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

The integral from 0 to 2 of (e^t+e^{-t}) is (e^4-1)/(e^2)

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

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

\frac{e ^{4} - 1}{e ^{2}}

Decimal

7.25372 \dots

Solution steps

\int_{0}^{2} e^{t} + e^{- t} d t

Rewrite using trig identities

= \int_{0}^{2} 2 cosh (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}^{2} cosh (t) d t

Use the common integral:

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

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

Compute the boundaries:

\frac{e ^{4} - 1}{2 e ^{2}}

= 2 \cdot \frac{e ^{4} - 1}{2 e ^{2}}

Simplify

= \frac{e ^{4} - 1}{e ^{2}}

\int_{e}^{e^{2}} ln (x) d x

n = 0 \sum \infty (- 1)^{2 n + 1}

\int \frac{1}{1 + 6 x} d x

\frac{d}{d x} ((x^{2} + 2)^{2} (x^{4} + 4)^{4})

5 x y \frac{d y}{d x} = 3

What is the integral from 0 to 2 of (e^t+e^{-t}) ?
The integral from 0 to 2 of (e^t+e^{-t}) is (e^4-1)/(e^2)