# How do you compute the value of #int t^2+3tdt# of [1,4]?

I found:

Now we need to subtract the values obtained:

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To compute the value of ( \int_{1}^{4} (t^2 + 3t) , dt ), you first need to find the antiderivative of the integrand, which is ( \frac{t^3}{3} + \frac{3t^2}{2} ). Then, evaluate this antiderivative at the upper and lower limits of integration (4 and 1, respectively), and subtract the result of the lower limit from the result of the upper limit.

So,

[ \int_{1}^{4} (t^2 + 3t) , dt = \left[ \frac{t^3}{3} + \frac{3t^2}{2} \right]_{1}^{4} ]

[ = \left( \frac{4^3}{3} + \frac{3(4)^2}{2} \right) - \left( \frac{1^3}{3} + \frac{3(1)^2}{2} \right) ]

[ = \left( \frac{64}{3} + \frac{48}{2} \right) - \left( \frac{1}{3} + \frac{3}{2} \right) ]

[ = \left( \frac{64}{3} + 24 \right) - \left( \frac{1}{3} + \frac{3}{2} \right) ]

[ = \left( \frac{64}{3} + \frac{72}{3} \right) - \left( \frac{2}{6} + \frac{9}{6} \right) ]

[ = \left( \frac{136}{3} \right) - \left( \frac{11}{6} \right) ]

[ = \frac{272}{6} - \frac{11}{6} ]

[ = \frac{261}{6} ]

[ = \frac{87}{2} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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