What is the integral from 1 to 3 of 2xe^{-4x} ?

The integral from 1 to 3 of 2xe^{-4x} is -(-5e^8+13)/(8e^{12)}

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integral from 1 to 3 of 2xe^{-4x}

\int_{1}^{3} 2 x e^{- 4 x} d x

- \frac{- 5 e ^{8} + 13}{8 e ^{12}}

Decimal Notation

0.01143 \dots

Solution steps

\int_{1}^{3} 2 x e^{- 4 x} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{3} x e^{- 4 x} d x

Apply u-substitution

: \int_{- 4}^{- 12} \frac{e ^{u} u}{16} d u

= 2 \cdot \int_{- 4}^{- 12} \frac{e ^{u} u}{16} d u

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

= 2 (- \int_{- 12}^{- 4} \frac{e ^{u} u}{16} d u)

Take the constant out:

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

= 2 (- \frac{1}{16} \cdot \int_{- 12}^{- 4} e^{u} u d u)

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 12}^{- 4}

= 2 (- \frac{1}{16} [e^{u} u - \int e^{u} d u]_{- 12}^{- 4})

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

= 2 (- \frac{1}{16} [e^{u} u - e^{u}]_{- 12}^{- 4})

Simplify

2 (- \frac{1}{16} [e^{u} u - e^{u}]_{- 12}^{- 4}) : - \frac{1}{8} [e^{u} u - e^{u}]_{- 12}^{- 4}

= - \frac{1}{8} [e^{u} u - e^{u}]_{- 12}^{- 4}

Compute the boundaries:

- \frac{5}{e ^{4}} + \frac{13}{e ^{12}}

= - \frac{1}{8} (- \frac{5}{e ^{4}} + \frac{13}{e ^{12}})

Simplify

= - \frac{- 5 e ^{8} + 13}{8 e ^{12}}

\int 3 x^{2} + 4 x + 1 d x

\frac{d y}{d x} + \frac{1}{x + 4} y = x^{- 2}

f (x) = \frac{9 x}{e ^{x}}

\int_{0}^{3} (2 x^{2} - 6 x) d x

x \to 0 lim (\frac{2}{x} - \frac{2}{∣ x ∣})

What is the integral from 1 to 3 of 2xe^{-4x} ?
The integral from 1 to 3 of 2xe^{-4x} is -(-5e^8+13)/(8e^{12)}