The diagram above shows the curve $\displaystyle y=2x^2-5x+5$ and straight line $\displaystyle y=5-x$.Find

(a)the points if intersection of the two graphs

i get the answer $\displaystyle (0,5)(2,3)$

(b)The area of the shaded region.

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- Mar 24th 2010, 02:57 AMlindrosDefinite integral problem
The diagram above shows the curve $\displaystyle y=2x^2-5x+5$ and straight line $\displaystyle y=5-x$.Find

(a)the points if intersection of the two graphs

i get the answer $\displaystyle (0,5)(2,3)$

(b)The area of the shaded region. - Mar 24th 2010, 04:03 AMStroodle
I've never really solved a problem like this before (or any integration problem actually) but would this work:

$\displaystyle \int_{0}^{3} 5-x\ dx-\left(\int_{0}^{3}5-x\ dx\ -\ \int_{0}^{3}2x^2-5x+5\ dx \right )$

Then do a similar thing for $\displaystyle x\in \left [ 3,5\right ]$ and add the results together. - Mar 24th 2010, 04:31 AMlindros
this is my step to get the answer but the answer provided not same with my answer

$\displaystyle \frac{1}{2}(5)(5) - \left(\int_{0}^{5}(2x^2-5x=5) dx$

$\displaystyle =\frac{25}{2}-(\frac 2{x^3}{3}-\frac{5x^2}{2})_0^5$

$\displaystyle =\frac{25}{2}-({2(5)^3}{3}-\frac{5(2)^2}{2}=5(5))-0$

$\displaystyle =\frac{25}{2}-(\frac{275}{6}$

$\displaystyle =\frac{-33}{1}{3}$

is that correct??lol,my latex is wrong - Mar 24th 2010, 04:55 AMsa-ri-ga-ma
You can do this way.

[Intg(5 - x)*dx between 0 to 5 -[ Inte(2x^2 -5x + 5)*dx] between 0 to 3.