What is the integral from 0 to 1 of (2e^x) ?

The integral from 0 to 1 of (2e^x) is 2(e-1)

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integral from 0 to 1 of (2e^x)

\int_{0}^{1} (2 e^{x}) d x

2 (e - 1)

Decimal

3.43656 \dots

Solution steps

\int_{0}^{1} 2 e^{x} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} e^{x} d x

Use the common integral:

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

= 2 [e^{x}]_{0}^{1}

Compute the boundaries:

e - 1

= 2 (e - 1)

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What is the integral from 0 to 1 of (2e^x) ?
The integral from 0 to 1 of (2e^x) is 2(e-1)