Is the integral always the area under the curve? It might be a stupid question but if I were to ask to compute the definite integral
$$\int_{\frac{- \pi}{2}}^{\frac{\pi}{2}} \sin(x) \ dx$$
then on plugging the values then I would get "$0$" as the answer. 
But if I were to find the area covered by the function then, should I integrate as following:
$$AREA =  \left|\int_{\frac{- \pi}{2}}^{0} \sin(x) \ dx\right| + \int_{0}^{\frac{\pi}{2}} \sin(x) \ dx = 2??$$
 A: Saying "the integral is the area under the curve" is a common misconception that needs qualification. More precisely:


*

*If $f(x)\ge 0$ on $(a,b)$, then the area under the curve is given by $\int_a^b f(x)\,dx$.

*More generally, with no qualifications on the sign of $y=f(x)$, we say that $\int_a^b f(x)\,dx$ represents the (net) signed area under the curve meaning that on subinterval(s) where $f(x)\leq 0$, we think of the integral as "negative area", and on subintervals where $f(x)\geq 0$, we think of the integral as "positive area". Then the integral over the whole interval is the sum of these two.


More precisely, if $f:I\to\mathbb R$ with $I=P\cup N$ and
$$
\begin{cases}
f(x)\geq 0, &x\in P,\\
f(x)\leq 0, &x\in N,
\end{cases}
$$
then
$$
\int_I f(x)\,dx=\underbrace{\int_P f(x)\,dx}_{\geq 0}+\underbrace{\int_N f(x)\,dx}_{\leq 0}.$$
In your example, 
$$\int_{-\pi/2}^{\pi/2}\sin x\,dx,$$
we can envision the problem like this

We see that $\sin x\leq 0$ on $[-\pi/2,0]$ and $\sin x\geq 0$ on $[0,\pi/2]$, thus
\begin{align}
\int_{-\pi/2}^{\pi/2}\sin x\,dx=\int_{-\pi/2}^0 \sin x\,dx+\int_0^{\pi/2}\sin x\,dx
=-1+1=0.
\end{align}
From the picture, the "net area" of the $-1$ in red and the $+1$ in green is of course $0$, and the calculus reflects that.
On the other hand, the total area of the shaded region (not signed area) is
$$
\int_{-\pi/2}^{\pi/2}|\sin x|\,dx=\int_{-\pi/2}^0 -\sin x\,dx+\int_0^{\pi/2}\sin x\,dx=-(-1)+1=2.
$$
In this case, we negated the $-1$ area in red to get a $+1$ and then added this to the $+1$ area in green to get a total area of $2$.
So be careful about saying "integral is the area under the curve" without qualification!
