What is the integral from 0 to ln(2) of (e^t-e^{-t}) ?

The integral from 0 to ln(2) of (e^t-e^{-t}) is 1/2

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integral from 0 to ln(2) of (e^t-e^{-t})

\int_{0}^{l n (2)} (e^{t} - e^{- t}) d t

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{0}^{l n (2)} e^{t} - e^{- t} d t

Rewrite using trig identities

= \int_{0}^{l n (2)} 2 sinh (t) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{l n (2)} sinh (t) d t

Use the common integral:

\int sinh (t) d t = cosh (t)

= 2 [cosh (t)]_{0}^{l n (2)}

Compute the boundaries:

\frac{1}{4}

= 2 \cdot \frac{1}{4}

Simplify

= \frac{1}{2}

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What is the integral from 0 to ln(2) of (e^t-e^{-t}) ?
The integral from 0 to ln(2) of (e^t-e^{-t}) is 1/2