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# Evaluate the integral. $\displaystyle \int^{1}_{-1} t(1 - t)^2 \,dt$

## $-\frac{4}{3}$

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to evaluate the definite integral from negative 1 to 1 of three times one minus t squared DT. We apply substitution and here we let U equal to one -T. And so D you is just negative DT or that negative do you? Is equal to DT? We want to change the balance from T to you. So if T is equal to negative one then we have U equal to one minus negative one, that's equal to two. And if T is one we have U equal to 1 -1, that's equal to zero. And so we rewrite our integral into the integral from 2-0 of tee times we have one minus T. That's you raised to the second power times DT which is negative. Do you? Now? We still want to change T here and you will do that using the substitution that we had, If U is equal to 1 -5. Then from here we can say that T is equal to one minus you. And so we changed that and we have the integral from 2-0 of one minus you times you squared times negative. Do you now note that if we have the integral from A to B of F of X dx this is the same as the negative of the integral from B to A of F of x dx. We're going to flip this and we have the negative of the integral from 0 to 2 of U squared minus You to the 3rd power times negative do you? Which is the same as the integral from 0 to 2 of you squared minus U. To the third power do you? And here we integrate and we get You raised to the 3rd power over 3-. You do the forest power over four. This evaluated from 0 to 2. Now if us to we have to to the third power over sweet -2 to the 4th power over four. This minus when you is zero Which becomes zero eventually. And so we have 8/3 minus 16/4 or four. And this gives us the value of 8 -12/3. Or that's just Negative for over three. So this is the value of the integral.

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