How do you evaluate the definite integral #int (5-16x^-3) dx# from #[1,2]#?

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Answer 1 · 2016-08-11T15:53:05

#int_1^2 (5-16x^-3)d x=1#

#int_1^2 (5-16x^-3)d x=?#

#int_1^2 (5-16x^-3)d x=|5x+16*1/2x^-2|_1^2#

#int_1^2 (5-16x^-3)d x=(5*2+8*2^-2)-(5*1+8*1^-2)#

#int_1^2 (5-16x^-3)d x=(10+8/4)-(5+8/1)#

#int_1^2 (5-16x^-3)d x=14-13#

#int_1^2 (5-16x^-3)d x=1#