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

The integral from 2 to 3 of 2e^{-2x} is 1/(e^4)-1/(e^6)

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

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

\frac{1}{e ^{4}} - \frac{1}{e ^{6}}

Decimal

0.01583 \dots

Solution steps

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

Apply u-substitution

: \int_{- 4}^{- 6} - \frac{1}{2} e^{u} d u

= 2 \cdot \int_{- 4}^{- 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}^{- 4} - \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}^{- 4} e^{u} d u))

Use the common integral:

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

= 2 (- (- \frac{1}{2} [e^{u}]_{- 6}^{- 4}))

Simplify

2 (- (- \frac{1}{2} [e^{u}]_{- 6}^{- 4})) : [e^{u}]_{- 6}^{- 4}

= [e^{u}]_{- 6}^{- 4}

Compute the boundaries:

\frac{1}{e ^{4}} - \frac{1}{e ^{6}}

= \frac{1}{e ^{4}} - \frac{1}{e ^{6}}

\int_{0}^{\infty} 3 e^{- s t} d t

\int_{0}^{1} (2 e^{x} - \frac{5}{e ^{x}})^{2} d x

\int_{10}^{x} 9 t^{2} d t

\int_{0}^{\infty} e^{- 2 a t} d t

\int_{- 1}^{1} x^{2} 4 - x^{2} d x

What is the integral from 2 to 3 of 2e^{-2x} ?
The integral from 2 to 3 of 2e^{-2x} is 1/(e^4)-1/(e^6)