What is the integral from 1 to R of xe^{-4x} ?

The integral from 1 to R of xe^{-4x} is 1/16 (-4Re^{-4R}-e^{-4R}+5/(e^4))

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integral from 1 to R of xe^{-4x}

\int_{1}^{R} x e^{- 4 x} d x

\frac{1}{16} (- 4 R e^{- 4 R} - e^{- 4 R} + \frac{5}{e ^{4}})

Solution steps

\int_{1}^{R} x e^{- 4 x} d x

Apply u-substitution

: \int_{- 4}^{- 4 R} \frac{e ^{u} u}{16} d u

= \int_{- 4}^{- 4 R} \frac{e ^{u} u}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{- 4}^{- 4 R} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 4}^{- 4 R}

= \frac{1}{16} [e^{u} u - \int e^{u} d u]_{- 4}^{- 4 R}

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

= \frac{1}{16} [e^{u} u - e^{u}]_{- 4}^{- 4 R}

Compute the boundaries:

- 4 R e^{- 4 R} - e^{- 4 R} + \frac{5}{e ^{4}}

= \frac{1}{16} (- 4 R e^{- 4 R} - e^{- 4 R} + \frac{5}{e ^{4}})

\frac{d y}{d x} = (y - 7)^{2}

\frac{d}{d x} (x (x^{2} + 1)^{3})

\frac{\partial}{\partial x} (8 x^{2} y + 5 x y^{3})

y = 7 x^{2} - 4 x, - 74

What is the integral from 1 to R of xe^{-4x} ?
The integral from 1 to R of xe^{-4x} is 1/16 (-4Re^{-4R}-e^{-4R}+5/(e^4))