Limit of Riemann Sum equal to integral

Question

I don't understand why Limit of a Riemann sum suddenly turns into a integral. It makes no sense both intuitively and geometrically to me. In case, differentiation it is easy understand though.

Answer 1

the Reimann sum is the definition of the integral.

I think what you are really saying is that you don't understand why the integral equals the anti-derivative.

And that is the "Fundamental Theorem of Calculus."

suppose $F(x) = \int_0^t f(t) dt$

That is, we have a curve $f(t)$ and we color everyting above the interval (0,x) on the $x$-axis to the bottom of the cuve. That area equals $F(x).$ Now we want to make that area just a tiny bit larger. So we make a very thin stripe just to the right of $x.$ How much has $F(x)$ changed? The area has changed by the length of the stripe $\times$ the thickness of the stripe. Lets call the thickness h.

$F(x+h) = F(x) + f(x)h$

Now what is the derivative of $F(x)$?

$\lim_\limits{h\to 0} \dfrac{F(x+h) - F(x)}{h} = \dfrac{f(x) h}{h} = f(x)$

Perhaps not rigorous, but it might help with the intuition.

For a more formal proof, I am sure your calculus book has one, and it should be easy enough to google it.

Answer 2

There is no reason why the limit of a Riemann sum turns into an integral: the (Riemann) integral of a function over an interval $[a,b]$ is by definition the limit of a Riemann sum. However, if you are aware of the geometric definition of the integral, as the area under a curve, we can see why this definition makes sense.
Say we want to compute the area under the curve described by the equation $y=f(x)$ on some interval $[a,b]$. One way to do this is to cut up our interval into small sub-intervals, and add up the area of rectangles with these sub-intervals as their base, and whose heights are approximately that of the graph of $f$. More formally, we can define a subdivision of $[a,b]$ by specifying $N+1$ points $a=x_0<x_1<x_2<..<x_{N-1}<x_N=b$. Under the right conditions, the value of $f$ on each of the $[x_i,x_{i+1}]$ will be sufficiently close to some $f(a_i)$ with $a_i\in [x_i,x_{i+1}]$, so that the area under the graph of $f$ on $[x_i,x_{i+1}]$ will be close to $f(a_i)\cdot(x_{i+1}-x_i)$. Then we may sum up all these approximate areas to get a total approximate area under the graph of $f$ on $[a,b]$ , that is, let $$S_N=\sum_{i=1}^{N}f(a_{i-1})\cdot(x_{i}-x_{i-1})$$ $S_N$ is precisely a Riemann sum associated with the subdivision $\sigma=(x_0,..,x_N)$. We may take our subdivision and choice of $a_i$'s in some convenient way, for instance make the $x_i$'s evenly spaced, take $f(a_i)$ to be the sup or the inf of $f$ on $[x_i,x_{i+1}]$, or any other choice we find useful, but under the right conditions, we find that the value of $S_N$ converges as the subdivision gets finer (that is, $\sup\{x_{i+1}-x_i,0\leq i\leq N\}$ gets small enough), regardless of the particular subdivisions or choice of $a_i$'s. If this is indeed the case, we say $f$ is (Riemann) integrable on $[a,b]$, and define $$\int_a^bf(x)dx=\lim_{n\mapsto \infty}S_n$$ Where $(S_n)$ is any sequence of Riemann sums whose corresponding subdivisions get finer and finer. In other words, if $\sigma=(x_0,..,x_N)$, let $\delta(\sigma)=\sup\{x_{i+1}-x_i,0\leq i\leq N\}$, and let $(S_n)$ have corresponding subdivisions $(\sigma_n)$. Then we require that $\lim_{n\mapsto \infty}\delta(\sigma_n))=0$.
Notice, the very notation of the integral informally reflects this idea of discrete summation. the $\int$ sign is an elongated S for "sum", and the $dx$ stands for an infinitesimal variation in $x$, that is, $x_{i+1}-x_i$ if the subdivision is "infinitely fine"