What is the integral from 0 to 2 of 6e^x ?

The integral from 0 to 2 of 6e^x is 6(e^2-1)

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integral from 0 to 2 of 6e^x

\int_{0}^{2} 6 e^{x} d x

6 (e^{2} - 1)

Decimal Notation

38.33433 \dots

Solution steps

\int_{0}^{2} 6 e^{x} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{2} e^{x} d x

Use the common integral:

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

= 6 [e^{x}]_{0}^{2}

Compute the boundaries:

e^{2} - 1

= 6 (e^{2} - 1)

\int_{0}^{1} x^{13} + 1 3^{x} d x

\int_{1}^{3} (\frac{1}{x})^{2} d x

\int_{- 1}^{1} (2 x + 1) d x

\int_{1}^{6} 2 x d x

\int_{2}^{6} x^{2} - 7 x + 6 d x

What is the integral from 0 to 2 of 6e^x ?
The integral from 0 to 2 of 6e^x is 6(e^2-1)