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I read that integration is the opposite of differentiation AND at the same time is a summation process to find the area under a curve. But I can't understand how these things combine together and actually an integral can be the same time those two things. If the integration is the opposite of differentiation, then the result of the integration should be the initial function from which we derived the derivative function. How can this be at the same time the area under the curve of the (which?) function?

trig
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2 Answers2

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What you are asking is, esencially, the fundamental theorem of calculus.

The process in analysis is such that first, for continuous functions (actually, a slightly larger class of functions, not only continuous, but that isn't all that important), we can calculate the area under their graphs by taking the limit of Riemann sums. That is, for a function $f$ defined on $[a,b]$, we have a strictly defined method with which to calculate the surface area under the graph of $f$. We use the notation

$$\int_a^bf(t)dt$$ to denote this surface area. (note: no "integration" has taken part so far. We only took some limits and defined a number which is the surface area).

Now, for a continuous function $f$ defined on $[a,b]$, we can define the function

$$F(x) = \int_a^x f(t)dt$$ which is equal to the area under the curve $\{(t, f(t)| x\in[a,x]\}$

For this function, the fundamental theorem states that the function $F$ is the antiderivative of $f$, meaning that $F'=f$, and it is obvious that $F(a) = \int_a^b f(t) dt$.

Note: What the theorem basically says is "the area under the function $f$ is an antiderivative of $f$".

This theorem has huge consequences, and the one you want is this one:

For a continuous function $f$, it is easy to show that if $F' = f$ and $G' = f$ (if $G$ and $G$ are antiderivatives of $f$), then $F(x) = G(x) + C$ for some constant $C\in\mathbb R$.

The consequence of this is that if you find any one antiderivative $G$ of $f$, you can immediatelly be sure that $$G(x) = \int_a^xf(t)dt + C,$$

meaning that $G(b) - G(a)$ is equal to $$\int_a^bf(t)dt + C - \int_a^a f(t)dt - C = \int_a^b f(t)dt,$$ precisely the surface area under $f$ on the interval $[a,b]$.

EDIT:

You said to give you an example using $f(x) = 3x^2$, so here you go:

I will look at $f$ on the interval $[0,1]$. We know that one antiderivative of $f$ is the function $g(x) = x^3 + 5$ (or any other function of the form $x^3 + C$, including $x^3$). Now, define the function $F(x)$ as:

$$F(x) = \int_0^xf(t)dt,$$ Where the definite integral is calculated using Riemann sums. We do not yet know how to calculate $F(x)$, but we know that $F$ is a well defined function. We also know that $F(0) = 0$, since there is no area under one single point.

What we want to evaluate is $F(1)$, since that is the area under $f$ on the interval $[0,1]$.

Now what do we know?

  1. Using the fundamental theorem of calculus, we know that $F$ is an antiderivative of $f$, i.e. we know that $F'(x) = f(x)$.
  2. We know that $g'(x) = f(x)$.
  3. We can also prove that if $F_1$ and $F_2$ are both antiderivatives of $f$, then they differ only by a constant, i.e. if $F_1' = F_2' = f$, then $F_1 = F_2 + C$. Strictly speaking, if $F_1' = F_2',$ then $$\exists C_1 \in\mathbb R\forall x\in[0,1]: F_1(x) = F_2(x) + C_1.$$

From these three points, we can conclude, without knowig what $F$ is, that $F$ differs from $g$ by only a constant, i.e. there exists some constant $C$ such that $g(x) = F(x) + C$. Now, what is the value $g(1) - g(0)$ equal to?

Well, we know that $g(1) = F(1) + C$ and $g(0) = F(0) + C$, so $$g(1)-g(0) = (F(1)+C) - (F(0) + C) = F(1) + C - F(0) + C = F(1) - F(0).$$

We also know that $F(0) = 0$, so

$$g(1) - g(0) = F(1) - F(0) = F(1) - 0 = F(1).$$

This means that if we calculate $g(1) - g(0)$ (and we can do that, since we know what $g$ is - remember, $g$ is any antiderivative of $f$!), we will actually calculate the number $F(1)$. But by definition, $F(1)$ is equal to the area under $f$ on the interval $[0,1]$, so we have, by plugging the values $1$ and $0$ into the indefinite integral of $f$, calculated the definite integral of $f$.

5xum
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  • can you give me an example please? for example let's say we have a function f(x)=3x^2. The antiderivative of this function is another function g(x) whose derivative is the function f(x), so the antiderivative is g(x)=x^3+c. So to calculate the area under the curve f(x)=3x^2, I somehow use the g(x)=x^3+c? 1) how do I actually get the area of the surface by using g(x)? 2) why g(x) gives the area of f(x)? is it something we just accept without proof? like an axiom? 3) what about c? does it matter to calculate it? and how do we calculate it? – trig Oct 30 '14 at 10:20
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    @trig See my edit. Read it. Try to understand. If anything is unclear, ask away. – 5xum Oct 30 '14 at 11:19
  • makes some sense, so the area under f at the interval [0,1] is really 1^3?? – trig Oct 30 '14 at 11:50
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    @trig Indeed, since $g(0) = 5$ and $g(1) = 6$, meaning $g(1) - g(0) = 1$. There is no axiom you need to accept without proof, only the fundamental theorem of calculus which you can find the proof for in most calculus textbooks. – 5xum Oct 30 '14 at 11:52
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Integration is not exactly the opposite as differentiation. And the interpretation of the integral as the area under a curve is good for positive functions only, otherwise you must keep your attention for signs. These things combine together by the fundamental theorem of calculus, which states that if $f$ is a continuous function, then: $$\int_a^b f(x) \ \mathrm{d}x = F(b) - F(a),$$ where $F$ is any primitive of $f$.

Ivo Terek
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