Question

graph the curves y=|x|-1, y= 1-x^2. identify the pints of of intersection and hence find the area enclosed by the 2 curves.

Answer 1

the intersection points are x=1 and x=-1 right?

Answer 2

if it's right. how to find the area?

Answer 3

krebante, cn u help me?

Answer 4

Lets find the points of intersection:\[|x| - 1 = 1-x^2 \Rightarrow |x| + x^2 = 2.\]If x is positive,\[x + x^2 = 2 \Rightarrow x = -2, 1\]since we are working with positive x, we discard -2 and the intersection for positive x is 1.
If x is negative,\[-x + x^2 = 2 \Rightarrow x = -1, 2\]and as before we discard 2 and the intersection for negative x is -1.

Answer 5

I'll show you how to find the area now.

Answer 6

ok

Answer 7

Lets start by noticing that \[|x| - 1 < 0\]if\[-1 \leq x \leq 1\]and\[1 - x^2 > 0\]if\[-1 \leq x \leq 1,\]so\[1-x^2 \geq |x| - 1.\]The area between two curves is the integral of the positive difference of these curves. Since \[1-x^2 \geq |x| - 1,\]that is\[A = \int_{-1}^1 \left(\left(1-x^2\right) - \left(|x| - 1\right)\right)dx = \int_{-1}^1 \left(2 - x^2 - |x| \right)dx = \frac{10}{3} - \int_{-1}^1 |x|dx.\]For the last integral,\[\int_{-1}^1 |x| dx= \int_{-1}^0 |x|dx + \int_0^1 |x|dx = \int_{-1}^0 (-x)dx + \int_0^1 xdx = 1.\]so, finally\[A = \frac{10}{3} - 1 = \frac{7}{3}.\]

Answer 8

ok.thanks a lot. God Bless u