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Calculate the area bounded by the curves

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

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Welcome to MSE! Is this homework? If so, please tag it as such. Dod you write the entire problem down (for example, limits of integration)? What have you tried on your own and where are you confused? These might be some of the reasons someone down-voted your question, so you might want to do some updates. Regards – Amzoti Jan 20 '13 at 19:07

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First note that the curves intersect at $(0,-1)$ and $(1,0)$. And the area you are interested in is the area between the two curves between these two points. Hence, we get the area between the two curves as $$\int_{x=0}^{x=1} \int_{y=x^2-1}^{y=x-1} dydx = \int_{x=0}^{x=1} (x-x^2) dydx = \left(\dfrac{x}2 - \dfrac{x^3}3 \right)_{x=0}^{x=1} = \dfrac12 - \dfrac13 = \dfrac16$$

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+1 nothing, but a graph, can illustrate a maths problem. – Babak S. Jan 20 '13 at 19:01

First, calculate the intersection of the functions.

$$ x - 1 = x^2 - 1 $$ $$ 0 = x^2 - x $$

We get: $$ x = 0, 1 $$

Now we integrate:

$$ \int_{0}^1 (x^2 - 1) - (x-1) \mathrm d x $$ $$ \int_{0}^1 (x^2 - x) \mathrm d x $$ $$ -\frac{1}{6} $$

Clearly the area must be positive, so we have: $$ \color{green}{\frac{1}{6}} $$

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Hint: Find the point of intersection of the two curves, calculate the area of each curve by integrating through the point of intersection, then subtract areas to get whats between them.

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