What is the integral from 0 to 2 of 6e^{-2t} ?

The integral from 0 to 2 of 6e^{-2t} is 3(1-1/(e^4))

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integral from 0 to 2 of 6e^{-2t}

\int_{0}^{2} 6 e^{- 2 t} d t

3 (1 - \frac{1}{e ^{4}})

Decimal Notation

2.94505 \dots

Solution steps

\int_{0}^{2} 6 e^{- 2 t} d t

Take the constant out:

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

= 6 \cdot \int_{0}^{2} e^{- 2 t} d t

Apply u-substitution

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

= 6 \cdot \int_{0}^{- 4} - \frac{1}{2} e^{u} d u

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

= 6 (- \int_{- 4}^{0} - \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

= 6 (- (- \frac{1}{2} \cdot \int_{- 4}^{0} e^{u} d u))

Use the common integral:

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

= 6 (- (- \frac{1}{2} [e^{u}]_{- 4}^{0}))

Simplify

6 (- (- \frac{1}{2} [e^{u}]_{- 4}^{0})) : 3 [e^{u}]_{- 4}^{0}

= 3 [e^{u}]_{- 4}^{0}

Compute the boundaries:

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

= 3 (1 - \frac{1}{e ^{4}})

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What is the integral from 0 to 2 of 6e^{-2t} ?
The integral from 0 to 2 of 6e^{-2t} is 3(1-1/(e^4))