Rombus area 

Here is the graph of the following equation :
[display]f(x)=\frac13x[/display] What is the area of greyed region ?
y=1/3 x
Dumy solution
We know that the equation is [math]f(x)=\frac{1}{3}x[/math]
The area under the curve is given by : [display]\int \frac{1}{3}x dx=\frac{1}{6}x^2[/display] Then the requested area is given by : [display]\int_3^7 \frac{1}{3}x dx=[\frac{1}{6}x^2]_3^7=\frac{1}{6}7^2-\frac{1}{6}3^2=\frac{1}{6} (49-9)=\frac{20}{3}[/display]
Smart solution (click to open)
The area of Trapezoid is : [display]\frac{(B+b)h}{2}[/display] In other words : [display]\frac{[f(7)+f(3)]4}{2}[/display] With actual values, it gives : [display]\frac{\frac{7+3}{3}4}{2}=\frac{7+3}{3}2=\frac{20}{3}[/display]