What is the integral from 0 to t of xe^{-x} ?

The integral from 0 to t of xe^{-x} is -te^{-t}-e^{-t}+1

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integral from 0 to t of xe^{-x}

\int_{0}^{t} x e^{- x} d x

- t e^{- t} - e^{- t} + 1

Solution steps

\int_{0}^{t} x e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x - \int - e^{- x} d x]_{0}^{t}

= [- e^{- x} x - \int - e^{- x} d x]_{0}^{t}

\int - e^{- x} d x = e^{- x}

= [- e^{- x} x - e^{- x}]_{0}^{t}

Compute the boundaries:

- t e^{- t} - e^{- t} + 1

= - t e^{- t} - e^{- t} + 1

x y^{^{'}} - y = 2 x^{2} - 2 x - 2

x y^{^{'}} - y = x, y (1) = 10

\int_{0}^{2} π (3 x^{2})^{2} d x

\frac{\partial}{\partial x} (4 x^{2} + 3 y^{4})

x \to 1 - lim (\frac{1}{x} - 1)

What is the integral from 0 to t of xe^{-x} ?
The integral from 0 to t of xe^{-x} is -te^{-t}-e^{-t}+1