What is the integral from-1 to 1 of 2e^x ?

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

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integral from-1 to 1 of 2e^x

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

2 (e - \frac{1}{e})

Decimal Notation

4.70080 \dots

Solution steps

\int_{- 1}^{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_{- 1}^{1} e^{x} d x

Use the common integral:

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

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

Compute the boundaries:

e - \frac{1}{e}

= 2 (e - \frac{1}{e})

f (x) = \frac{2}{x - 3}, at x = 5

x \to \infty lim (\frac{( e ^{x} )}{2 x ^{2}})

\int cos (3 x) sin^{2} (3 x) d x

f (t) = \frac{4 t}{4 + t}

\int \frac{9}{1 - x ^{2}} d x

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