Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

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

$$A'(x)=f(x)$$

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).

share|improve this question
3  
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. –  Willie Wong Mar 5 '12 at 11:35
    
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 $y=A'(x)=f(x)$? –  Aesir Mar 5 '12 at 12:12
add comment

1 Answer

up vote 9 down vote accepted

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:

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

share|improve this answer
    
Thanks for this, I am reading / working through it now. –  Aesir Mar 6 '12 at 18:23
1  
Great answer, Peter. –  Gigili May 6 '12 at 11:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.