How do you evaluate the definite integral #int|x^3+64| dx# from #[-8,0]#?

Question

Emilia Aguilar · Accepted Answer

#896#

#f(x) = x^3+64# changes sign for #x = -4# so

#int_{-8}^0abs(f(x))dx = -int_{-8}^{-4}f(x)dx+int_{-4}^0f(x)dx#

but

#int_a^bf(x)dx = 64 b + b^4/4-(64 a + a^4/4)# then

#int_{-8}^0 abs(f(x))dx=704 + 192 = 896#

Evelyn Agard · Answer

undefined To evaluate the definite integral of |x^3 + 64| from -8 to 0, we need to break the interval [-8, 0] into two parts: [-8, -4] and [-4, 0]. In each part, we find the integral separately and then sum the results.

For the interval [-8, -4], since x^3 + 64 is positive in this range, |x^3 + 64| simplifies to x^3 + 64. Thus, the integral becomes the integral of (x^3 + 64) dx from -8 to -4.

For the interval [-4, 0], since x^3 + 64 is negative in this range, |x^3 + 64| simplifies to -(x^3 + 64). Thus, the integral becomes the integral of -(x^3 + 64) dx from -4 to 0.

Now, we calculate each integral separately:

For the interval [-8, -4]:
∫(x^3 + 64) dx from -8 to -4
= [(1/4)x^4 + 64x] from -8 to -4
= [(1/4)(-4)^4 + 64(-4)] - [(1/4)(-8)^4 + 64(-8)]
= (64 - 256) - (1024 - 512)
= -192 - 512
= -704

For the interval [-4, 0]:
∫-(x^3 + 64) dx from -4 to 0
= -∫(x^3 + 64) dx from -4 to 0
= -[ (1/4)x^4 + 64x ] from -4 to 0
= -[ (1/4)(0)^4 + 64(0) - ((1/4)(-4)^4 + 64(-4)) ]
= -[ 0 - (64 - 256) ]
= -(256 - 64)
= -192

Finally, we sum the results of both integrals:
-704 (from [-8, -4]) + (-192) (from [-4, 0]) = -704 - 192 = -896.

Therefore, the value of the definite integral of |x^3 + 64| from -8 to 0 is -896.