# Thread: Area bounded by curves(check answer)

1. ## Area bounded by curves(check answer)

Hello.
Calculate the area bounded by these curves:
$y=-x^2+4x-3$
$y=4x-3$
$x=2$

My answer is $\frac{77}{12}$ . I went like this:
For the area above x
$S_1 = \int\limits_{\frac{3}{4}}\limits^{2}{(4x-3)}dx-\int\limits_{1}\limits^{2}{(-x^2+4x-3)dx}$
and for the area under x
$
S_2 = -\Bigg(\int\limits_{0}\limits^{1}{(-x^2+4x-3)dx}-\int\limits_{0}\limits^{\frac{3}{4}}{(4x-3)dx}\Bigg)
$

So $S = S_1+S_2$ Someone tell me if the answer is correct cause I don't have the result.

Thanks

2. the parabola and line intersect at x = 0, and , except at x = 0,
$4x-3 > -x^2 + 4x - 3$

$A = \int_0^2 (4x - 3) - (-x^2 + 4x - 3) \, dx = \frac{8}{3}$

3. I'm confused. Why you didn't need to separate the area, area above x-axis and area under x-axis(cause under x it will have a -)

4. it does not make a difference.

if f(x) > g(x), then the area between them is $A = \int_a^b f(x) - g(x) \, dx$ no matter where f and g reside relative to the x-axis.

here is a simple analogy ... consider two points (1, 5) and (1, 2) ... the distance between the two points is 5 - 2, correct?

now, what about (1,5) and (1,-2) ... isn't the distance 5 - (-2) ?

it's the same idea with the functions.