What is the integral from 0 to 4 of 60000te^{-t} ?

The integral from 0 to 4 of 60000te^{-t} is 60000(-5/(e^4)+1)

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integral from 0 to 4 of 60000te^{-t}

\int_{0}^{4} 60000 t e^{- t} d t

60000 (- \frac{5}{e ^{4}} + 1)

Decimal

54505.30833 \dots

Solution steps

\int_{0}^{4} 60000 t e^{- t} d t

Take the constant out:

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

= 60000 \cdot \int_{0}^{4} t e^{- t} d t

Apply Integration By Parts

: [- e^{- t} t - \int - e^{- t} d t]_{0}^{4}

= 60000 [- e^{- t} t - \int - e^{- t} d t]_{0}^{4}

\int - e^{- t} d t = e^{- t}

= 60000 [- e^{- t} t - e^{- t}]_{0}^{4}

Compute the boundaries:

- \frac{5}{e ^{4}} + 1

= 60000 (- \frac{5}{e ^{4}} + 1)

\int x \frac{1}{x} d x

\int csc^{3} (x) cot^{5} (x) d x

\frac{\partial}{\partial x} (y ln (x) + x y^{2})

\int (\frac{1}{x ^{5}} - x^{5} - \frac{1}{3}) d x

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

What is the integral from 0 to 4 of 60000te^{-t} ?
The integral from 0 to 4 of 60000te^{-t} is 60000(-5/(e^4)+1)