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∫t2⋅0.15e−0.15(t−0.5)dt\int t^2\cdot0.15e^{-0.15\left(t-0.5\right)}dt

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Final answer to the exercise

0.15−0.15t2e(−0.15t+0.075)+0.15(−2e(−0.15t+0.075)t(−0.15)2)+0.15(2e(−0.15t+0.075)(−0.15)3)+C0\frac{0.15}{-0.15}t^2e^{\left(-0.15t+0.075\right)}+0.15\left(\frac{-2e^{\left(-0.15t+0.075\right)}t}{{\left(-0.15\right)}^2}\right)+0.15\left(\frac{2e^{\left(-0.15t+0.075\right)}}{{\left(-0.15\right)}^{3}}\right)+C_0

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∫t2·0.15e−0.15(t−0.5)dt
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