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How do you find the definite integral for: #(2-x^2)dx# for the intervals [2,-1]?

1 Answer

May 23, 2016

#-3#

Explanation:

Definite integral would be

#int_2^-1 (2-x^2)dx#
#=[2x-x^3 /3]_2^-1#

#=(-2+1/3)-(4-8/3)#

#=-6+9/3#

#=-3#

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