1. ## area of triangle

I have no idea how to do this can someone teach me step by step please?!

using integration, find the area of the triangle with vertices (0,0)(2,1)(-1,6)

2. Given a function $\displaystyle f(x)$, $\displaystyle \int_a^b f(x) dx$ gives us the area between the function and the x-axis, from a to b.

Naturally, when we have two functions $\displaystyle f(x)$ and $\displaystyle g(x)$, where $\displaystyle f(x)$ is always bigger than $\displaystyle g(x)$, then $\displaystyle \int_a^b f(x) - g(x) dx$ gives us the area between the two functions.

So, we have 3 points $\displaystyle (0,0), (2,1), (-1,6)$. We can find three lines from these points, and each line will cross two of the points. These will be the sides of the triangle. We can obtain these with simple algebra

$\displaystyle L_1(x) = \frac{1}{2}x$ crosses (0,0) and (2,1)

$\displaystyle L_2(x) = -6x$ crosses (0,0) and (-1,6)

$\displaystyle L_3(x) = \frac{-5}{3}x + \frac{13}{3}$ crosses (2,1) and (-1,6)

Now if we draw out the triangle on the xy plane, we see that $\displaystyle L_3$ is above $\displaystyle L_1$ when $\displaystyle 0\leq x \leq 2$

So the area between the two lines is given by $\displaystyle \int_0^2 L_3(x) - L_1(x) dx$.

Similarly, $\displaystyle L_3$ is about $\displaystyle L_2$ when $\displaystyle -1 \leq x \leq 0$ and the area between the two lines is given by $\displaystyle \int_{-1}^0 L_3(x) - L_2(x) dx$.

I'll leave you to solve these integrals. Now we see that by our setup, the sum of the two areas, is the area of the triangle. So the area of the triangle is $\displaystyle \int_0^2 L_3(x) - L_1(x)dx + \int_{-1}^0 L_3(x) - L_2(x)dx$

3. Hi gearshifter,

Drawing a rectangle from (-1,0) to (2,6) creates 3 triangles around yours. The surface of your triangle can be written as the surface of the rectangle minus the surface of the three triangles around your triangle:

18-[3+1+7.5] = 6.5

This can be done for all possible triangles!

In your case the surfaces of the three triangles around the main traingle can be calcluated using integrals:

-6x integrated from -1 to 6 = 3
triangle 1: 3

0.5x integrated from 0 to 2 = 1
triangle 2: 1

4,33-(5/3)x integrated from -1 to 2 = 10.5
triangle 3: 3*6-10.5=7.5

This method is not as elegant as one given by Haven, but hopefully better understandable.

4. If you have time left you can try the Gauss method:

The surface of an figure can be calculated by multiplying the following factors per line. The summation over all lines leads to the surface.

a) The length of the line
b) The average x-coordinate
c) The x value of the normal on the line

First line ([0,0] [2,1])
a1=5^0.5=2.236
b1=1
c1=cos(atan(2))=0.447

second line ([2,1] [-1,6])
a2=34^0.5=5.831
b2=0.5
c2=cos(atan(3/5))=0.857

third line ([-1,6] [0,0])
a3=37^0.5=6.083
b3=-0.5
c3=-cos(atan(1/6))=-0.986
(NOTICE the normal vector has a negative x-value)

surface = a1*b1*c1 + a2*b2*c2 + a3*b3*c3 = 6.5

This method is far too complicated for this case, but can be usefull for more complicated geometries.