What is the integral of 50+3.5xe^{-0.01x^2} ?

The integral of 50+3.5xe^{-0.01x^2} is 50x-175e^{-0.01x^2}+C

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integral of 50+3.5xe^{-0.01x^2}

\int 50 + 3.5 x e^{- 0.01 x^{2}} d x

50 x - 175 e^{- 0.01 x^{2}} + C

Solution steps

\int 50 + 3.5 x e^{- 0.01 x^{2}} 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 50 d x + \int 3.5 x e^{- 0.01 x^{2}} d x

\int 50 d x = 50 x

\int 3.5 x e^{- 0.01 x^{2}} d x = - 175 e^{- 0.01 x^{2}}

= 50 x - 175 e^{- 0.01 x^{2}}

Add a constant to the solution

= 50 x - 175 e^{- 0.01 x^{2}} + C

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

y^{^{'}} = \frac{( 3 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2} )}{x ^{5} y}

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

\int_{0}^{π} (sin^{2} (x)) d x

What is the integral of 50+3.5xe^{-0.01x^2} ?
The integral of 50+3.5xe^{-0.01x^2} is 50x-175e^{-0.01x^2}+C