What is the integral from 0 to y of 15e^{-2x-3y} ?

The integral from 0 to y of 15e^{-2x-3y} is -15/2 (e^{-5y}-e^{-3y})

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integral from 0 to y of 15e^{-2x-3y}

\int_{0}^{y} 15 e^{- 2 x - 3 y} d x

- \frac{15}{2} (e^{- 5 y} - e^{- 3 y})

Solution steps

\int_{0}^{y} 15 e^{- 2 x - 3 y} d x

Take the constant out:

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

= 15 \cdot \int_{0}^{y} e^{- 2 x - 3 y} d x

Apply u-substitution

: \int_{- 3 y}^{- 5 y} - \frac{1}{2} e^{u} d u

= 15 \cdot \int_{- 3 y}^{- 5 y} - \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

= 15 (- \frac{1}{2} \cdot \int_{- 3 y}^{- 5 y} e^{u} d u)

Use the common integral:

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

= 15 (- \frac{1}{2} [e^{u}]_{- 3 y}^{- 5 y})

Simplify

15 (- \frac{1}{2} [e^{u}]_{- 3 y}^{- 5 y}) : - \frac{15}{2} [e^{u}]_{- 3 y}^{- 5 y}

= - \frac{15}{2} [e^{u}]_{- 3 y}^{- 5 y}

Compute the boundaries:

e^{- 5 y} - e^{- 3 y}

= - \frac{15}{2} (e^{- 5 y} - e^{- 3 y})

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\int_{1}^{2} \frac{1}{t + 1} d t

What is the integral from 0 to y of 15e^{-2x-3y} ?
The integral from 0 to y of 15e^{-2x-3y} is -15/2 (e^{-5y}-e^{-3y})