What is the integral from-3 to 1 of 2e^x ?

The integral from-3 to 1 of 2e^x is 2(e-1/(e^3))

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integral from-3 to 1 of 2e^x

\int_{- 3}^{1} 2 e^{x} d x

2 (e - \frac{1}{e ^{3}})

Decimal Notation

5.33698 \dots

Solution steps

\int_{- 3}^{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_{- 3}^{1} e^{x} d x

Use the common integral:

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

= 2 [e^{x}]_{- 3}^{1}

Compute the boundaries:

e - \frac{1}{e ^{3}}

= 2 (e - \frac{1}{e ^{3}})

x \to \infty lim (0.7 x^{2})

\frac{\partial}{\partial x} (x^{a} (\frac{y}{x})^{1 - a})

\frac{\partial}{\partial y} (ln (y) + 2 x y^{3})

n = 0 \sum \infty n! x^{n}

\int 2 sin (t) cos (t) e^{- c o s (t)} d t

What is the integral from-3 to 1 of 2e^x ?
The integral from-3 to 1 of 2e^x is 2(e-1/(e^3))