What is the integral from 0 to 3 of-e^{-2y} ?

The integral from 0 to 3 of-e^{-2y} is -(e^6-1)/(2e^6)

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integral from 0 to 3 of-e^{-2y}

\int_{0}^{3} - e^{- 2 y} d y

- \frac{e ^{6} - 1}{2 e ^{6}}

Decimal

- 0.49876 \dots

Solution steps

\int_{0}^{3} - e^{- 2 y} d y

Take the constant out:

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

= - \int_{0}^{3} e^{- 2 y} d y

Apply u-substitution

= - \int_{0}^{- 6} - \frac{1}{2} e^{u} d u

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

= - (- \int_{- 6}^{0} - \frac{1}{2} e^{u} d u)

Take the constant out:

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

= - (- (- \frac{1}{2} \cdot \int_{- 6}^{0} e^{u} d u))

Use the common integral:

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

= - (- (- \frac{1}{2} [e^{u}]_{- 6}^{0}))

Simplify

= - \frac{1}{2} [e^{u}]_{- 6}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{6}}

= - \frac{1}{2} (1 - \frac{1}{e ^{6}})

Simplify

= - \frac{e ^{6} - 1}{2 e ^{6}}

What is the integral from 0 to 3 of-e^{-2y} ?
The integral from 0 to 3 of-e^{-2y} is -(e^6-1)/(2e^6)