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Question
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Sum
Solution
`int_1^2 (x - 1)/x^2 "d"x = int_1^2 (x/x^2 - 1/x^2) "d"x`
= `int_1^2 (1/x - x^-2) "d"x`
= `int_1^2 1/x "d"x - int_1^2 x^-2 "d"x`
= `[log|x|]_1^2 - [((x^(2 + 1))/(-2 + 1))]^2`
= `{log|2| - log|1|} - {1/x}_1^2`
= `[log 2 - 0] + [1/2 - 1/1]`
= `log 2 + [(1 - 2)/2]`
= `log 2 - 1/2`
= `1/2 [2 log 2 - 1]`
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Definite Integrals
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