What is the integral from 0 to x of 10e^{-2x-3y} ?

The integral from 0 to x of 10e^{-2x-3y} is -10/3 (e^{-5x}-e^{-2x})

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integral from 0 to x of 10e^{-2x-3y}

\int_{0}^{x} 10 e^{- 2 x - 3 y} d y

- \frac{10}{3} (e^{- 5 x} - e^{- 2 x})

Solution steps

\int_{0}^{x} 10 e^{- 2 x - 3 y} d y

Take the constant out:

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

= 10 \cdot \int_{0}^{x} e^{- 2 x - 3 y} d y

Apply u-substitution

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

= 10 \cdot \int_{- 2 x}^{- 5 x} - \frac{1}{3} e^{u} d u

Take the constant out:

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

= 10 (- \frac{1}{3} \cdot \int_{- 2 x}^{- 5 x} e^{u} d u)

Use the common integral:

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

= 10 (- \frac{1}{3} [e^{u}]_{- 2 x}^{- 5 x})

Simplify

10 (- \frac{1}{3} [e^{u}]_{- 2 x}^{- 5 x}) : - \frac{10}{3} [e^{u}]_{- 2 x}^{- 5 x}

= - \frac{10}{3} [e^{u}]_{- 2 x}^{- 5 x}

Compute the boundaries:

e^{- 5 x} - e^{- 2 x}

= - \frac{10}{3} (e^{- 5 x} - e^{- 2 x})

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What is the integral from 0 to x of 10e^{-2x-3y} ?
The integral from 0 to x of 10e^{-2x-3y} is -10/3 (e^{-5x}-e^{-2x})