What is the integral from 0 to 3 of 3e^{-x} ?

The integral from 0 to 3 of 3e^{-x} is 3(1-1/(e^3))

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integral from 0 to 3 of 3e^{-x}

\int_{0}^{3} 3 e^{- x} d x

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

Decimal

2.85063 \dots

Solution steps

\int_{0}^{3} 3 e^{- x} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{3} e^{- x} d x

Apply u-substitution

= 3 \cdot \int_{0}^{- 3} - e^{u} d u

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

= 3 (- \int_{- 3}^{0} - e^{u} d u)

Take the constant out:

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

= 3 (- (- \int_{- 3}^{0} e^{u} d u))

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

1 - \frac{1}{e ^{3}}

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

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