Expressing Riemann sums as definite integral 
I'm confused about the last part. Using the definition of an integral, I know that $\Delta$x=$\frac{3}{2n}$ but what will $a$ be here?. $\frac 1 2$ or $\frac {-3} {2} $?
I was thinking I could manipulate the expression so that I have the same expression in both the brackets, of the form $a$ + $\Delta$ x $i$ and then compare that with the definition for find a, b and f(x). The problem seems to be that the resultant a, b and f(x) will depend entirely on how I manipulate the expression, which is arbitrary. So how do I go about this? 
 A: In these examples, you first decide what $x$ should be as a function of $i$. In all the present cases you may choose $\displaystyle x_i=x_i(n) = \frac{i}{n}$. The increment (which should be small when $n$ is large) is  the difference between $x_i$ for two successive values of $i$, here:
  $$ \Delta x = x_{i+1}-x_i = \frac{i+1}{n} - \frac{i}{n}=\frac1n $$
To find the integration bounds look at the first and last values of $x_i$: $x_1=\frac{1}{n}$ and $x_n = \frac{n}{n}=1$. You should take the $n\rightarrow \infty$ limit in those to get the bounds, i.e. $a=\lim \frac{1}{n} = 0$ and $b=1$. So whenever your function has the form:
 $$ \sum_{i=1}^n f(\frac{i}{n}) \frac{1}{n}  =\sum_{i=1}^n f(x_i) \Delta x $$
for suitable $f$ then the corresponding Riemann integral form is $$\int_0^1 f(x)dx$$
This gives the form in the post of Salahaman.
You can, however also, as you are hinting at, choose $x_i=x_i(n) = \frac{3i}{2n}$ which gives the increment $\Delta x = x_{i+1}-x_i = \frac{3}{2n}$. To find the bounds proceed as above: $a = \lim_{n\rightarrow \infty} x_1 (n) = \lim_n \frac{3}{2n}=0$ and $b=\lim_n \frac{3n}{2n} = \frac{3}{2}$. The sum in (d) then corresponds to the integral:
  $$ \int_0^{3/2} (x+ \frac{1}{2}) \tan ( x - \frac{3}{2}) dx $$
There is indeed some arbitrariness as seen above. All should give the same value for the integral, although not necessarily the same expression. They are related by a change of variables transformation.
A: hint for d)
The sum can be written as
$$\sum_{i=1}^nf (x_i)(x_{i+1}-x_i) $$
$$=\sum_{i=1}^n f (\frac {i}{n})(\frac {i+1}{n}-\frac {i}{n})$$
with
$$f (x)=\frac {3}{2}(\frac {3}{2}x+\frac {1}{2})\tan \Bigl(\frac {3}{2}(x-1)\Bigr) $$
The limit is $$\int_0^1 f (x)dx $$
