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

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

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

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

\frac{2 ( e ^{6} - 1 )}{3 e ^{6}}

Decimal

0.66501 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{- 6} - \frac{1}{3} e^{u} d u

= 2 \cdot \int_{0}^{- 6} - \frac{1}{3} e^{u} d u

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

= 2 (- \int_{- 6}^{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

= 2 (- (- \frac{1}{3} \cdot \int_{- 6}^{0} e^{u} d u))

Use the common integral:

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

= 2 (- (- \frac{1}{3} [e^{u}]_{- 6}^{0}))

Simplify

2 (- (- \frac{1}{3} [e^{u}]_{- 6}^{0})) : \frac{2}{3} [e^{u}]_{- 6}^{0}

= \frac{2}{3} [e^{u}]_{- 6}^{0}

Compute the boundaries:

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

= \frac{2}{3} (1 - \frac{1}{e ^{6}})

Simplify

= \frac{2 ( e ^{6} - 1 )}{3 e ^{6}}

\frac{d}{d x} ((x^{\frac{1}{3}} - 1)^{3})

x y^{^{'}} = y + 3 x^{2} sin (x), y (π) = 0

\int_{2}^{4} \frac{1}{x ^{2} - 25} d x

x \to 0 lim ((4 x)^{7 x})

- 2 x^{2} + x, - 2, 1

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