Context: High School question.
Find the surface area between the curve of the function $y=6-3x^{2}$ and the function $y=3x$ in the interval $[0,2]$
My approach:
-We must find the points of intersection first:
$$6-3x^{2}=3x$$$$x^{2}+x-2=0$$$$x=-2,x=1$$
Then we must evaluate these integrals to find the area:
$$A=\left | \int_{0}^{1}(-3x^{2}-3x+6)dx \right |+\left | \int_{1}^{2}(-3x^{2}-3x+6)dx \right |$$
My question is why couldn't we just evaluate this integral as usual to find the area?
$$\left | \int_{0}^{2}(-3x^{2}-3x+6)dx \right|$$
$\endgroup$ 14 Answers
$\begingroup$Because the area is the integral of (function which is above) minus (function which is below). In other words, it is equal to$$\int_0^1\lvert6-3x^2-3x\rvert\,\mathrm dx.$$But$$\lvert6-3x^2-3x\rvert=\begin{cases}6-3x^2-3x&\text{ if }x\in[0,1]\\3x-(6-3x^2)&\text{ otherwise.}\end{cases}$$
$\endgroup$ $\begingroup$We want to add up the two pieces of area.
If you integrate them as one term. You are actually computing the difference between the two areas.
\begin{align} \left|\int_0^2-3x^2+6-3x\, dx\right|&=\left|\int_0^1-3x^2+6-3x\, dx+\int_1^2-3x^2+6-3x\, dx\right|\\ &=\left|\int_0^1-3x^2+6-3x\, dx-\int_1^2 3x-(-3x^2+6)\, dx\right|\\ &=\left|\int_1^2 3x-(-3x^2+6)\, dx-\int_0^1-3x^2+6-3x\, dx\right|\\ \end{align}
Net area is shown integrated from difference of two curves$ y_2-y_1=-3x^{2}-3x+6 $ starting at $(0,6)$.
At first the area integal $ -x^3-\dfrac{3 x^2}{2}+6 x $ is positive, maximum at $x=1,$ then at $x\approx 1.81174 $ zero when two areas balance out and thereafter it is negative.
At $x=2$ net evaluated area is $ A= -2$ shown.
$\endgroup$ $\begingroup$It is $$\int_{-2}^16-3x^2-3xdx$$ the result is given by $$\frac{27}{2}$$ You can add a constant $C$ to both curves and you must not care if the curves are below the axes or not.
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