What is #int_(-4)^(1) |x^2-4|dx #?

Question

What is #int_(-4)^(1)  |x^2-4|dx #?

Paisley Johnson · Accepted Answer

#int_(-4)^1 abs(x^2-4)dx = 59/3#

Consider the function:

#f(x) = x^2-4 = (x+2)(x-2)#

we know that a second polynomial with positive leading coefficient is negative in the interval between its roots and negative outside the interval, so:

#f(x) > 0 # for #x in(-oo,-2) uu (2,+oo)#

#f(x) < 0 # for # x in (-2,2)#

Using the additivity of the definite integral we can split it as:

#int_(-4)^1 abs(x^2-4)dx = int_(-4)^-2abs(x^2-4)dx + int_(-2)^1 abs(x^2-4)dx#

#int_(-4)^1 abs(x^2-4)dx = int_(-4)^-2 (x^2-4)dx - int_(-2)^1 (x^2-4)dx#

Now:

#int_(-4)^-2 (x^2-4)dx = [x^3/3-4x]_(-4)^(-2) = -8/3+8+64/3-16 = 32/3#

#int_(-2)^1 (x^2-4)dx = [x^3/3-4x]_(-2)^1 = 1/3-4+8/3-8 = -9#

So:

#int_(-4)^1 abs(x^2-4)dx = 32/3+9 = 59/3#

Aurora Alvarado · Answer

undefined To evaluate the integral $\int_{-4}^{1} |x^2 - 4| \, dx$, we need to split the integral into intervals where $|x^2 - 4|$ changes its behavior. Since $|x^2 - 4|$ changes sign at $x = -2$ and $x = 2$, we partition the interval of integration accordingly:

$\int_{-4}^{1} |x^2 - 4| \, dx = \int_{-4}^{-2} (4 - x^2) \, dx + \int_{-2}^{2} (x^2 - 4) \, dx + \int_{2}^{1} (x^2 - 4) \, dx$

After integrating each part separately, we get:

$\int_{-4}^{1} |x^2 - 4| \, dx = \frac{112}{3}$