Calculate the integral from 0 to 4 of |x^2 - 1|. Thanks!
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.
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
To calculate the integral from 0 to 4 of |x^2 - 1|, you need to split the integral into two separate cases because the absolute value function changes the behavior of the expression inside it depending on whether it is positive or negative.
Case 1: (x^2 - 1) ≥ 0
In this case, the absolute value function does not affect the expression, so we can simply integrate (x^2 - 1) from 0 to 4.
To find the integral, you can use the power rule for integration.
∫(x^2 - 1) dx = (1/3)x^3 - x + C
Then, you can evaluate the antiderivative at the upper and lower limits:
(1/3)(4^3) - 4 - [(1/3)(0^3) - 0] = (64/3) - 4 = 52/3
Case 2: (x^2 - 1) < 0
In this case, the expression inside the absolute value becomes negative. To handle this, we need to change the sign of the quantity inside the absolute value. So, we integrate -(x^2 - 1) from 0 to 4.
∫-(x^2 - 1) dx = -((1/3)x^3 - x) + C = -[(1/3)x^3 - x] + C
Again, evaluate the antiderivative at the upper and lower limits:
-[(1/3)(4^3) - 4] - [-((1/3)(0^3) - 0)] = -[(64/3) - 4] - [(0 - 0)] = -52/3
Finally, add the results of both cases together, since we are dealing with an absolute value:
52/3 + (-52/3) = 0
Therefore, the integral from 0 to 4 of |x^2 - 1| is equal to 0.