What is the integral of 2x^1e^{-y} ?

The integral of 2x^1e^{-y} is e^{-y}x^2+C

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integral of 2x^1e^{-y}

\int 2 x^{1} e^{- y}

e^{- y} x^{2} + C

Solution steps

\int 2 x^{1} e^{- y} d x

Take the constant out:

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

= 2 e^{- y} \cdot \int x^{1} d x

Apply the Power Rule

: \frac{x ^{2}}{2}

= 2 e^{- y} \frac{x ^{2}}{2}

Simplify

2 e^{- y} \frac{x ^{2}}{2} : e^{- y} x^{2}

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

Add a constant to the solution

= e^{- y} x^{2} + C

\int \frac{x + 9}{x ^{2} + 16} d x

\frac{d}{d x} (9 x^{2} - 2)

\int \frac{6 x ^{2} - 4 x - 25}{x - 2} d x

h \to 0 lim (\frac{( 1 + h ) - 1}{h})

x \to - 2 lim (x^{2} - 3)