Here is the graph of the following equation :
[display]f(x)=\frac13x[/display]
What is the area of greyed region ?
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]
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]
