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

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

$7/3--7/3$

$14/3$

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