I know we say \(\displaystyle \lim\limits_{n \to \infty} \sum\limits_{i=1}^n f\big( x_{i} \big) \cdot \Delta x = \int_a^b f\big( x_{i} \big) dx \ \ \ \ \ \ \ \ \ \ \text{ where } \Delta x = \frac{b-a}{n} \) but my problem is how do we relate these both
This is not exactly correct. The "limit" on the left has to be taken in a specific way- we cannot just write it as a limit of a function.
* The summation notation is solved by adding terms n times
The point of the Riemann sums is to show that we
can define the "area under the curve" for figures other than the basic "rectangle", "parallelogram", and "circle" of classical geometry.
* Whereas integral is solved by taking Anti-Derivative
They are solved by different methods then how are they related?
I know the area under curve can be found by summation BUT I don't have understanding of how that same area can be found by taking antiderivative
I'm confused and seriously need you guys help
First, do you agree that the summation, which is a summation of areas of rectangles approximating the region under the curve, is an approximation to the actual area, and that the limit, both taking more rectangles
and making the rectangles smaller, gives the area itself? That should be fairly intuitive.
The proof of the other part, that this area is given by the
anti-derivative is given in most Calculus text-books. I'll try to reproduce it here:
First, draw the graph. (It might help to actually do that yourself.) To start with, take y= f(x) such that f(x)> 0 and that f be continuous between, say, a and b. Now,
define A(x) to be the area under that curve from a to some value x> a (the "Riemann sums" construction assures us that we can do this). Let \(\displaystyle \Delta x\) be some small positive number. Then, by definition, \(\displaystyle A(x+ \Delta x)\) is the area from a to \(\displaystyle x+ \Delta x\) and so \(\displaystyle A(x+ \Delta x)- A(x)\) is the area under the curve from x to \(\displaystyle x+ \Delta x\). That will
almost be a very thin rectangle. "Almost" because the top boundary is the curve y= f(x) rather than a horizontal line. If we let \(\displaystyle y_0\) be the
smallest value of f(x) and \(\displaystyle y_1\) be the largest value of f(x) between \(\displaystyle x\) and \(\displaystyle x+ \Delta x\) then by the
intermediate value theorem there exist some \(\displaystyle y_0\le \overline{y}\le y_1\) such that the area under the curve, between x and \(\displaystyle x+ \Delta x\) is
equal to the area of the rectangle with base from x to \(\displaystyle x+ \Delta x\) and height \(\displaystyle \overline{y}\). That is, the area is given by \(\displaystyle A(x+\Delta x)- A(x)= (\Delta x)(\overline{y})\). In that case, we have \(\displaystyle \frac{A(x+ \Delta x)- A(x)}{\Delta x}= \overline{y}\).
Now, take the limit as \(\displaystyle \Delta x\) goes to 0. On the left of that last equation, we have \(\displaystyle \lim_{\Delta x\to 0}\frac{A(x+ \Delta x)- A(x)}{\Delta x}= \frac{dA}{dx}\). On the right, since we always have \(\displaystyle \overline{y}= f(x')\) for some x'
between x and \(\displaystyle x+ \Delta x\), as \(\displaystyle \Delta x\) goes to 0, x' goes to x and \(\displaystyle \overline{y}= f(x')\) goes to f(x). That is, in the limit, that last equation becomes
\(\displaystyle \frac{dA}{dx}= f(x)\).
The derivative of the area function
really is f(x) so the area
really is the "anti-derivative" of f(x).
