Area of Curve using Direct Formula for the area of Triangle

Question

For the line $f(x)=3x+6$ on $[0,6]$ what is the area of the curve using the direct formula for the area of triangles?

Can someone please tell me how to go about this and help me work through it?

Answer 1

This is one way by dividing the area into two triangles.

$A = \frac{1}{2}(6)(24) + \frac{1}{2}(6)(6) = 90$

Answer 2

In the above diagram (not drawn to scale) area of top gray triangle and bottom rectangle respectively ( for any $x$ ) are computed as:

$$ \frac12\cdot x \cdot 3x +6x = \frac{3x^2}{2} + 6x $$

The same result is also obtained by integration of the function $f(x)$ between $x$ limits $(0,x).$ The rectangle can be looked upon as two right triangles made by drawing a diagonal across serving as a hypotenuse.

Answer 3

The area $[ODBC]$ under the curve $y=3x+6$ on the interval $x\in[0,6]$ can be expressed as the difference of the areas of triangles $[ADB]$ and $[AOC]$:

\begin{align} [ODBC]&=[ADB]-[AOC] \\ &= \tfrac12\cdot|AD|\cdot|DB| - \tfrac12\cdot|AC|\cdot|OC| = \tfrac12\cdot8\cdot 24 - \tfrac12\cdot 2 \cdot6 =90 . \end{align}