What is the integral from 0 to t of y(t-w)e^{-4w} ?

The integral from 0 to t of y(t-w)e^{-4w} is y(-(t(e^{-4t}-1))/4-1/16 (-4te^{-4t}-e^{-4t}+1))

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integral from 0 to t of y(t-w)e^{-4w}

\int_{0}^{t} y (t - w) e^{- 4 w} d w

y (- \frac{t ( e ^{- 4 t} - 1 )}{4} - \frac{1}{16} (- 4 t e^{- 4 t} - e^{- 4 t} + 1))

Solution steps

\int_{0}^{t} y (t - w) e^{- 4 w} d w

Take the constant out:

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

= y \cdot \int_{0}^{t} (t - w) e^{- 4 w} d w

Expand

(t - w) e^{- 4 w} : t e^{- 4 w} - e^{- 4 w} w

= y \cdot \int_{0}^{t} t e^{- 4 w} - e^{- 4 w} w d w

Apply the Sum Rule:

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

= y (\int_{0}^{t} t e^{- 4 w} d w - \int_{0}^{t} e^{- 4 w} w d w)

\int_{0}^{t} t e^{- 4 w} d w = - \frac{t ( e ^{- 4 t} - 1 )}{4}

\int_{0}^{t} e^{- 4 w} w d w = \frac{1}{16} (- 4 t e^{- 4 t} - e^{- 4 t} + 1)

= y (- \frac{t ( e ^{- 4 t} - 1 )}{4} - \frac{1}{16} (- 4 t e^{- 4 t} - e^{- 4 t} + 1))

\int_{0}^{1} y e^{- y^{2}} d y

\int_{- 4}^{1} 2 - x d x

\int_{0}^{5} (x - 5)^{2} d x

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

\int_{1}^{10} 1

What is the integral from 0 to t of y(t-w)e^{-4w} ?
The integral from 0 to t of y(t-w)e^{-4w} is y(-(t(e^{-4t}-1))/4-1/16 (-4te^{-4t}-e^{-4t}+1))