# int_0^1(cosh(t))dt Evaluate the integral.

### Textbook Question

Chapter 5, 5.3 - Problem 38 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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We have to evaluate the integral \int_{0}^{1}cosh(t)dt 

We know that the integral of cosh(t) = sinh(t) . Therefore we can write,

\int_{0}^{1}cosh(t)dt=[sinh(t)]_{0}^{1}

=sinh(1)-sinh(0)

=sinh(1)

= 1.175

scisser | (Level 3) Honors

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The antiderivative of cosh(t)=sinh(t)

Therefore,

int_0^1(cosh(t))dt=sinh(t)|_0^1

Plug in the upper value and subtract the lower limit

=sinh(1)-sinh(0)

=1.17