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

The integral from 0 to t of e^{-u}(t-u)^2 is t^2-2t-2e^{-t}+2

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

\int_{0}^{t} e^{- u} (t - u)^{2} d u

t^{2} - 2 t - 2 e^{- t} + 2

Solution steps

\int_{0}^{t} e^{- u} (t - u)^{2} d u

Apply u-substitution

: \int_{0}^{- t} - e^{v} (t + v)^{2} d v

= \int_{0}^{- t} - e^{v} (t + v)^{2} d v

Take the constant out:

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

= - \int_{0}^{- t} e^{v} (t + v)^{2} d v

Expand

e^{v} (t + v)^{2} : t^{2} e^{v} + 2 t e^{v} v + e^{v} v^{2}

= - \int_{0}^{- t} t^{2} e^{v} + 2 t e^{v} v + e^{v} v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int_{0}^{- t} t^{2} e^{v} d v + \int_{0}^{- t} 2 t e^{v} v d v + \int_{0}^{- t} e^{v} v^{2} d v)

\int_{0}^{- t} t^{2} e^{v} d v = t^{2} (e^{- t} - 1)

\int_{0}^{- t} 2 t e^{v} v d v = 2 t (- t e^{- t} - e^{- t} + 1)

\int_{0}^{- t} e^{v} v^{2} d v = t^{2} e^{- t} - 2 (- t e^{- t} - e^{- t}) - 2

= - (t^{2} (e^{- t} - 1) + 2 t (- t e^{- t} - e^{- t} + 1) + t^{2} e^{- t} - 2 (- t e^{- t} - e^{- t}) - 2)

Simplify

- (t^{2} (e^{- t} - 1) + 2 t (- t e^{- t} - e^{- t} + 1) + t^{2} e^{- t} - 2 (- t e^{- t} - e^{- t}) - 2) : t^{2} - 2 t - 2 e^{- t} + 2

= t^{2} - 2 t - 2 e^{- t} + 2

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\int_{0}^{10} x^{2} - 5 d x

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