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Evaluate the integral ∫(8 to 0) t² sin(2t) dt.

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Final answer:

To evaluate the integral ∫(8 to 0) t² sin(2t) dt, we can use integration by parts. The formula ∫ u dv = uv - ∫ v du is applied. After performing the necessary substitutions and integrating, we can simplify the expression to find the exact value of the integral.

Step-by-step explanation:

To evaluate the integral ∫(8 to 0) t² sin(2t) dt, we can use integration by parts. Let's use the formula ∫ u dv = uv - ∫ v du, where u = t² and dv = sin(2t) dt. Differentiate u to get du = 2t dt, and integrate dv to get v = -cos(2t)/2. Substituting these values into the integration by parts formula, we have:

∫(8 to 0) t² sin(2t) dt = [(-t² cos(2t))/2] - ∫(-cos(2t)/2)(2t dt)

Now we can integrate the second term using u-substitution. Let u = 2t, then du = 2 dt. The new integral becomes:

-∫(8 to 0) cos(u)/4 du = [-sin(u)/4]

Substituting u back in terms of t, we get the final result:

[(-t² cos(2t))/2] - [-sin(2t)/4]

Simplifying this expression further will give you the exact value of the integral.

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