What is the integral from 1 to 2 of-2e^{4t-6} ?

The integral from 1 to 2 of-2e^{4t-6} is -(e^4-1)/(2e^2)

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integral from 1 to 2 of-2e^{4t-6}

\int_{1}^{2} - 2 e^{4 t - 6} d t

- \frac{e ^{4} - 1}{2 e ^{2}}

Decimal Notation

- 3.62686 \dots

Solution steps

\int_{1}^{2} - 2 e^{4 t - 6} d t

Take the constant out:

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

= - 2 \cdot \int_{1}^{2} e^{4 t - 6} d t

Apply u-substitution

: \int_{- 2}^{2} e^{u} \frac{1}{4} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

= - \frac{e ^{4} - 1}{2 e ^{2}}

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What is the integral from 1 to 2 of-2e^{4t-6} ?
The integral from 1 to 2 of-2e^{4t-6} is -(e^4-1)/(2e^2)