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AITS & Test SerieshardNUMERICAL

Question

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Answer: 14.00

💡 Solution & Explanation

 n |x| sin x n dx e 1 e     n |x| 0 e dx.                   1 x x 0 50 1 n e n e 1 n 1 1 n 1 ! 3 17 ! So | a b | 14  

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