What is the integral from 0 to 3 of (e^t-e^{-t}) ?

The integral from 0 to 3 of (e^t-e^{-t}) is (e^6+1)/(e^3)-2

Popular >

integral from 0 to 3 of (e^t-e^{-t})

\int_{0}^{3} (e^{t} - e^{- t}) d t

\frac{e ^{6} + 1}{e ^{3}} - 2

Decimal

18.13532 \dots

Solution steps

\int_{0}^{3} e^{t} - e^{- t} d t

Rewrite using trig identities

= \int_{0}^{3} 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}^{3} sinh (t) d t

Use the common integral:

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

= 2 [cosh (t)]_{0}^{3}

Compute the boundaries:

\frac{e ^{6} + 1}{2 e ^{3}} - 1

= 2 (\frac{e ^{6} + 1}{2 e ^{3}} - 1)

Simplify

= \frac{e ^{6} + 1}{e ^{3}} - 2

\int_{- 1}^{1} sin^{7} (x) cos (x) d x

\int_{0}^{4 (1 - x)} 6 (1 - x - \frac{y}{4}) d y

\int_{0}^{1} \frac{6 x}{1 + x ^{2}} d x

\int_{- 2}^{2} (x - 1) d x

\int_{0}^{4} (3 x^{2} - 4) d x

What is the integral from 0 to 3 of (e^t-e^{-t}) ?
The integral from 0 to 3 of (e^t-e^{-t}) is (e^6+1)/(e^3)-2