What is the integral of (1+0.5t)e^{-0.004t} ?

The integral of (1+0.5t)e^{-0.004t} is -125e^{-0.004t}t-31500e^{-0.004t}+C

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integral of (1+0.5t)e^{-0.004t}

\int (1 + 0.5 t) e^{- 0.004 t} d t

- 125 e^{- 0.004 t} t - 31500 e^{- 0.004 t} + C

Solution steps

\int (1 + 0.5 t) e^{- 0.004 t} d t

Expand

(1 + 0.5 t) e^{- 0.004 t} : e^{- 0.004 t} + 0.5 e^{- 0.004 t} t

= \int e^{- 0.004 t} + 0.5 e^{- 0.004 t} t d t

Apply the Sum Rule:

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

= \int e^{- 0.004 t} d t + \int 0.5 e^{- 0.004 t} t d t

\int e^{- 0.004 t} d t = - 250 e^{- 0.004 t}

\int 0.5 e^{- 0.004 t} t d t = 31250 (- 0.004 e^{- 0.004 t} t - e^{- 0.004 t})

= - 250 e^{- 0.004 t} + 31250 (- 0.004 e^{- 0.004 t} t - e^{- 0.004 t})

Simplify

- 250 e^{- 0.004 t} + 31250 (- 0.004 e^{- 0.004 t} t - e^{- 0.004 t}) : - 125 e^{- 0.004 t} t - 31500 e^{- 0.004 t}

= - 125 e^{- 0.004 t} t - 31500 e^{- 0.004 t}

Add a constant to the solution

= - 125 e^{- 0.004 t} t - 31500 e^{- 0.004 t} + C

\frac{\partial}{\partial x} (5 x^{2} - 2 y)

\int (x^{\frac{3}{2}} + 8 x + 9) d x

f (x) = - 3 x^{2} + 6 x + 4, at x = 1

x \to 0 lim (\frac{ln ( x + 2 )}{x})

\frac{d y}{d x} = \frac{y}{y - x}

What is the integral of (1+0.5t)e^{-0.004t} ?
The integral of (1+0.5t)e^{-0.004t} is -125e^{-0.004t}t-31500e^{-0.004t}+C