What is the integral from 0 to 1 of 2xe^{1-x} ?

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

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integral from 0 to 1 of 2xe^{1-x}

\int_{0}^{1} 2 x e^{1 - x} d x

2 (- 2 + e)

Decimal Notation

1.43656 \dots

Solution steps

\int_{0}^{1} 2 x e^{1 - 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}^{1} x e^{1 - x} d x

Apply Integration By Parts

: [- e^{1 - x} x - \int - e^{1 - x} d x]_{0}^{1}

= 2 [- e^{1 - x} x - \int - e^{1 - x} d x]_{0}^{1}

\int - e^{1 - x} d x = e^{1 - x}

= 2 [- e^{1 - x} x - e^{1 - x}]_{0}^{1}

Compute the boundaries:

- 2 + e

= 2 (- 2 + e)

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