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I have just finished a course on calculus. So I started pondering on the fundamental theorem of calculus and the relation between integration and differentiation . We all know that integration is just the inverse of differentiation. When I look at the graph of a function and look at what the derivative means and the integral between the interval [a,b] means they seem to be logically way too apart from each other. But based on calculations we just say that integration is just the inverse of differentiation. So is this fact only true when we look at calculations or is there even some geometric relation between the two processes ?

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  • $\begingroup$ I think this answer will clear up your concerns. $\endgroup$ – Andrey Kaipov Apr 19 '15 at 8:27
  • $\begingroup$ @AndreyKaipov fine i get it but how is it related to get the tangent line in a given diagram of a curve $\endgroup$ – user210387 Apr 19 '15 at 8:29
  • $\begingroup$ For a graphical intuition you can see at my answer to math.stackexchange.com/questions/1189262/… $\endgroup$ – Emilio Novati Apr 19 '15 at 8:59
  • $\begingroup$ @EmilioNovati what about indefinite integrals we dont have any intervals there.... $\endgroup$ – user210387 Apr 19 '15 at 9:25
  • $\begingroup$ In the figure the area start from $0$ and can terminate at any value $x$. But for an indefinite integral you can choose an other starting point $a$, and the difference between the two areas is a constant (the area between $0$ and $a$) and this represents the fact that there are many primitives functions that differ for a constant. $\endgroup$ – Emilio Novati Apr 19 '15 at 10:00
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The inverse of differentiation is actually: $$\int \limits_0^x \frac{\mathrm{d}f(t)}{\mathrm{d}t} \mathrm{d}t = f(x)$$

This means that as long as $\frac{\mathrm{d}f(t)}{\mathrm{d}t}$ is positive, the integral keeps getting bigger, so the function is rising, when $\frac{\mathrm{d}f(t)}{\mathrm{d}t}$ becomes negative the integral starts becoming smaller so the function is descending.

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  • $\begingroup$ What does this mean geometrically@Rugen Heidbuchel $\endgroup$ – user210387 Apr 19 '15 at 8:30
  • $\begingroup$ @Rememberme The integral is the surface area under the function. Here you start at 0, and take the surface area under the differential until x. When we advance from zero (to let's say right) to x, the integral rises when the differential is positive. The definition of the differential says that f rises when the differential is positive. You can see that this is exactly the same behavior, so the formulation above makes sense. $\endgroup$ – Rugen Heidbuchel Apr 19 '15 at 8:39
  • $\begingroup$ Okay lets take an example........i have got the tangent line of a curve between some interval,now what does the inverse process on the tangent line mean or what is the stuff that i have to do to the tangent line in order to find the integral @Rugen Heidbuchel $\endgroup$ – user210387 Apr 19 '15 at 8:42
  • $\begingroup$ @Rememberme To find the integral you would have to look at the tangent for every point between 0 and the x you're looking at. The slope of the tangent is the differential's value. When you divide the interval [0,x] into subintervals you would have to take the slope of the tangent of the midpoint of every subinterval (any other point in the interval would also work). You would then multiply that value by the length of the subinterval. When you do this for infinitely small subintervals you get the integral, which would result in f(x)... $\endgroup$ – Rugen Heidbuchel Apr 19 '15 at 9:01
  • $\begingroup$ @Rememberme Note that the length of the subintervals for infinitely small subintervals becomes dt, so the integral is the sum from 0 to x, of the product of the differential and dt. $\endgroup$ – Rugen Heidbuchel Apr 19 '15 at 9:04
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I hope that tis two pictures can help.

enter image description here

In the first is represented the area $F(x)$ under the graph of the function $y=f(x)$, starting from $x=0$. When we pass from a point $x$ to $x+dx$ the area increments of a quantity that is approximated by the pink rectangle, i.e. $f(x)dx=F(x+dx)-F(x)=dF(x)$. We can put this intuition in a rigorous statement using the limit notion, as: $$ \lim_{x \rightarrow 0}\dfrac{dF(x)}{dx}=\lim_{x \rightarrow 0}\dfrac{F(x+dx)-F(x)}{dx} =f(x). $$ And tis means that $f(x)$ is the derivative of $F(x)$.

Note that if we start from a point $a \ne 0$ the difference from the two areas is the blue area in the figure, that is a constant value $C$ , and this implies that we have a new primitive function of $f(x)$ : $G(x)=F(x)+C$, in accord with the fact that two functions that differ for a constant have the same derivative $f(x)$.

enter image description here

The second picture express the same result using the relation between a function and his derivative. Starting from $x=a$, where the value of the function is $F(a)$, we can reach the value of the function in $x=t$ adding increments $dF$ to any value $f(x)$ that we reach. This can be expressed as: $$ F(t)=F(a)+ \sum _a^t dF(x) $$

Now note that $dF(x)$ can be approximated as $dF(x)=F'(x)dx$, where $F'(x)$ is the derivative of $F(x)$ at the point $x$, and, using the limit , we have: $$ F(t)=F(a)+ \lim_{dx \rightarrow 0}\sum _a^t dF(x)=\int_a^t F'(x)dx $$

Note that, also in this case $F(t)$ is a primitive of a function $f(x)=F'(x)$ , defined less a constant $F(a)$.

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