What is the integral from 1 to 3 of 4e^{x-1} ?

The integral from 1 to 3 of 4e^{x-1} is (4(e^3-e))/e

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integral from 1 to 3 of 4e^{x-1}

\int_{1}^{3} 4 e^{x - 1} d x

\frac{4 ( e ^{3} - e )}{e}

Decimal Notation

25.55622 \dots

Solution steps

\int_{1}^{3} 4 e^{x - 1} d x

Take the constant out:

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

= 4 \cdot \int_{1}^{3} e^{x - 1} d x

e^{x - 1} = e^{- 1} e^{x}

= 4 \cdot \int_{1}^{3} e^{- 1} e^{x} d x

Take the constant out:

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

= 4 e^{- 1} \cdot \int_{1}^{3} e^{x} d x

Apply exponent rule:

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

= 4 \cdot \frac{1}{e} \cdot \int_{1}^{3} e^{x} d x

Use the common integral:

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

= 4 \cdot \frac{1}{e} [e^{x}]_{1}^{3}

Simplify

4 \cdot \frac{1}{e} [e^{x}]_{1}^{3} : \frac{4}{e} [e^{x}]_{1}^{3}

= \frac{4}{e} [e^{x}]_{1}^{3}

Compute the boundaries:

e^{3} - e

= \frac{4}{e} (e^{3} - e)

Simplify

= \frac{4 ( e ^{3} - e )}{e}

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What is the integral from 1 to 3 of 4e^{x-1} ?
The integral from 1 to 3 of 4e^{x-1} is (4(e^3-e))/e