# Integration math solution

• Apr 30th 2008, 05:16 AM
urmiprincess
Integration math solution
Find the first-quadrant area bounded by the axes and the functions

f(x) = 0.5x + 2; g(x) = 2x - 4
• Apr 30th 2008, 06:27 AM
Soroban
Hello, urmiprincess!

Did you make a sketch?

Quote:

Find the first-quadrant area bounded by the axes and the functions:

. . $\displaystyle \begin{array}{cccc}f(x) &= & \frac{1}{2}x +2 & {\color{blue}[1]}\\ g(x) &=& 2x - 4 & {\color{blue}[2]}\end{array}$

Code:

          |                      B           |                    ...o (4,4)           |              ...*:::* :           |        ...*:::::::*  :           |  ...*:::::::::::*    :         A o2::::::::::::::*      :           |:::::::::::::*        :           |:::::::::D:*          : E       - - o - - - - o - - - - - - + -         O|      * 2            4           |    *           |  *           | *         C *-4           |

$\displaystyle \text{Line }{\color{blue}[1]} \:=\:AB$
$\displaystyle \text{Line }{\color{blue}[2]} \:=\:CB$
. . They intersect at $\displaystyle B(4,4)$

We want the area of quadrilateral $\displaystyle ABDO$
. . Drop perpendicular $\displaystyle BE$

Trapezoid $\displaystyle ABEO$ has: .$\displaystyle h = OE = 4,\;b_1 = AO = 2,\;b_2 = BE = 4$
. . Its area is: .$\displaystyle \frac{4}{2}(2 + 4) \:=\:12$

Right triangle BED has: .$\displaystyle b = DE = 2,\;h = BE = 4$
. . Its area is: .$\displaystyle \frac{1}{2}(2)(4) \:=\:4$

Therefore: .$\displaystyle \text{Area}(ABDO) \;=\;12 - 4 \:=\:\boxed{8}$