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

The integral from-1 to 1 of e^{2x-2} is (e^4-1)/(2e^4)

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integral from-1 to 1 of e^{2x-2}

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

\frac{e ^{4} - 1}{2 e ^{4}}

Decimal Notation

0.49084 \dots

Solution steps

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

Apply u-substitution

: \int_{- 4}^{0} e^{u} \frac{1}{2} d u

= \int_{- 4}^{0} e^{u} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 4}^{0} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{- 4}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{4}}

= \frac{1}{2} (1 - \frac{1}{e ^{4}})

Simplify

= \frac{e ^{4} - 1}{2 e ^{4}}

\int cot (t) d t

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\int sec^{2} (1 - 4 x) d x

\frac{d y}{d x} = e^{x - y}, y (0) = ln (8)

x \to 0 lim ((1 + 6 x)^{\frac{3}{x}})

What is the integral from-1 to 1 of e^{2x-2} ?
The integral from-1 to 1 of e^{2x-2} is (e^4-1)/(2e^4)