What is the integral from 1 to 3 of 7e^{x+2} ?

The integral from 1 to 3 of 7e^{x+2} is 7e^5-7e^3

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integral from 1 to 3 of 7e^{x+2}

\int_{1}^{3} 7 e^{x + 2} d x

7 e^{5} - 7 e^{3}

Decimal

898.29335 \dots

Solution steps

\int_{1}^{3} 7 e^{x + 2} d x

Take the constant out:

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

= 7 \cdot \int_{1}^{3} e^{x + 2} d x

e^{x + 2} = e^{2} e^{x}

= 7 \cdot \int_{1}^{3} e^{2} e^{x} d x

Take the constant out:

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

= 7 e^{2} \cdot \int_{1}^{3} e^{x} d x

Use the common integral:

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

= 7 e^{2} [e^{x}]_{1}^{3}

Compute the boundaries:

e^{3} - e

= 7 e^{2} (e^{3} - e)

Simplify

= 7 e^{5} - 7 e^{3}

What is the integral from 1 to 3 of 7e^{x+2} ?
The integral from 1 to 3 of 7e^{x+2} is 7e^5-7e^3