# Can you evaluate the integral by interpreting it in terms of areas?

May 28, 2018

${\int}_{- 4}^{3} 1 - x \mathrm{dx} = \frac{21}{2}$

#### Explanation:

Let's sketch the graph of $f \left(x\right) = 1 - x$. Add the vertical lines $x = - 4$ and $x = 3$.

Integrating is just finding the area under a curve, so since we have two triangle, we just find their areas, add them up and we're all done.

The first triangle has base length of $5$ units and height $5$ units. The area is therefore $\frac{5 \cdot 5}{2} = \frac{25}{2}$.

The second triangle has base length $2$ and height $- 2$. Therefore the area will be $\frac{2 \cdot - 2}{2} = - 2$

Therefore ${\int}_{- 4}^{3} 1 - x \mathrm{dx} = \frac{25}{2} + \left(- 2\right) = \frac{21}{2}$

Hopefully this helps!