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

The integral from-1 to 2 of e^{-3x} is (e^9-1)/(3e^6)

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

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

\frac{e ^{9} - 1}{3 e ^{6}}

Decimal Notation

6.69435 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{- 6} - \frac{1}{3} e^{u} d u

= \int_{3}^{- 6} - \frac{1}{3} e^{u} d u

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

= - \int_{- 6}^{3} - \frac{1}{3} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{- 6}^{3} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{3} [e^{u}]_{- 6}^{3})

Simplify

= \frac{1}{3} [e^{u}]_{- 6}^{3}

Compute the boundaries:

e^{3} - \frac{1}{e ^{6}}

= \frac{1}{3} (e^{3} - \frac{1}{e ^{6}})

Simplify

= \frac{e ^{9} - 1}{3 e ^{6}}

\int_{1}^{6} \frac{4 ( ln ( x ) ) ^{3}}{x} d x

\int_{0}^{3} 3^{s} d s

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

\int_{0}^{1} \frac{x ^{2}}{( 1 + x ) ^{4}} d x

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

What is the integral from-1 to 2 of e^{-3x} ?
The integral from-1 to 2 of e^{-3x} is (e^9-1)/(3e^6)