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

The integral from 0 to 2 of te^{-t} is -3/(e^2)+1

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

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

- \frac{3}{e ^{2}} + 1

Decimal Notation

0.59399 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

- \frac{3}{e ^{2}} + 1

= - \frac{3}{e ^{2}} + 1

\frac{x}{x + 11}

y^{^{''}} + y = (sin (x))^{2}

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

y^{^{'}} = 3 x - y

y^{^{'}} = \frac{x ^{2} y ^{4} - x ^{2}}{y ^{3}}

What is the integral from 0 to 2 of te^{-t} ?
The integral from 0 to 2 of te^{-t} is -3/(e^2)+1