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

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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.

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