Definition of the Riemann sum prove that:
$$
\int_a^b f(x)\,dx = \lim_{n\to\infty} h\bigg(f(a)+f(a+h)+\ldots+f(a+(n-1)h)\bigg);\, h =\frac{b-a}{n}
$$
I know that the right side comes from Riemann sum and it gives the net signed area under the curve, but how the left-hand side represents the signed area under the curve, and how they both evaluate to the same value? I am not able to relate the two.
 A: We're just using the definition of the Riemann Integral here.
Let $[a,b]$ be a closed interval such that $f$ is Riemann-Integrable. Define a partition:
$$a = x_0 < x_1 < x_2 < \ldots < x_{n-1} < x_n = b$$
so that:
$$\forall i \in \{1,2,3,\ldots,n \}: h = x_{i}-x_{i-1} = \frac{b-a}{n}$$
Now, for each interval $[x_{i-1},x_i]$, we pick a point $\xi_i$ and define the sum:
$$S = \sum_{i=1}^{n} f(\xi_i)h$$
If we take the limit of the above sum as the norm of the partition goes to $0$, we get the Riemann Integral of $f$ on the interval $[a,b]$. In this case, the norm of the partition is just $h$ so we simply let $h \to 0$.
Now, very conveniently, I'm going to define $\xi_i = a+(i-1)h$. That is, notice that it's just going to be the left endpoint of the intervals $[x_{i-1},x_i]$. So, we have the sum:
$$S = \sum_{i=1}^{n} f(a+(i-1)h)h = h \cdot \sum_{i=1}^{n} f(a+(i-1)h)$$
Then, we can see that:
$$\int_{a}^{b} f(x) \ dx = \lim_{h \to 0} \left[ h \left(\sum_{i=1}^{n} f(a+(i-1)h) \right) \right]$$
as was desired. It's sort of weird to ask why anything gives the area under any curve because what I've done above makes no mention of the area under any graph. If you want to see the definition applied to the geometric problem of finding the area under a graph, then you should definitely draw some pictures to get some more intuition for how this works.
A: Your question is why the following is true
$$
\left( \int f(x)dx \right)\Bigg|_a^b =  \int_a^b f(x)dx
$$
that is why the difference between the values of an antiderivative (indefinite integral) denoted as $\int f(x)dx$, which is a function of $x$, equals the definite integral, denoted as $\int_a^b f(x)dx$? That statement is the Fundamental Theorem of Calculus
Simply explained: The limit of a Riemann sum (if it exists) is called the definite integral. The difference between (or the sum of) two definite integrals is again a definite integral (that should be intuitive). Now a definite integral as a function of its the upper limit is an antiderivative of the integrated function
$$ \Phi_c(x) = \int_c^x f(t)dt$$
and the difference between two values of an antiderivate (two definte integrals), LHS, is then a definite integral, RHS.
Additional:
Every continuous real-valued function (of one variable) $f$ has an antiderivative, and that is (assuming the definite integral is already defined) the definite integral from some constant lower limit $a$ to a variable upper limit $x$, let it be denoted
$$ \Phi_c(x) = \int_c^x f(t)dt$$
the prove that this is an antiderivative is by the mean value theorem and by the property of the definite integral that
$$
\int_a^c f(t)dt +\int_c^b f(t)dt = \int_a^b f(t)dt 
$$
the derivative of $\Phi_c(x)$ wrt $x$ is then
$$
\lim_{h\rightarrow0}\frac{\int_c^{x+h} f(t)dt - \int_c^x f(t)dt}{h}=\lim_{h\rightarrow0}\frac{\int_x^{x+h} f(t)dt}{h}=\lim_{h\rightarrow0}\frac{h\ f(x+\tilde h)}{h}= f(x)
$$
where $0 \leq \tilde h \leq h$. So the difference between antiderivatives
