What is the integral from 0 to 4 of 2te^{-t} ?

The integral from 0 to 4 of 2te^{-t} is 2(-5/(e^4)+1)

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integral from 0 to 4 of 2te^{-t}

\int_{0}^{4} 2 t e^{- t} d t

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

Decimal

1.81684 \dots

Solution steps

\int_{0}^{4} 2 t e^{- t} d t

Take the constant out:

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

= 2 \cdot \int_{0}^{4} t e^{- t} d t

Apply Integration By Parts

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

= 2 [- e^{- t} t - \int - e^{- t} d t]_{0}^{4}

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

= 2 [- e^{- t} t - e^{- t}]_{0}^{4}

Compute the boundaries:

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

= 2 (- \frac{5}{e ^{4}} + 1)

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

\frac{\partial}{\partial x} (\frac{y}{x} ln (x))

n = 1 \sum \infty (cos (2.6))^{n}

\int (5 x + 1)^{\frac{1}{2}} d x

y^{^{''}} + 4 y = 3 cos (2 t)

What is the integral from 0 to 4 of 2te^{-t} ?
The integral from 0 to 4 of 2te^{-t} is 2(-5/(e^4)+1)