What is the integral from 0 to 7 of (e^t+e^{-t}) ?

The integral from 0 to 7 of (e^t+e^{-t}) is (e^{14}-1)/(e^7)

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integral from 0 to 7 of (e^t+e^{-t})

\int_{0}^{7} (e^{t} + e^{- t}) d t

\frac{e ^{14} - 1}{e ^{7}}

Decimal Notation

1096.63224 \dots

Solution steps

\int_{0}^{7} e^{t} + e^{- t} d t

Rewrite using trig identities

= \int_{0}^{7} 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}^{7} cosh (t) d t

Use the common integral:

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

= 2 [sinh (t)]_{0}^{7}

Compute the boundaries:

\frac{e ^{14} - 1}{2 e ^{7}}

= 2 \cdot \frac{e ^{14} - 1}{2 e ^{7}}

Simplify

= \frac{e ^{14} - 1}{e ^{7}}

x \to 0 + lim ((1 + x)^{\frac{1}{x}})

y = 3 e^{- x}, (0, 3)

\frac{d y}{d x} + \frac{3 x y}{x ^{2} + 1} = \frac{6 x}{x ^{2} + 1}

\int \frac{1}{4 + 8 x} d x

\int 7 x^{\frac{2}{5}} + 3 x^{- \frac{4}{5}} d x

What is the integral from 0 to 7 of (e^t+e^{-t}) ?
The integral from 0 to 7 of (e^t+e^{-t}) is (e^{14}-1)/(e^7)