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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?

$$\left | \int_{0}^{2}(-3x^{2}-3x+6)dx \right|$$

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  • $\begingroup$ The integral $\int_{-1}^1 x \mathrm{d}x$ is equal to $0$ but this is not the area under the curve $y=x$ between $x=-1$ and $x=1$. $\endgroup$ – Peter Foreman May 4 at 15:07
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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$$\lvert6-3x^2-3x\rvert=\begin{cases}6-3x^2-3x&\text{ if }x\in[0,1]\\3x-(6-3x^2)&\text{ otherwise.}\end{cases}$$

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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}

enter image description here

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  • $\begingroup$ But why then if we want find the area between two curves like $y=6x-x^{2}$ and $y=x^{2}-2x$ we take the integral without worrying about this, although some parts of the latter function are below the x-axis? $\endgroup$ – Positron12 May 4 at 16:51
  • $\begingroup$ the important thing is not whether they are above or below the $x$-axis, but whether the relative position of the two curve swaps. $\endgroup$ – Siong Thye Goh May 4 at 16:55
  • $\begingroup$ Could you explain further please. $\endgroup$ – Positron12 May 4 at 16:56
  • $\begingroup$ the area between two curve between $f$ and $g$ can be obtained via $\int |f-g| \, dx$. To remove the absolute value inside the integral, we can split the integral into the region where $f \ge g$ and the region where $g \ge f$ and then sum them up. We can't just move the absolute value outside the integral directly. $\endgroup$ – Siong Thye Goh May 4 at 17:00
  • $\begingroup$ why wasn't the absolute value needed for other problems? i.e when should I put it? $\endgroup$ – Positron12 May 4 at 17:02
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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.

enter image description here

At $x=2$ net evaluated area is $ A= -2$ shown.

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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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