What is the integral from 0 to 4 of 2xe^{-x} ?

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

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integral from 0 to 4 of 2xe^{-x}

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

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

Decimal Notation

1.81684 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

\int_{0}^{1} \frac{1}{x ^{2} + 4 x + 3} d x

\int_{0}^{1} 1 + (2 y)^{2} d y

\int_{- 5}^{3} x^{2} + x - 12 d x

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

\int_{0}^{5.4} 378 - 17 t^{2} d t

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