Hello,

I'm in minor confusion why my answer is wrong as opposed towards the book's answer.

I'm asked to find the area enclosed by the equations,

$\displaystyle y_{1} = (2x -1)\ and\ y_{2} = (3 + 2x - x^2) $

$\displaystyle So\ the\ point\ (s)\ of\ intersection\ implies: (2x - 1) = (3 + 2x - x^2) $

$\displaystyle \Rightarrow x = \pm2 $

If I compute the area under each equation, and also consider a sign test on both integrals seperately, then:-

$\displaystyle A_{a} = \int ^{2}_{-2} (3 + 2x - x^2)\ dx = \left[(3)(2) + 2^2 - \frac{2^3}{3}\right] - \left[(3)(-2) + 2^2 + \frac{(2)^3}{3}\right] $

$\displaystyle A_{a} = \frac{20}{3}\ units $

$\displaystyle \Rightarrow A_{a}\ is\ (+ve) $

and,

$\displaystyle A_{b} = \int ^{2}_{-2} (2x -1)\ dx = \left(2^2 -2\right) - \left(2^2 + 2\right) = (-4)\ units $

$\displaystyle \Rightarrow A_{b}\ is\ (-ve) $

Now if I consider another sign test,

$\displaystyle Area\,A_{c},\ enclosed\ by\ both\ equations\ = A_{a} \pm A_{b} \Rightarrow A_{c} = (+) + (-)\ or\ (+) - (-) \Rightarrow A_{c}\ is\ (-ve)\ or\ (+ve) $

The total area available under either equation, with limit $\displaystyle x = \pm2 $, is $\displaystyle A_a $ . Then the area $\displaystyle A_{c} $, enclosed by $\displaystyle y_{1} $ and $\displaystyle y_{2} $ will be less than $\displaystyle A_{a} $. So in order for $\displaystyle A_{b} $ to be subtracted from $\displaystyle A_{a} $ it must have its sign reversed because it's negative.

So,

$\displaystyle A_{c} = A_{a} + A{b} = \frac{20}{3} + (-4) = \frac{8}{3}\ units $

However, the book has answer of $\displaystyle \frac{32}{3}\ units $

Therefore,

$\displaystyle A_{c} = \frac{20}{3} - (-4) = \frac{32}{3}\ units $

The enclosed area $\displaystyle A_{c} $ is greater than $\displaystyle A_a $

I don't understand why my answer is wrong?