What is the integral from 0 to 10 of 20te^{-0.6t} ?

The integral from 0 to 10 of 20te^{-0.6t} is 54.59159…

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integral from 0 to 10 of 20te^{-0.6t}

\int_{0}^{10} 20 t e^{- 0.6 t} d t

54.59159 \dots

Solution steps

\int_{0}^{10} 20 t e^{- 0.6 t} d t

Take the constant out:

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

= 20 \cdot \int_{0}^{10} t e^{- 0.6 t} d t

Apply u-substitution

: \int_{0}^{- 6} \frac{5 e ^{u} u}{1.8} d u

= 20 \cdot \int_{0}^{- 6} \frac{5 e ^{u} u}{1.8} d u

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

= 20 (- \int_{- 6}^{0} \frac{5 e ^{u} u}{1.8} d u)

Take the constant out:

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

= 20 (- \frac{5}{1.8} \cdot \int_{- 6}^{0} e^{u} u d u)

Simplify

= 20 (- 2.77777 \dots \cdot \int_{- 6}^{0} e^{u} u d u)

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 6}^{0}

= 20 (- 2.77777 \dots [e^{u} u - \int e^{u} d u]_{- 6}^{0})

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

= 20 (- 2.77777 \dots [e^{u} u - e^{u}]_{- 6}^{0})

Simplify

= - 55.55555 \dots [e^{u} u - e^{u}]_{- 6}^{0}

Compute the boundaries:

- 1 + \frac{7}{e ^{6}}

= - 55.55555 \dots (- 1 + \frac{7}{e ^{6}})

Simplify

= 54.59159 \dots

\int \frac{x + 3}{( x + 1 ) ( x + 2 )} d x

- X^{3} + X, X = - 1, Y = - \frac{1}{2}

\frac{d}{d x} (arccos (\frac{x}{2}))

f (- 1) = 4 x^{2} + 4 x - 3

f (x) = (\frac{x}{x + 1})^{2}

What is the integral from 0 to 10 of 20te^{-0.6t} ?
The integral from 0 to 10 of 20te^{-0.6t} is 54.59159…