Please check my answers. Thank you.

Evaluate the given definite integral using the fundamental theorem of calculus. Round to 2 decimal places if needed.

$\displaystyle \int_{-4}^0 (2x + 6)^4\, dx$

$\displaystyle = \frac{1}{2} \frac{(2x + 6)^5}{5} |_0^{-4}$

$\displaystyle = [\frac{1}{10}(2(0) + 6)^5] - [\frac{1}{10}(2(-4) + 6)^5]$

$\displaystyle = 780.8$

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Find the area of the region R that lies under the given curve y = f(x) over the indicated interval $\displaystyle a \leq x \leq b$. Round to the 6th decimal places. Under $\displaystyle y = x^5e^{-x^6}$, over $\displaystyle 0 \leq x \leq 1$.

$\displaystyle \int_1^0 x^5e^{-x^6}\, dx$

$\displaystyle u = x^6$

$\displaystyle du = 6x^5\, dx$

$\displaystyle = \frac{1}{6} \int_1^0 e^{-u}\, du$

$\displaystyle = [-\frac{1}{6}e^{-1}] - [-\frac{1}{6}e^0]$

$\displaystyle = 0.105353$

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A study indicates that t months from now the population of a town will be growing at a rate of $\displaystyle P'(t) = 8 + 2t^{\frac{2}{3}}$ people per month. By how much will the population growth of the town increase over the next 6 months? Round to the nearest integer.

$\displaystyle \int_6^0 8 + 2t^{\frac{2}{3}}\, dt$

$\displaystyle = 8t + \frac{6}{5}t^{\frac{5}{3}} |_6^0$

$\displaystyle = [8(6) + \frac{6}{5}(6)^{\frac{5}{3}}]$

$\displaystyle = 72$

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The number of bacteria present after t minutes of an experiment was$\displaystyle Q(t) = 2100e^{0.04t}$. What was the average number of bacteria present during the first 2 minutes of the experiment. Round to 1 decimal place.

$\displaystyle V = \frac{1}{2-0} \int_2^0 2100e^{0.04t}\, dt$

$\displaystyle = 1050 \int_2^0 e^{0.04t}\, dt $

$\displaystyle = 1050 (25e^{0.08} - 25e^{0})$

$\displaystyle = 2186.3$