i - x_{i-1})\,f(m_i).$

If we define $f(x)$ as the derivative of another function $F(x)$, the summand can be simplified using the Mean Value Theorem. In other words $\displaystyle F'(x) = f(x) \,\,\,\,\,\textrm{or}\,\,\,\,\, F(x) = \int{f(x)\,dx}$.

Remembering from the Mean Value Theorem that $(b-a)\,f'(c) = f(b) - f(a)$ for some $c \in (a,b)$, that means by making $n \to \infty$, the midpoint of each interval will become the only point in between each subinterval, and thus $\displaystyle (x_i - x_{i-1})\,f(m_i) = F(x_i) - F(x_{i-1}) $.

Substituting into the formula for the area gives

\[ \displaystyle \begin{align*}
A &=\lim_{n \to \infty}\sum_{i = 1}^n \,(x_i - x_{i-1})\,f(m_i)\\
&= \lim_{n \to \infty}\sum_{i = 1}^n\, \left[F(x_i) - F(x_{i-1})\right]\\
&= \lim_{n \to \infty} \left\{\, \left[F(x_1) - F(x_0)\right]+\left[F(x_2) - F(x_1)\right]+ \left[ F(x_3) - F(x_2) \right] + \dots + \left[ F(x_n) - F(x_{n-1}) \right] \right\} \\
&= \lim_{n \to \infty} \left[ F(x_n) - F(x_0) \right] \\
&= F(x_n) - F(x_0) \\
&= F(b) - F(a) \end{align*}\]


So, in order to exactly evaluate the area underneath a curve $\displaystyle f(x) $ between $\displaystyle x = a $ and $\displaystyle x = b $, one needs to evaluate an antiderivative of $\displaystyle f(x) $ at $\displaystyle x= b $, and then subtract the value of the same antiderivative at $\displaystyle x = a $. In other words, if $\displaystyle F(x) $ is an antiderivative of $\displaystyle f(x) $, then:

\[ \displaystyle \begin{align*} A &= \int_a^b{f(x)\,dx} \\ &= F(b) - F(a) \end{align*} \]
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Theorem 1: The Extreme Value Theorem (given without proof due to how obvious it seems).

The Extreme Value Theorem states that for any function $\displaystyle f(x) $ defined over an interval $ \displaystyle x \in [\alpha, \beta]$, it must reach its maximum and minimum at some points in $\displaystyle x \in [\alpha, \beta]$
... so that this function must have at least a maximun and a minimum in $\displaystyle x \in [-1,1]$?...

$\displaystyle f(x)= \begin{cases} \frac{1}{x}\ \text{if }\ -1\le x \le1\ , x \ne 0 \\ 0\ \text{if}\ x=0 \end{cases}$ (1)

Kind regards

$\chi$ $\sigma$
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
... so that this function must have at least a maximun and a minimum in $\displaystyle x \in [-1,1]$?...

$\displaystyle f(x)= \begin{cases} \frac{1}{x}\ \text{if }\ -1\le x \le1\ , x \ne 0 \\ 0\ \text{if}\ x=0 \end{cases}$ (1)

Kind regards

$\chi$ $\sigma$
Don't be picky :p I obviously meant defined and continuous on the entire interval.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
The integral calculus tutorial is a work in progress. The very next post, which is in draft form at the moment, covers much of this.

I'll tell you what: why don't you start on a Multi-variable Calculus tutorial? That way, we won't be duplicating effort.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
The integral calculus tutorial is a work in progress. The very next post, which is in draft form at the moment, covers much of this.

I'll tell you what: why don't you start on a Multi-variable Calculus tutorial? That way, we won't be duplicating effort.
Because multivariable calculus isn't really my strong suit...
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Because multivariable calculus isn't really my strong suit...
Well, there's nothing like teaching a subject for packing down your own knowledge of it. I've certainly found that to be the case.
 

chisigma

Well-known member
Feb 13, 2012
1,704
The integral calculus tutorial is a work in progress. The very next post, which is in draft form at the moment, covers much of this...
I read and appreciate very much the integral calculus tutorial You wrote in calculus section, especially for the fact that it is 'educationally' very effective. If I can do a suggestion it would be very useful to insert at the end a section dedicated to Riemann and Lebesgue integration, a 'duality' that has never been for me full clear...

Kind regards

$\chi$ $\sigma$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
I read and appreciate very much the integral calculus tutorial You wrote in calculus section, especially for the fact that it is 'educationally' very effective. If I can do a suggestion it would be very useful to insert at the end a section dedicated to Riemann and Lebesgue integration, a 'duality' that has never been for me full clear...

Kind regards

$\chi$ $\sigma$
I'll keep that in mind. I have an explanation or two that might help to make sense out of it all.