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Find the area between the curves: $$y=x+1$$

$$y=-3x+13$$ $$y=-\frac{1}{3}x-\frac{1}{3}$$

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I see this when I graph it, and it's not easy for me to understand what variable I should integrate with. Any suggestions?

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    Calculate the distances between the points, and use Heron's formula. – 2017-02-09
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    It doesn't matter , both require splitting the area. – 2017-02-09
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    You can use cross product instead, I think that's a bit more pleasant than splitting the region, integrating, etc. – 2017-02-09
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    Unforunately this is for calculus and not linear algebra so I cannot – 2017-02-09

3 Answers 3

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$$S=\int_{-1}^5(f_H(x)-f_L(x))dx,$$ $$f_L(x) = f_3(x) = -\frac13x-\frac13,$$ $$f_H(x) = \begin{cases} f_1(x) = x+1, \text{ for } x\in(-1,3)\\ f_2(x) = -3x+13, \text{ for } x\in(3,5). \end{cases} $$ $$S=\int_{-1}^3(f_H(x)-f_L(x))dx + \int_{3}^5(f_H(x)-f_L(x))dx,$$ $$S=\int_{-1}^3(f_1(x)-f_3(x))dx + \int_{3}^5(f_2(x)-f_3(x))dx.$$

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    What is the meaning of $f_H, f_L$? – 2017-02-09
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    High and Low bounds of curvilinear trapezoid. – 2017-02-09
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$$ I_1 = \int^0_{-1} (x+1)-\left(\frac{-1}{3}x-\frac{1}{3}\right)dx$$

$$ I_2=\int^3_{0}(x+1)-\left(\frac{-1}{3}x-\frac{1}{3}\right)dx$$

$$ I_3=\int^5_3 (-3x+13)-\left( \frac{-1}{3}x-\frac{1}{3} \right)dx$$

$$ A_{total}=I_1+I_2+I_3=?$$

Also notice that $I_1+I_2$ can be simplified to

$I_{1} + I_2=\int^3_{-1}(x+1)-\left(\frac{-1}{3}x-\frac{1}{3}\right)dx$

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Partition the triangular region $R$ into $R_1$ and $R_2$ where

$$R_1=\left\{(x,y)\in\Bbb R^2:-1\leq x\leq 3 \text{ and }\frac{-x}{3}-\frac{1}{3}\leq y\leq x+1\right\}$$ and $$R_2=\left\{(x,y)\in\Bbb R^2:3\leq x\leq 5\text{ and } \frac{-x}{3}-\frac{1}{3}\leq y\leq -3x+13\right\}.$$ Denoting by $A_1$ and $A_2$ the corresponding areas of $R_1$ and $R_2$, we get

$$A_1=\int_{-1}^3\left[(x+1)-\left(\frac{-x}{3}-\frac{1}{3}\right)\right]dx$$ and $$A_2=\int_{3}^5\left[(-3x+13)-\left(\frac{-x}{3}-\frac{1}{3}\right)\right]dx.$$ The required area therefore is $$A=A_1+A_2.$$ Can you do the computations of $A_1$ and $A_2$?