What is the integral of 1000te^{-0.5t} ?

The integral of 1000te^{-0.5t} is 4000(-0.5e^{-0.5t}t-e^{-0.5t})+C

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integral of 1000te^{-0.5t}

\int 1000 t e^{- 0.5 t} d t

4000 (- 0.5 e^{- 0.5 t} t - e^{- 0.5 t}) + C

Solution steps

\int 1000 t e^{- 0.5 t} d t

Take the constant out:

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

= 1000 \cdot \int t e^{- 0.5 t} d t

Apply u-substitution

: \int \frac{2}{0.5} e^{u} u d u

= 1000 \cdot \int \frac{2}{0.5} e^{u} u d u

Take the constant out:

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

= 1000 \cdot \frac{2}{0.5} \cdot \int e^{u} u d u

Simplify

= 1000 \cdot 4 \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= 1000 \cdot 4 (e^{u} u - \int e^{u} d u)

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

= 1000 \cdot 4 (e^{u} u - e^{u})

Substitute back

u = - 0.5 t

= 1000 \cdot 4 (e^{- 0.5 t} (- 0.5 t) - e^{- 0.5 t})

Simplify

= 4000 (- 0.5 e^{- 0.5 t} t - e^{- 0.5 t})

Add a constant to the solution

= 4000 (- 0.5 e^{- 0.5 t} t - e^{- 0.5 t}) + C

y^{^{''}} - 4 y^{^{'}} + 4 y = 120 x^{2} e^{2 x}

x \to - 4 lim (\frac{x ^{2} + 5}{x + 4})

3 x^{2} y^{^{'}} = y^{^{'}} + 3 x e^{- y}

\frac{d}{d z} (- z)

\int \frac{1}{x ^{2} + 4 x + 6} d x

What is the integral of 1000te^{-0.5t} ?
The integral of 1000te^{-0.5t} is 4000(-0.5e^{-0.5t}t-e^{-0.5t})+C