Why is the integral of $x$ from $-1$ to $1$ $x^2/2$ rather than $x^2$? Why is $\int_{-1}^1 x =\frac{x^2}{2}$ rather than $x^2$?
 A: Consider $f(x)=x$ for x equals 0 to a, it is a right-angled triangle (the blue region in figure below). 

The area of the square (having sides equal a) is $a^2$. Since there are two identical triangles each filling half the area, the area under the curve of f(x)=x for x=0 to a is $\frac{a^2}{2}$, which is precisely what you get from computing the definite integral, since a definite integral calculates area under the curve.
Another example, consider f(x)=x from a to b, with a>=0. The definite integral from a to b calculated the area of the blue region illustrated below. It can be reduced to the previous case as the area of the bigger triangle from 0 to b minus the area of the smaller triangle from 0 to a. Hence $\frac{b^2}{2} -\frac{a^2}{2}$.

Perhaps I should also mention, when the area is below the x-axis, such as f(x) for x=a to 0 (a<0), a change of sign is needed, i.e $\frac{-a^2}{2}$.
A: Integration is the opposite of differentiation.
If you differentiate  x^2/2 you get x (you bring down the power of 2, which cancels with the 1/2 out front) which means that when you integrate x you must get x^2/2.
A: First, the integral of $x$ from $-1$ to $1$ is a number, not a function. Definite integrals give you the net signed area between the $X$-axis and the function. So the integral of $x$ from $-1$ to $1$ is neither $x^2$  nor $x^2/2$. 
Second, the indefinite integral of $x$ is a family of functions, namely, the family of all functions $F(x)$ such that $F'(x)=x$; that is, the family of antiderivatives of $x$. Since the indefinite integral is a family of functions and not a single function, the indefinite integral of $x$ is neither $x^2/2$  nor $x^2$. 
Now, by the Fundamental Theorem of Calculus, if you are trying to compute the definite integral of a function $f(x)$ that is continuous on $[a,b]$, and you are lucky enough to have any antiderivative $F(x)$ of $f(x)$, then you can compute the definite integral as the Total Change of $F(x)$ on $[a,b]$. That is,
$$\int_a^b f(x)\,dx = F(b)-F(a).$$
So, in order to compute 
$$\int_{-1}^1 x\,dx$$
you have several choices:


*

*Find the net signed area geometrically. This is very easy in this case, because all you have is two triangles that cancel each other, so the net signed area is $0$. That is, $\int_{-1}^1 x\,dx = 0$.

*Use the definition of the integral as a limit of Riemann sums. Again, not hard to do: the function $y=x$ is increasing, the left hand and right hand sums are easy to compute, and the limit turns out to be $0$. For example, the left hand Riemann sum with $n$ intervals, each of equal length $\frac{2}{n}$, is:
\begin{align*}
\mathrm{LHS}(n) &= \frac{-n}{n}\left(\frac{2}{n}\right) + \frac{-n+2}{n}\left(\frac{2}{n}\right) + \cdots + \frac{n-2}{n}\left(\frac{2}{n}\right)\\
&= \frac{2}{n^2}\Bigl(-n + (-n+2) + (-n+4)+\cdots+\bigl(-n+(2n-2)\bigr)\Bigr)\\
&= \frac{2}{n^2}\Bigl( -n(n) + 2\bigl(0 + 1 + 2 + \cdots + (n-1)\bigr)\Bigr)\\
&= \frac{2}{n^2}(-n^2 + n^2 - n) = \frac{-2n}{n^2} = -\frac{2}{n}.
\end{align*}
Thus, 
$$\lim\limits_{n\to\infty}\mathrm{LHS}(n) = \lim_{n\to\infty}\left(-\frac{2}{n}\right) = 0.$$
The right hand sums also converge to $0$, so the integral is equal to $0$. 

*Use the Fundamental Theorem of Calculus and find some function $F(x)$ whose derivative is $x$ on $[-1,1]$. Now, $F(x)=x^2$  does not work because $F'(x) = 2x\neq x$. But $\mathcal{F}(x) = \frac{1}{2}x^2$  does work because $\mathcal{F}'(x) = \frac{1}{2}(2x) = x$. That means (by the Constant Function Theorem) that
$$\int x\,dx = \frac{1}{2}x^2 + C,$$
(the family of functions that are the antiderivatives of $x$ are the vertical translates of $\frac{1}{2}x^2$) so you can use any one of them (say, the one with $C=0$) to compute $\int_{-1}^1x\,dx$. 
