Question

How do you evaluate the integral (x^2+6) when the limits are 2 and -2

Answer 1

First, take the basic interval. Raise the power of each part and then divide by the new exponent: (x^2+1)/(2+1) + (6^0+1)/(0+1)
This will give you (x^3)/3 + 6x
Then, to evaluate it with the given limits, evaluate it from the lower bound (-2) and subtract that from the evaluation of the upper bound (2)

Answer 2

Cool, thanks!

Answer 3

\({(-2)^3\over 3}+6(-2)-({(2)^3 \over3}+6(2))\)

\(({-8 \over 3}-12)-( {8 \over 3} +12)\)

\( {-16 \over 3} -24={-88 \over 3}\)