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

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

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

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

e^{- x} (x - 1)

Solution steps

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

Take the constant out:

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

= 2 e^{- x} \cdot \int_{1}^{x} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{1}^{x}

= 2 e^{- x} [\frac{y ^{2}}{2}]_{1}^{x}

Compute the boundaries:

\frac{x}{2} - \frac{1}{2}

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

Simplify

= e^{- x} (x - 1)

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\int_{\frac{3}{2}}^{2} \frac{1}{- x ^{2} + 4 x - 3} d x

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