Calculating the area bounded by $y=x-2$ and the $x$-axis on the interval $[-1,3]$. Integration gives a value that conflicts with geometry.

algebra-precalculuscalculusdefinite integralsintegration

I am trying to calculate the area bounded by the equation $y=x-2$ along the x-axis between the interval $[-1, 3 ]$.

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Evaluating the following integral give me the result of $-4$

$$\int_{-1}^{3}(x-2) \mathbf{d}x$$

But, when I calculate area, using triangle's formula 1/2*b*h, the result comes out to be 5.

Area of bigger triangle = 1/2 * 3 * 3 = 9/2

Area of smaller triangle = 1/2 * 1 * 1 = 1/2

Total area = 9/2 + 1/2 = 5

Since the area is always positive, I am not taking account of + or - signs. If I take that into account the answer comes out to be correct (same as integration).

What am I doing wrong? And why do I need to account for sign changes?

Best Answer

There is a concept of area, which is always positive, and a concept of algebraic or oriented area, which can take both signs.

With an integral like $\displaystyle\int f(x)\,dx$, which can be negative when $f$ is negative, you obtain an algebraic area. If you want to compute it geometrically, you need to identify the positively oriented and negatively oriented zones and take the difference of their areas.

If you don't want to deal with algebraic areas, compute the integral $\displaystyle\int |f(x)|\,dx$ instead.

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