Confused about calculating the area under the curve What is the area under the curve of the following function?
$f(x) = x² + 2x -3$
$x=-4$
$x=2$
Please, I'd like to see an image.
Here is the graphic:
https://www.desmos.com/calculator/oe0ja17spg
 A: The area enclosed between the $x$ axis and $f(x) = x^2 + 2x - 3$ over $\left[-4, 2 \right]$ is given by: $$ A = \int_{-4}^{-3} f(x) \text{ d}x + \left| \ \int_{-3}^{1} f(x) \text{ d}x \ \right| + \int_{1}^{2} f(x) \text{ d}x $$ since the sign of the integral over $\left[ -3, 1 \right]$ is negative.


As a small addendum, consider the following integral $$ \int_{a}^{b} \left| f(x) \right| \text{ d}x $$ supposing that $f(x) \leqslant 0$ over some interval $[c,d] \subset [a,b]$ and non-negative elsewhere $$\begin{aligned} \implies  \int_{a}^{b} \left| f(x) \right| \text{ d}x & = \int_{a}^{c} f(x) \cdot \underbrace{\mathrm{sgn}(f(x))}_{+1} \text{ d}x + \int_{c}^{d} f(x) \cdot \underbrace{\mathrm{sgn}(f(x))}_{-1} \text{ d}x + \int_{d}^{b} f(x) \cdot \underbrace{\mathrm{sgn}(f(x))}_{+1} \text{ d}x \\ & = \int_{a}^{c} f(x) \text{ d}x - \underbrace{\int_{c}^{d} f(x) \text{ d}x}_{\text{Which is negative}} + \int_{d}^{b} f(x) \text{ d}x \\ & = A \end{aligned} $$ using $\left| f(x) \right| = f(x) \cdot \mathrm{sgn}(f(x)) $ on the respective domains over which we're integrating.

A: 
$$ area=s_1+|s_2|+s_3=\\\int_{-4}^{-3} (x^2+2x-3)dx +\\ |\int_{-3}^{1} (x^2+2x-3)dx |+\\\int_{1}^{2} (x^2+2x-3)dx$$
