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Find the area enclosed by two curves.

$y=x^2$ and $y=-x^2+x+1$

My answer:

$\frac{-1}{2} \le x \le 1$

$x^2 \le y \le -x^2+x+1$

$\int_{}^{} \int_{}^{}dxdy = \int_{- \frac{1}{2} }^{1} ( \int_{x^2}^{-x^2+x+1}dy)dx$

$P= \int_{ \frac{-1}{2} }^{1}\left( (-x^2+x+1)-(x^2)\right)= \frac{9}{8}$

Is that correct, because the answer in my excercice book is $\frac{ \sqrt{5} }{6}$

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    I got the same as you, you I'd say the book is wrong, you are right...probably. – 2017-01-11
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    How about that one: Find the volume of the solid enclosed by : $z=7-x^2$, $z=-2$, $y=-1$, $y=4$. My answer is 180, the answer in the book is again totally different.. – 2017-01-11
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    Yep, I get the same $\;180\;$ : $$\int_{-3}^3\int_{-1}^4\int_{-2}^{7-x^2}dzdydx=180$$ Are you sure you're looking at the solutions in the correct chapter, sections...*book* ? – 2017-01-11
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    I am, I checked multiple times to be sure, to be precise it is not a book per se, these are the materials our lecturer prepared for us, attaching allegedly "correct" answers. Ugh. – 2017-01-11
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    Well, those notes at least will make a nice bonfire...:) – 2017-01-11

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