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

The integral from-2 to 0 of 1/(e^{-2x)} is (e^4-1)/(2e^4)

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integral from-2 to 0 of 1/(e^{-2x)}

\int_{- 2}^{0} \frac{1}{e ^{- 2 x}} d x

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

Solution steps

\int_{- 2}^{0} \frac{1}{e ^{- 2 x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int_{- 2}^{0} e^{2 x} 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_{0}^{1} e^{x} + 1 d x

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

\int_{0}^{2 π} 2 d x

\int_{0}^{1} x (x + 1) d x

\int_{0}^{π} 2 d x

What is the integral from-2 to 0 of 1/(e^{-2x)} ?
The integral from-2 to 0 of 1/(e^{-2x)} is (e^4-1)/(2e^4)