What is the integral from-2 to 2 of 1/4 e^{tx} ?

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

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integral from-2 to 2 of 1/4 e^{tx}

\int_{- 2}^{2} \frac{1}{4} e^{t x} d x

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

Solution steps

\int_{- 2}^{2} \frac{1}{4} e^{t x} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int_{- 2}^{2} e^{t x} d x

Apply u-substitution

: \int_{- 2 t}^{t \cdot 2} \frac{e ^{u}}{t} d u

= \frac{1}{4} \cdot \int_{- 2 t}^{t \cdot 2} \frac{e ^{u}}{t} d u

Take the constant out:

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

= \frac{1}{4} \cdot \frac{1}{t} \cdot \int_{- 2 t}^{t \cdot 2} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{4} \cdot \frac{1}{t} [e^{u}]_{- 2 t}^{t \cdot 2}

Simplify

\frac{1}{4} \cdot \frac{1}{t} [e^{u}]_{- 2 t}^{t \cdot 2} : \frac{[ e ^{u} ] _{- 2 t}^{2 t}}{4 t}

= \frac{[ e ^{u} ] _{- 2 t}^{2 t}}{4 t}

Compute the boundaries:

e^{2 t} - e^{- 2 t}

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

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