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Integrate: x^2 sqr rot (4x^3 - 16)dx from 2 to 5
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\[\int_2^5 x^2\sqrt{4x^3-16}~dx\] Set \(u=4x^3-16\), so that \(du=12x^2~dx\), or \(\dfrac{du}{12}=x^2~dx\). \[\frac{1}{12}\int_{16}^{484} \sqrt{u}~du\]
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