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

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

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integral from-3 to 1 of (e^{x+3})

\int_{- 3}^{1} (e^{x + 3}) d x

e^{4} - 1

Decimal Notation

53.59815 \dots

Solution steps

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

e^{x + 3} = e^{3} e^{x}

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

Use the common integral:

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

= e^{3} [e^{x}]_{- 3}^{1}

Compute the boundaries:

e - \frac{1}{e ^{3}}

= e^{3} (e - \frac{1}{e ^{3}})

Simplify

= e^{4} - 1

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