Hello.

Calculate the area bounded by these curves:

$\displaystyle y=-x^2+4x-3$

$\displaystyle y=4x-3$

$\displaystyle x=2$

My answer is $\displaystyle \frac{77}{12}$ . I went like this:

For the area above x

$\displaystyle S_1 = \int\limits_{\frac{3}{4}}\limits^{2}{(4x-3)}dx-\int\limits_{1}\limits^{2}{(-x^2+4x-3)dx}$

and for the area under x

$\displaystyle

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

Thanks