What is the integral of e^{-λx} ?

The integral of e^{-λx} is -1/λ e^{-λx}+C

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integral of e^{-λx}

\int e^{- λ x} d x

- \frac{1}{λ} e^{- λ x} + C

Solution steps

\int e^{- λ x} d x

Apply u-substitution

: \int - \frac{e ^{u}}{λ} d u

= \int - \frac{e ^{u}}{λ} d u

Take the constant out:

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

= - \frac{1}{λ} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{1}{λ} e^{u}

Substitute back

u = - λ x

= - \frac{1}{λ} e^{- λ x}

Add a constant to the solution

= - \frac{1}{λ} e^{- λ x} + C

f (x) = x^{2} - 6

x \to - 1 + lim (\frac{x}{x ^{2} + 1})

x \to - 1 lim (6 x^{2} + 5 x - 1)

\frac{d}{d x} (csc^{11} (x))

y = 7^{x}