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

1 Answer

Jul 29, 2016

#896#

Explanation:

#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#

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