a) Find the area bounded by the curve y² - 3x + 3 = 0 and the line x = 4.
b) Find the area bounded by the curve y = 4x - x² and the line x = 0 and y = 4.
1. Transcribe this equation of the relation into equations of functions:
$\displaystyle f(x) = \sqrt{3} \cdot \sqrt{x-1}$ or $\displaystyle f(x) = -\sqrt{3} \cdot \sqrt{x-1}$
2. Calculate the zero(s) of the function (or relation): y = 0
$\displaystyle -3x +3 = 0~\iff~ x = 1$
3. Therefore the area is:
$\displaystyle a=2\cdot \sqrt{3} \cdot \int_1^4\left(\sqrt{x-1}\right)dx = 2\cdot \sqrt{3} \cdot \left[\frac23 \cdot (x-1)^{\frac32}\right]_1^4$ = $\displaystyle 2\cdot \sqrt{3} \cdot \frac23 \cdot 3^{\frac32} = 12$
1. Calculate the difference of functions:
$\displaystyle g(x)=4-(-x^2+4x)$
2. Calculate the zeros of g:
$\displaystyle g(x) = x^2-4x+4 = (x-2)^2$ thus x = 2 is the only zero.
3. Calculate the area:
$\displaystyle a = \int_0^2\left( (x-2)^2\right)dx=\left[\frac13 (x-2)^3\right]_0^2 = \frac83$