How do you evaluate the definite integral #int (t^2+2)dt# from [-1,1]?

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Answer 1 · 2018-06-25T05:32:21

#int_-1^1(t^2+2)dt=14/3#

#int_-1^1(t^2+2)dt#

By power rule we can simply add one to our power, and divide by the resulting power such that

#(t^2+2)rarr(1/3t^3+2t)#

Subbing in our bounds gives

#(1/3(1)^3+2(1))-(1/3(-1)^3+2(-1))#

#7/3--7/3#

#14/3#