To find the area between y=x and y= x^2 − 2 on the interval [0,4], set up the integrals to find the area.

Question

To find the area between y=x and y= x^2 − 2 on the interval [0,4], set up the integrals to find the area.

Answer 1

x=x^2 -2

x^2 -x -2 = 0

(x-2)(x+1) = 0

x =2 is a point where the two lines intersect. x=-1 is the other point of intersection but it's outside the interval

the only area really in the interval [0,4] is just [0,2)

There's 3 subareas. Calculate the area in each then add them up.

You can't do one integral as it would treat areas above the x axis as positive and subtract

areas below the x-axis

one integral is of x^2-2 evaluated from 0 to sqr2. That gives (4/3)sqr2

another area is triangular shaped between the y=x line and the x axis, from 0 to sqr2 = 1

last area is below the y=x line and the parabola, the integral of x-x^2+2 evaluated

between sqr2 and 2. 7/3 - (4/3)sqr2

the 3 areas sum to 10/3 or 3 1/3

there's a 4th area that might be included, the integral of where the parabola is above the line y=x

from x=2 to x=4. If you include that, that's 8 2/3 for a grand total area of 12