What is the integral of (e^{-x}-e^{2x})/2 ?

The integral of (e^{-x}-e^{2x})/2 is 1/2 (-e^{-x}-1/2 e^{2x})+C

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integral of (e^{-x}-e^{2x})/2

\int \frac{e ^{- x} - e ^{2 x}}{2} d x

\frac{1}{2} (- e^{- x} - \frac{1}{2} e^{2 x}) + C

Solution steps

\int \frac{e ^{- x} - e ^{2 x}}{2} d x

Take the constant out:

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

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

= \frac{1}{2} (\int e^{- x} d x - \int e^{2 x} d x)

\int e^{- x} d x = - e^{- x}

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

= \frac{1}{2} (- e^{- x} - \frac{1}{2} e^{2 x})

Add a constant to the solution

= \frac{1}{2} (- e^{- x} - \frac{1}{2} e^{2 x}) + C

\frac{d}{d x} (\frac{x - 3}{x + 3})

\frac{d}{d x} (\frac{m}{x})

f (t) = \frac{2 t}{t ^{2} + 1}

f (x) = \frac{1}{x ^{2}}, at x = 5

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

What is the integral of (e^{-x}-e^{2x})/2 ?
The integral of (e^{-x}-e^{2x})/2 is 1/2 (-e^{-x}-1/2 e^{2x})+C