What is the integral from-1 to 0 of 3e^{5t-8} ?

The integral from-1 to 0 of 3e^{5t-8} is (3(e^5-1))/(5e^{13)}

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integral from-1 to 0 of 3e^{5t-8}

\int_{- 1}^{0} 3 e^{5 t - 8} d t

\frac{3 ( e ^{5} - 1 )}{5 e ^{13}}

Solution steps

\int_{- 1}^{0} 3 e^{5 t - 8} d t

Take the constant out:

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

= 3 \cdot \int_{- 1}^{0} e^{5 t - 8} d t

Apply u-substitution

: \int_{- 13}^{- 8} e^{u} \frac{1}{5} d u

= 3 \cdot \int_{- 13}^{- 8} e^{u} \frac{1}{5} d u

Take the constant out:

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

= 3 \cdot \frac{1}{5} \cdot \int_{- 13}^{- 8} e^{u} d u

Use the common integral:

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

= 3 \cdot \frac{1}{5} [e^{u}]_{- 13}^{- 8}

Simplify

3 \cdot \frac{1}{5} [e^{u}]_{- 13}^{- 8} : \frac{3}{5} [e^{u}]_{- 13}^{- 8}

= \frac{3}{5} [e^{u}]_{- 13}^{- 8}

Compute the boundaries:

\frac{1}{e ^{8}} - \frac{1}{e ^{13}}

= \frac{3}{5} (\frac{1}{e ^{8}} - \frac{1}{e ^{13}})

Simplify

= \frac{3 ( e ^{5} - 1 )}{5 e ^{13}}

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What is the integral from-1 to 0 of 3e^{5t-8} ?
The integral from-1 to 0 of 3e^{5t-8} is (3(e^5-1))/(5e^{13)}