What is the integral from 1 to sqrt(z of)ye^{y^2-z} ?

The integral from 1 to sqrt(z of)ye^{y^2-z} is 1/2 e^{-z}(e^z-e)

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integral from 1 to sqrt(z of)ye^{y^2-z}

\int_{1}^{z} y e^{y^{2} - z} d y

\frac{1}{2} e^{- z} (e^{z} - e)

Solution steps

\int_{1}^{z} y e^{y^{2} - z} d y

Apply u-substitution

: \int_{1}^{z} \frac{e ^{u - z}}{2} d u

= \int_{1}^{z} \frac{e ^{u - z}}{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_{1}^{z} e^{u - z} d u

e^{u - z} = e^{- z} e^{u}

= \frac{1}{2} \cdot \int_{1}^{z} e^{- z} 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} e^{- z} \cdot \int_{1}^{z} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

e^{z} - e

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

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What is the integral from 1 to sqrt(z of)ye^{y^2-z} ?
The integral from 1 to sqrt(z of)ye^{y^2-z} is 1/2 e^{-z}(e^z-e)