Question #0993b

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Answer 1 · 2017-05-05T11:31:05

$\frac{16}{3}$ $16/3$ sq. units

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Given: $y^{2} = 2 x + 3$ $y^2=2x+3$ and $y = x$ $y = x$

Here is a graph of the two equations:

graph{(2x+3-y^2)(y-x)=0 [-10, 10, -5, 5]}

Solving these equations:

$y^{2} = 2 x + 3 [1]$ $y^2=2x+3" [1]"$
$y = x [2]$ $y = x" [2]"$

Substituting equation [2] into equation [1]:

$x^{2} = 2 x + 3$ $x^2=2x+3$

$x^{2} - 2 x - 3 = 0$ $x^2 - 2x - 3=0$

$(x - 3) (x + 1) = 0$ $(x-3)(x+1)=0$

$x = 3 and x = - 1$ $x = 3 and x = -1$

$:. "Area" = int_-1^3(y - (y^2-3)/2)dy$

$= {:y^2/2-y^3/6+3/2y]_-1^3$

$= 16/3 "units"^2$