How do you evaluate the definite integral $int (x^2-4)dx$ from [2,4]?

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Answer 1 · 2017-02-22T17:41:55

The answer is $=32/3$

We need

$intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

$int_2^4(x^2-4)dx$

$=[x^3/3-4x]_2^4$

$=(64/3-16)-(8/3-8)$

$=64/3-8/3-16+8$

$=56/3-8$

$=32/3$

How do you evaluate the definite integral int (x^2-4)dxint (x^2-4)dx from [2,4]?