Calculate the integral from 0 to 4 of |x^2 - 1|. Thanks!

Answer 1

first, you have to get the value of x where |x^2 - 1| becomes zero,
thus x = -1 and 1
since the boundaries given are only from 0 to 4, we consider only what happens to the graph at x = 1 (because 1 is within the boundaries)
first, draw or imagine the graph. the graph of x^2-1 is a parabola that opens upward, and has a vertex at (0,-1).

since the given is in absolute value, |x^2 - 1| , all points below the x-axis (or points with negative y-coordinate) will be reflected along the x-axis,, now, you have a positive value of the integral. since we reflected the points, the original equation mus have the opposite signs. thus,
at 0 to 1 , |x^2-1| must be positive, and
at 1 to 4 , |x^2-1| is positive (according to the graph). therefore:

integral (1-x^2) from 0 to 1 +
integral (x^2-1) from 1 to 4

i think you know how to integrate this,,

hope this helps.

Answer 2

Graph y = x^2 - 1 from 0 to 4
There will be a part from 0 to 1 which is below the x-axis and then from 1 to 4 will be above the x-axis

so "flip u" that little part , (reflect it in the x-axis)
so now your would have
integral [1 - x^2] from 0 to 1 + integral[x^2-1] from 1 to 4
= [x - x^3/3] from 0 to 1 + [x^3/3 - x) from 1 to 4
= (1 - 1/3 - 0 ) + (64/3 - 4 - (1/3 - 1) )
= 1 - 1/3 + 18
= 56/3