Asked by gregory
definite integral of (e^(8x))cosh(8x)dx from 0 to 1/16
-I substitued the definition of cosh in, i.e. (e^(8x) + e^(-8x)/2 but i cant get the "e's" to combine and i know the answer is e/32
-I substitued the definition of cosh in, i.e. (e^(8x) + e^(-8x)/2 but i cant get the "e's" to combine and i know the answer is e/32
Answers
Answered by
Reiny
you had
∫ e^(8x) (e^(8x) + e^(-8x) )/2 dx
which is
= ∫ ( e^(16x) + e^0 )/2 dx
= (1/2)(1/16)e^(16x) + (1/2)x | from 0 to 1/16
= (1/2)(1/16) e^1 + 1/32 - (1/2)(1/16)e^0 - 0
= (1/32)e
or
= e/32
∫ e^(8x) (e^(8x) + e^(-8x) )/2 dx
which is
= ∫ ( e^(16x) + e^0 )/2 dx
= (1/2)(1/16)e^(16x) + (1/2)x | from 0 to 1/16
= (1/2)(1/16) e^1 + 1/32 - (1/2)(1/16)e^0 - 0
= (1/32)e
or
= e/32
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