How do you evaluate #\int _ { 2} ^ { x} ( 3t ^ { 2} - 1) d t #?

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Answer 1 · 2017-10-26T13:24:37

#x^3 - x - 6#

Your antiderivative is going to be #t^3 - t#

Evaluate that at t = 2:

#2^3-2 = 6#

Evaluate at t = x:

#x^3 - x#

...and subtract the lower limit value from the upper:

#x^3 - x - 6#
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