What is the integral of e^{1/2 x-3} ?

The integral of e^{1/2 x-3} is 2e^{1/2 (x-6)}+C

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integral of e^{1/2 x-3}

\int e^{\frac{1}{2} x - 3} d x

2 e^{\frac{1}{2} (x - 6)} + C

Solution steps

\int e^{\frac{1}{2} x - 3} d x

e^{\frac{1}{2} x - 3} = e^{- 3} e^{\frac{1}{2} x}

= \int e^{- 3} e^{\frac{1}{2} x} d x

Take the constant out:

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

= e^{- 3} \cdot \int e^{\frac{1}{2} x} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{3}} \cdot \int e^{\frac{1}{2} x} d x

Apply u-substitution

: \int e^{u} \cdot 2 d u

= \frac{1}{e ^{3}} \cdot \int e^{u} \cdot 2 d u

Take the constant out:

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

= \frac{1}{e ^{3}} \cdot 2 \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{e ^{3}} \cdot 2 e^{u}

Substitute back

u = \frac{1}{2} x

= \frac{1}{e ^{3}} \cdot 2 e^{\frac{1}{2} x}

Simplify

\frac{1}{e ^{3}} \cdot 2 e^{\frac{1}{2} x} : 2 e^{\frac{1}{2} (x - 6)}

= 2 e^{\frac{1}{2} (x - 6)}

Add a constant to the solution

= 2 e^{\frac{1}{2} (x - 6)} + C

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What is the integral of e^{1/2 x-3} ?
The integral of e^{1/2 x-3} is 2e^{1/2 (x-6)}+C