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

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

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integral from 0 to 1 of e^{-3x}

\int_{0}^{1} e^{- 3 x} d x

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

Decimal

0.31673 \dots

Solution steps

\int_{0}^{1} e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{- 3} - \frac{1}{3} e^{u} d u

= \int_{0}^{- 3} - \frac{1}{3} e^{u} d u

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

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

Use the common integral:

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

= - (- \frac{1}{3} [e^{u}]_{- 3}^{0})

Simplify

= \frac{1}{3} [e^{u}]_{- 3}^{0}

Compute the boundaries:

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

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

Simplify

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

\frac{\partial}{\partial x} (\frac{1}{y} - 2 x sin (y))

\frac{d}{d x} (\frac{( e ^{x} - 1 )}{x})

y^{^{''}} - y^{^{'}} - 2 y = - 2 t + 6 t^{2}

\frac{d}{d x} (e^{s i n^{2} (x)})

y = \frac{2}{x}, y = 8 x, y = \frac{1}{2} x

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