What is the integral from 0 to 1 of te^{-1} ?

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

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integral from 0 to 1 of te^{-1}

\int_{0}^{1} t e^{- 1} d t

\frac{1}{2 e}

Decimal Notation

0.18393 \dots

Solution steps

\int_{0}^{1} t e^{- 1} d t

Take the constant out:

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

= e^{- 1} \cdot \int_{0}^{1} t d t

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot \int_{0}^{1} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{1}

= \frac{1}{e} [\frac{t ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{e} \cdot \frac{1}{2}

Simplify

= \frac{1}{2 e}

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\frac{d}{d x} (\frac{1}{f ( x )})

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