What is the integral from 0 to 1 of e^x+e^{-x} ?

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

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integral from 0 to 1 of e^x+e^{-x}

\int_{0}^{1} e^{x} + e^{- x} d x

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

Decimal Notation

2.35040 \dots

Solution steps

\int_{0}^{1} e^{x} + e^{- x} d x

Rewrite using trig identities

= \int_{0}^{1} 2 cosh (x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} cosh (x) d x

Use the common integral:

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

= 2 [sinh (x)]_{0}^{1}

Compute the boundaries:

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

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

Simplify

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

\frac{d}{d t} (ln ((\frac{a + b t}{a - b t})^{\frac{1}{2}}))

arctan (9 t)

\int_{0}^{2} x^{2} e^{- 9 x} d x

\frac{d}{d x} (6 (sec (x) - tan (x))^{\frac{3}{2}})

x^{2} - 11, 10 x, [- 11, 0] y = 0

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