What is the integral from-1 to 1 of 1/(e^{5t)} ?

The integral from-1 to 1 of 1/(e^{5t)} is (e^{10}-1)/(5e^5)

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integral from-1 to 1 of 1/(e^{5t)}

\int_{- 1}^{1} \frac{1}{e ^{5 t}} d t

\frac{e ^{10} - 1}{5 e ^{5}}

Decimal

29.68128 \dots

Solution steps

\int_{- 1}^{1} \frac{1}{e ^{5 t}} d t

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{5 t}} = e^{- 5 t}

= \int_{- 1}^{1} e^{- 5 t} d t

Apply u-substitution

= \int_{5}^{- 5} - \frac{1}{5} e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 5}^{5} - \frac{1}{5} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{5} \cdot \int_{- 5}^{5} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{5} [e^{u}]_{- 5}^{5})

Simplify

= \frac{1}{5} [e^{u}]_{- 5}^{5}

Compute the boundaries:

e^{5} - \frac{1}{e ^{5}}

= \frac{1}{5} (e^{5} - \frac{1}{e ^{5}})

Simplify

= \frac{e ^{10} - 1}{5 e ^{5}}

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What is the integral from-1 to 1 of 1/(e^{5t)} ?
The integral from-1 to 1 of 1/(e^{5t)} is (e^{10}-1)/(5e^5)