What is the integral of 0.5e^{-0.5(x-1)} ?

The integral of 0.5e^{-0.5(x-1)} is -e^{-0.5(x-1)}+C

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integral of 0.5e^{-0.5(x-1)}

\int 0.5 e^{- 0.5 (x - 1)} d x

- e^{- 0.5 (x - 1)} + C

Solution steps

\int 0.5 e^{- 0.5 (x - 1)} d x

Take the constant out:

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

= 0.5 \cdot \int e^{- 0.5 (x - 1)} d x

Apply u-substitution

: \int - \frac{1}{0.5} e^{u} d u

= 0.5 \cdot \int - \frac{1}{0.5} e^{u} d u

Take the constant out:

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

= 0.5 (- \frac{1}{0.5} \cdot \int e^{u} d u)

Simplify

= 0.5 (- 2 \cdot \int e^{u} d u)

Use the common integral:

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

= 0.5 (- 2 e^{u})

Substitute back

u = - 0.5 (x - 1)

= 0.5 (- 2 e^{- 0.5 (x - 1)})

Simplify

= - e^{- 0.5 (x - 1)}

Add a constant to the solution

= - e^{- 0.5 (x - 1)} + C

\frac{d}{d x} (\frac{1}{x ^{4} + x ^{2} + 1})

x \to - 2 lim (x + 2)

n = 0 \sum \infty \frac{3}{3 ^{n}}

\frac{x ^{2}}{x ^{2} - 9}

x \to 0 lim (\frac{( 3 e ^{x} - 3 )}{x})

What is the integral of 0.5e^{-0.5(x-1)} ?
The integral of 0.5e^{-0.5(x-1)} is -e^{-0.5(x-1)}+C