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How do you evaluate the definite integral #int (x-2) dx# from [-1,0]?

1 Answer

May 26, 2017

#int_(-1)^0 (x-2)dx = - 5/2#

Explanation:

#int_(-1)^0 (x-2)dx = int_(-1)^0 (x-2)d(x-2)#

#int_(-1)^0 (x-2)dx = [(x-2)^2/2]_(-1)^0 #

#int_(-1)^0 (x-2)dx = (0-2)^2/2 - ((-1)-2)^2/2#

#int_(-1)^0 (x-2)dx = 2 - 9/2#

#int_(-1)^0 (x-2)dx = - 5/2#

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