What is the integral of x(4-10e^{-0.2x}) ?

The integral of x(4-10e^{-0.2x}) is 2x^2-250(-0.2e^{-0.2x}x-e^{-0.2x})+C

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integral of x(4-10e^{-0.2x})

\int x (4 - 10 e^{- 0.2 x}) d x

2 x^{2} - 250 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x}) + C

Solution steps

\int x (4 - 10 e^{- 0.2 x}) d x

Expand

x (4 - 10 e^{- 0.2 x}) : 4 x - 10 e^{- 0.2 x} x

= \int 4 x - 10 e^{- 0.2 x} x d x

Apply the Sum Rule:

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

= \int 4 x d x - \int 10 e^{- 0.2 x} x d x

\int 4 x d x = 2 x^{2}

\int 10 e^{- 0.2 x} x d x = 250 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x})

= 2 x^{2} - 250 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x})

Add a constant to the solution

= 2 x^{2} - 250 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x}) + C

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What is the integral of x(4-10e^{-0.2x}) ?
The integral of x(4-10e^{-0.2x}) is 2x^2-250(-0.2e^{-0.2x}x-e^{-0.2x})+C