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

The integral from 0 to ln(5) of e^{2x-y} is 12e^{-y}

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integral from 0 to ln(5) of e^{2x-y}

\int_{0}^{l n (5)} e^{2 x - y} d x

12 e^{- y}

Solution steps

\int_{0}^{l n (5)} e^{2 x - y} d x

Apply u-substitution

: \int_{- y}^{- y + 2 l n (5)} e^{u} \frac{1}{2} d u

= \int_{- y}^{- y + 2 l n (5)} e^{u} \frac{1}{2} 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_{- y}^{- y + 2 l n (5)} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{- y}^{- y + 2 l n (5)}

Compute the boundaries:

24 e^{- y}

= \frac{1}{2} \cdot 24 e^{- y}

Simplify

= 12 e^{- y}

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What is the integral from 0 to ln(5) of e^{2x-y} ?
The integral from 0 to ln(5) of e^{2x-y} is 12e^{-y}