1

i do get that Riemann sums is sum of infinite triangles with with infinitely small length. But definite integral is completely different you are taking anti derivative of f(x) at b and subtract anti derivative of f(x) at point a.

Explanation one that fails to make sense is this: definite integral is just notation it's same thing as limit of Riemann sums. And i was fooled till all of the sudden and quietly my class started using antiderivatives..symbol were same so it took a week for me confront my confusion.

Explanation two is this answer...but it fails to explain or just assume either that F'(x) = f(x) Or... if proving later then it assumes definite integral equals area. Basically doesn't prove one without not relying other which happen to be also be in need of proof or explanation to me.

So in short can you explain/proof how or why:

  • Definite Integral ( the difference of antiderivatives ) equals area F(x)=A(X)
  • F'(x)=f(x)

if you can show me proof of one without relying on other, i.e. F'(x) = f(x), without relying of F(x) = A(x) then it will be enough.

I have been trying to understand this for 40+ days..tutor, professor, 3 textbooks, and google all failed me.

tinlyx
  • 1,534
Muhammad Umer
  • 705
  • 4
    What you are asking about is called the Fundamental Theorem of Calculus. Maybe the Wikipedia article will help. – André Nicolas Mar 24 '14 at 20:42
  • yes i have read that article like 5 times...it assumes F'(x)=f(x) to show antiderivative = area..and it uses the same fact, F(x) = A(x) to show f(x)=F'(x)...so it uses A to proof B. And then uses B to proof A. – Muhammad Umer Mar 24 '14 at 20:51
  • The argument is that if $f$ is nicely behaved and $F(x)=\int_a^x f(t),dt$, then $F(x+h)-F(x)\approx hf(x)$. Here by $\int_a^x f(t),dt$ we mean the Riemann integral, no antiderivatives, no circularity. – André Nicolas Mar 24 '14 at 21:35
  • isn't Riemann integral just antiderivative of B - antiderivative of A – Muhammad Umer Mar 25 '14 at 01:50
  • 2
    Just a clarification: the definite integral is written $\int_a^b f(x)~dx$, and defined as the Riemann sum $\lim_{n\to\infty} \sum_{i=1}^n f(x_i)(x_i - x_{i-1})$. The indefinite integral is written $\int f~dx$ and is defined as $F$ such that $F' = f$. The confusing bit that can never be emphasized enough is that indefinite and definite integrals are defined wildly differently, and the similar notation is not justified until you prove the Fundamental Theorem of Calculus, which says: $\int_a^b f(x)~dx = F(b) - F(a)$, where $F = \int f~dx$, tying together definite and indefinite integrals. – Henry Swanson Mar 25 '14 at 01:54
  • 1
    @MuhammadUmer: No, it is defined entirely differently. Roughly speaking, we divide the interval from $a$ to $b$ into $n$ subintervals, using division points $x_0=a,x_1,\dots,x_n=b$. We calculate $\sum_1^n f(x_i)(x_{i+1}-x_i)$, and take the limit as $n\to\infty$ and the $x_{i+1}-x_i$ approach $0$. That limit is the Riemann integral. It is closely connected to an intuitive way of defining area. – André Nicolas Mar 25 '14 at 01:55
  • guys thanks a lot...i feel like i am somewhere now..but i need time to rearrange stuff in my mind. So i'll come back tomorrow with my understanding so far. – Muhammad Umer Mar 25 '14 at 02:01

2 Answers2

2

Given a continuous function $f:\>[a,b]\to{\mathbb R}$ we can define its integral over $[a,b]$ in various ways as a limit of Riemann sums. Denoting this limit by $$\int_{[a,b]} f(x)\>{\rm d}x$$ we have $$\int_{[a,b]} f(x)\>{\rm d}x=\lim_{\ldots}\>\sum_{k=1}^N f(\xi_k)(x_k-x_{k-1})\ ,$$ where I don't bother now about the exact definition. At any rate, when $f(x)\geq0$ on $[a,b]$ we can interpret this integral "intuitively" as "area under the curve $y=f(x)$".

Of course we want to compute this integral when $f(x)$ is some given expression in the variable $x$. In order to obtain a general procedure Newton had the following fundamental idea: We consider momentarily the upper limit $b$ as variable, i.e., we are looking at the area function $$A(u):=\int_{[a,u]} f(x)\>{\rm d}x\ .$$ A more or less geometric limit argument then shows that this area function has the property $$A'(u)=f(u)$$ for all $u$ in the range at hand. This means that $A(\cdot)$ is an antiderivative of $f$, and it is then not difficult to show that $$\int_{[a,b]} f(x)\>{\rm d}x=F(b)-F(a)\ ,\tag{1}$$ where $F$ can be any antiderivative, or: primitive, of $f$ on $[a,b]$.

The difference on the right hand side of $(1)$ does only depend on the given $f$, and not on the particular primitive $F$ chosen. We therefore want a notation for this difference that exhibits only $f$, as well as $a$ and $b$. The established notation is, as we all know, $$\int_a^b f(x)\>dx\ .\tag{2}$$ The expression $(2)$ has the following meaning: Compute, by whichever means, a primitive $F(x)$ of the function term $f(x)$, and return the difference $F(b)-F(a)$.

In this way the fundamental theorem of calculus assumes the (only seemingly tautological) form $$\int_{[a,b]} f(x)\>{\rm d}x=\int_a^b f(x)\>dx\ .$$ On the left hand side we have a limit of Riemann sums, and on the right hand side a difference $F(b)-F(a)$ for some primitive of $f$.

Christian Blatter
  • 232,144
0

One direction of what you're trying to understand is not too hard: If $F(x) = \int_a^x f(y) dy$ then when you take the derivative of $F$ at point $x$, the derivative is intuitively equal to the rate of change in the area under the curve $f$ that you get when you increase $x$ by a little bit. But if $f$ is continuous, then if you change $x$ a little bit then it doesn't change the value of $f(x)$ by much, so the change in area when you move from $x$ to $x + \delta$ is simply close to the area of the rectangle $\delta f(x)$ if $\delta$ is small. Thus the derivative of $F(x)$ is $\lim_{\delta \to 0} \delta f(x) / \delta = f(x)$. So if you are integrating a function $f$ to get area under the curve $F$, then $f$ is the derivative of $F$.

user2566092
  • 26,450