What is the integral from 0 to 1 of 3e^{-2x} ?

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

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integral from 0 to 1 of 3e^{-2x}

\int_{0}^{1} 3 e^{- 2 x} d x

\frac{3 ( e ^{2} - 1 )}{2 e ^{2}}

Decimal

1.29699 \dots

Solution steps

\int_{0}^{1} 3 e^{- 2 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}^{1} e^{- 2 x} d x

Apply u-substitution

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

= 3 \cdot \int_{0}^{- 2} - \frac{1}{2} e^{u} d u

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

= 3 (- \int_{- 2}^{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

= 3 (- (- \frac{1}{2} \cdot \int_{- 2}^{0} e^{u} d u))

Use the common integral:

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

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

Simplify

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

= \frac{3}{2} [e^{u}]_{- 2}^{0}

Compute the boundaries:

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

= \frac{3}{2} (1 - \frac{1}{e ^{2}})

Simplify

= \frac{3 ( e ^{2} - 1 )}{2 e ^{2}}

n = 1 \sum \infty \frac{7 \cdot n !}{n ^{n}}

\frac{\partial}{\partial x} (- cos (4 x))

t \to 0 lim (\frac{tan ( 6 t )}{sin ( 2 t )})

\int \frac{a x + b}{p x + q} d x

x (\frac{d y}{d x}) - y = x

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