Solve $\int^2_{-1} (1-x)dx$ by thinking in terms of area?

Question

On our last quiz in Calculus 1, my professor asked us to solve $\int^2_{-1} (1-x)dx$ by thinking in terms of area. I don't know what he means by, "thinking in terms of area". I can solve it myself, but I didn't get full points, because I wasn't able to ask him what he meant.

How can I solve this "by thinking in terms of area"?

$$\int^2_{-1} (1-x)dx$$

$$=\int^2_{-1} 1 - \int^2_{-1}x$$

$$=x - \frac{x^2}{2}+C$$

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$$\underset{x\to -1}\lim (x - \frac{x^2}{2}) = -1 - \frac{1}{2}=-\frac{3}{2}$$

$$\underset{x\to 2}\lim (x - \frac{x^2}{2}) = 2 - \frac{4}{2}=0$$

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$$\int^2_{-1} (1-x)dx=0-(-\frac{3}{2})=\frac{3}{2}$$

Answer 1

Integration computes the area between the graph and the axis. The red and blue region will cancel out.

What is left is the purple region which has area $1^2+\frac12 = \frac32$

Remark: Btw, $\lim_{x \to -1}(x-\frac{x^2}2)= - \frac32$.

Answer 2

That integral is the area of the triangle whose vertices are $(-1,2)$, $(-1,0)$, and $(1,0)$ minus the area of the triangle whose vertices are $(1,0)$, $(2,0)$, and $(2,-1)$. So,$$\int_{-1}^21-x\,\mathrm dx=2-\frac12=\frac32.$$

Answer 3

Consider

1): $\displaystyle{\int_{-1}^{2}}1 dx $:

The area of a rectangle from $x=-2$ to $x=1$ on the $x-$axis of height $1$, i.e.

$A_1 =3\cdot 1=3.$

2) Consider $y=x$ :

$\displaystyle{\int_{-1}^{2}}(x)dx $:

Negative contribution from $x=-1$ to $x=0$ on the $x$-axis:

$(1/2)(-1)(1)=-1/2;$

Positive contribution from $x=0$ to $x=2$ on the $x-$axis.

$(1/2)(2)(2)=2$;

Adding negative and positive contributions:

$A_2= 2 -(1/2)=3/2;$

Difference: 1)-2) :

$ A_1-A_2= 3-(3/2)=3/2$;