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

The integral from 0 to ln(u) of 2e^{-2y} is -1/(u^2)+1

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

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

- \frac{1}{u ^{2}} + 1

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{0}^{l n (u)} e^{- 2 y} d y

Apply u-substitution

: \int_{0}^{- 2 l n (u)} - \frac{1}{2} e^{v} d v

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

2 (- \frac{1}{2} [e^{v}]_{0}^{- 2 l n (u)}) : - [e^{v}]_{0}^{- 2 l n (u)}

= - [e^{v}]_{0}^{- 2 l n (u)}

Compute the boundaries:

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

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

Simplify

= - \frac{1}{u ^{2}} + 1

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