What is the integral from 120 to 150 of 0.1281*x*e^{-x/(100)} ?

The integral from 120 to 150 of 0.1281*x*e^{-x/(100)} is 134.25119…

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integral from 120 to 150 of 0.1281xe^{-x/(100)}

\int_{120}^{150} 0.1281 \cdot x \cdot e^{- \frac{x}{100}} d x

134.25119 \dots

Solution steps

\int_{120}^{150} 0.1281 x e^{- \frac{x}{100}} d x

Take the constant out:

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

= 0.1281 \cdot \int_{120}^{150} x e^{- \frac{x}{100}} d x

Apply Integration By Parts

: [- 100 e^{- \frac{x}{100}} x - \int - 100 e^{- \frac{x}{100}} d x]_{120}^{150}

= 0.1281 [- 100 e^{- \frac{x}{100}} x - \int - 100 e^{- \frac{x}{100}} d x]_{120}^{150}

\int - 100 e^{- \frac{x}{100}} d x = 10000 e^{- \frac{x}{100}}

= 0.1281 [- 100 e^{- \frac{x}{100}} x - 10000 e^{- \frac{x}{100}}]_{120}^{150}

Compute the boundaries:

- \frac{25000}{e ^{\frac{3}{2}}} + \frac{22000}{e ^{\frac{6}{5}}}

= 0.1281 (- \frac{25000}{e ^{\frac{3}{2}}} + \frac{22000}{e ^{\frac{6}{5}}})

Simplify

= 134.25119 \dots

4 y + y^{^{''}} = 0

(\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 9 x^{3})

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

\frac{d ^{2} y}{d t ^{2}} + 10 \frac{d y}{d t} + 25 y = 0

\frac{d}{d x} (\frac{x ^{3} y + x y ^{3}}{x ^{2} + y ^{2}})

What is the integral from 120 to 150 of 0.1281*x*e^{-x/(100)} ?
The integral from 120 to 150 of 0.1281*x*e^{-x/(100)} is 134.25119…