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Answer to Question #110051 in Calculus for Nimra

Question #110051

²∫¹ sinx dx


1
Expert's answer
2020-04-17T18:54:25-0400

"\\int_1^2 \\sin(x)dx=" -cos(2)-(-cos(1))=-cos(2)+cos(1)=cos(1)-cos(2).










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