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

The integral from 1 to 3 of 2e^{-2x} is 1/(e^2)-1/(e^6)

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integral from 1 to 3 of 2e^{-2x}

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

\frac{1}{e ^{2}} - \frac{1}{e ^{6}}

Decimal

0.13285 \dots

Solution steps

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

Apply u-substitution

= 2 \cdot \int_{- 2}^{- 6} - \frac{1}{2} e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{- 6}^{- 2} - \frac{1}{2} e^{u} d u)

Take the constant out:

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

= 2 (- (- \frac{1}{2} \cdot \int_{- 6}^{- 2} e^{u} d u))

Use the common integral:

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

= 2 (- (- \frac{1}{2} [e^{u}]_{- 6}^{- 2}))

Simplify

2 (- (- \frac{1}{2} [e^{u}]_{- 6}^{- 2})) : [e^{u}]_{- 6}^{- 2}

= [e^{u}]_{- 6}^{- 2}

Compute the boundaries:

\frac{1}{e ^{2}} - \frac{1}{e ^{6}}

= \frac{1}{e ^{2}} - \frac{1}{e ^{6}}

What is the integral from 1 to 3 of 2e^{-2x} ?
The integral from 1 to 3 of 2e^{-2x} is 1/(e^2)-1/(e^6)