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

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

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integral from 0 to 2 of 2e^x

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

2 (e^{2} - 1)

Decimal Notation

12.77811 \dots

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

e^{2} - 1

= 2 (e^{2} - 1)

(3 t - 1) (2 t - 2)^{- 1}

x \to 2 lim (3 x^{2} - 4 x + 5)

\frac{d}{d x} (\frac{x ^{2} + 4}{x ^{2} - 4})

\int \frac{30 x ^{5}}{2 + 5 x ^{6}} d x

\int \frac{x ^{2} - x + 4}{x ^{3} + 2 x} d x

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