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Evaluate the integral? : # int_(-1)^1 coshx dx #

Answer 1

# int_(-1)^1 coshx dx = e-1/e " "(~~2.35040) #

#coshx# is an even function so its is symmetrical about #Oy# graph{cosh x [-10, 10, -5, 5]} So, # int_(-1)^1 coshx dx=2 int_0^1 coshx dx #

If we go back to the definition of #cosh x=1/2(e^x+e^-x)# we have:

# int_(-1)^1 coshx dx = 2 int_0^1 1/2(e^x+e^-x) dx # # " " = int_0^1 (e^x+e^-x) dx # # " " = [e^x-e^-x]_0^1 # # " " = (e^1-e^-1) - (e^0-e^0) # # " " = e-1/e " "(~~2.35040) #

We could also use the fact that #int coshx=sinhx + C#

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Answer 2

Adam Adkins

The integral of cosh(x) from -1 to 1 is equal to 2.

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Answer 3

Paisley Johnson

The integral of ( \cosh(x) ) from -1 to 1 is ( \sinh(1) - \sinh(-1) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.