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Evaluate the integral.$\int_{-1}^{1} t(1-t)^{2} d t$

$-\frac{4}{3}$

Calculus 1 / AB

Calculus 2 / BC

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Integration Techniques

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Okay, so we can rewrite this Integral using that distributed properties to get T d t minus seem to go from negative 1 to 1 off to t squared D T plus the integral from negative +121212121 off T cubed DT. So if you take the integral using the power rule of each of the easy get T squared divided by two and we're gonna evaluate that from negative front one multiplied by two over three times T cubed from negative 2 1 2 3 1 to 1, plus a tea to the power of four, divided by four again from negative 4 4 1 to 1. So if you plug in the numbers here, you get one over to minus naked, one squared again, Just one over two by this to third, uh, minus. So it becomes plus to third plus one over four minus negative. One to the before is just one over four. Then here you have your final answer is equal to negative for over three

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