What is the integral from 0 to 3 of e^{-2x} ?

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

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integral from 0 to 3 of e^{-2x}

\int_{0}^{3} e^{- 2 x} d x

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

Decimal Notation

0.49876 \dots

Solution steps

\int_{0}^{3} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 6} - \frac{1}{2} e^{u} d u

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

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

= - \int_{- 6}^{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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

f (x) = \frac{sin ( x ^{2} )}{ln ( 2 x + 8 )}

\frac{d}{d n} (ln (n^{5} + 5 n))

\int_{0}^{3} \frac{3}{x ( x - 3 )} d x

(7, - 2), (9, 4)

\int \frac{1 - ln ( x )}{x} d x

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