1. ## Area

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.

2. Originally Posted by cazimi
a) Find the area bounded by the curve y² - 3x + 3 = 0 and the line x = 4.

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1. Transcribe this equation of the relation into equations of functions:

$f(x) = \sqrt{3} \cdot \sqrt{x-1}$ or $f(x) = -\sqrt{3} \cdot \sqrt{x-1}$

2. Calculate the zero(s) of the function (or relation): y = 0

$-3x +3 = 0~\iff~ x = 1$

3. Therefore the area is:

$a=2\cdot \sqrt{3} \cdot \int_1^4\left(\sqrt{x-1}\right)dx = 2\cdot \sqrt{3} \cdot \left[\frac23 \cdot (x-1)^{\frac32}\right]_1^4$ = $2\cdot \sqrt{3} \cdot \frac23 \cdot 3^{\frac32} = 12$

3. Originally Posted by cazimi
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b) Find the area bounded by the curve y = 4x - x² and the line x = 0 and y = 4.
1. Calculate the difference of functions:

$g(x)=4-(-x^2+4x)$

2. Calculate the zeros of g:

$g(x) = x^2-4x+4 = (x-2)^2$ thus x = 2 is the only zero.

3. Calculate the area:

$a = \int_0^2\left( (x-2)^2\right)dx=\left[\frac13 (x-2)^3\right]_0^2 = \frac83$