What is the integral from 0 to b of 9e^x ?

The integral from 0 to b of 9e^x is 9(e^b-1)

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integral from 0 to b of 9e^x

\int_{0}^{b} 9 e^{x} d x

9 (e^{b} - 1)

Solution steps

\int_{0}^{b} 9 e^{x} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{b} e^{x} d x

Use the common integral:

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

= 9 [e^{x}]_{0}^{b}

Compute the boundaries:

e^{b} - 1

= 9 (e^{b} - 1)

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What is the integral from 0 to b of 9e^x ?
The integral from 0 to b of 9e^x is 9(e^b-1)