I am trying to understand the relationship between differentiation and integration. Differentiation has been introduced to me by this diagram:

enter image description here

Which displays that the derivative of a point $x$ on a continuous function $f(x)$ is the gradient of a line which is a tangent to that particular point, as shown in the diagram.

This can be written as

$$\lim_{h \to 0} \left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)=f'\left(x\right)$$

and this gives the gradient of the point $(x, f(x))$.

Next examine this diagram:

enter image description here

The area under the continuous function from $f(x)$ from $a$ to $x$ is $A(x)$, shaded pink. The area of $A(a)$ is $0$ and the area from $a$ to $b$ is $A(b)$. Following this convention the area of $x$ to $x+h$ is


This section has with $h$ and the area is close to that of a rectangle so it can be said that

$$A\left(x+h\right)-A\left(x\right)\approx hf(x)$$

If you divide this through by $h$ you get an equation which shows that the derivative of $f(x)$ is $A'(x)$.

$$\lim_{h \to 0} \left(\frac{A\left(x+h\right)-A\left(x\right)}{h}\right)=f\left(x\right)$$

and hence


This is where I have a problem, earlier it was stated that the derivative is the gradient of a line which is a tangent to a point on a continuous function. Here however, we have an area, which encloses many points, which are not a line which is a tangent to a continuous function. So I do not see how you can find the derivative of an area because it is not a point that you can find the gradient of a line which is a tangent to it.?

(Images taken from course texts provided by the Open University, Chapters C1 & C2 of the course MST121).

  • 3
    $\begingroup$ Try plotting the curve $y = A(x)$. The "area so far" for every given $x$ is a number $A(x)$. This function has a graph. $A'(x)$ is precisely the slope of the tangent to this graph. $\endgroup$ – Willie Wong Mar 5 '12 at 11:35
  • $\begingroup$ Sorry, I don't fully understand this. If $A(x)$ is the area under the function, then how can I plot $y=A(x)$? As $y=A'(x)=f(x)$? $\endgroup$ – Aesir Mar 5 '12 at 12:12

Important: the derivative has a geometrical interpretation as you state, but that doesn't mean it is the definition of the derivative.

Try and see it this way: when we integrate (in Riemann's way), we are making a "continuous sum" of a continuous function $f(x)$ with respect to the infinitesimal quantity $dx$ over an interval $(a,x)$ - we are taking the "product" of $f$ and $dx$ over the infinitely many real values in $(a,x)$ and "summing" them up, getting a new function $F(x)$.

That is, suppose we define

$$F(x) = \int\limits_a^x f(t) dt$$

Then we can put this geometrically in the following diagram:

enter image description here

We're interested in finding $F'(x)$, so we construct our difference:

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{F\left( {x + \Delta x} \right) - F\left( x \right)}}{{\Delta x}}$$

We can write that expression as

$$\frac{1}{{\Delta x}}\left( {\int\limits_a^{x + \Delta x} {f\left( t \right)dt} - \int\limits_a^x {f\left( t \right)dt} } \right)$$

But using some theorems from definite integration we have

$$\frac{1}{{\Delta x}}\left( {\int\limits_x^{x + \Delta x} {f\left( t \right)dt} + \int\limits_a^x {f\left( t \right)dt} - \int\limits_a^x {f\left( t \right)dt} } \right) = \frac{1}{{\Delta x}}\int\limits_x^{x + \Delta x} {f\left( t \right)dt} $$

We also now that if $f(x)$ is continuous (which we assumed is), we have

$$\frac{1}{{\Delta x}}\int\limits_x^{x + \Delta x} {f\left( t \right)dt} = \frac{1}{{\Delta x}}f\left( \xi \right)\Delta x = f\left( \xi \right)$$

for some $\xi$ in $[x,x+\Delta x]$

Now taking the limit produces

$$F'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{F\left( {x + \Delta x} \right) - F\left( x \right)}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} f\left( \xi \right)$$

But we have that $\xi \in \left[ {x,x + \Delta x} \right]$, which means that

$$\mathop {\lim }\limits_{\Delta x \to 0} f\left( \xi \right) = f\left( x \right)$$

What have we done? We have retrieved the original function $f$ which we summed along with $dx$ over $(a,x)$ to obtain $F(x)$. So what does this tells us? That differentiation in the "operational" sense, reverts the process of integration, just like multiplication "reverts" the process of division.

I'm not a tacher or tutor or anything of the sort, so maybe you can get better answers from such people, but I hope you understand what I intended to explain.

Another imporant consequence of this is the following:

We have proven that if we have a function $f(x)$ and defined $F(x)$ as

$$F\left( x \right) = \int\limits_a^x {f\left( t \right)dt} $$

then $F'(x) = f(x)$, that is, $F$ is a primitive of $f$. But we know that two primitives of $f$ will only differ by a constant term, so, let $G$ be another primitive. We have that

$$\int\limits_a^x {f\left( t \right)dt} - G\left( x \right) = C$$

But then putting $x=a$ gives

$$ - G\left( a \right) = C$$

We get a new result. If $G$ is a primitive of $f$, then the following holds:

$$\int\limits_a^x {f\left( t \right)dt} = G\left( x \right) - G\left( a \right)$$

  • $\begingroup$ Thanks for this, I am reading / working through it now. $\endgroup$ – Aesir Mar 6 '12 at 18:23
  • 2
    $\begingroup$ Great answer, Peter. $\endgroup$ – Gigili May 6 '12 at 11:52

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