What is the integral of 2e^{-t}t ?

The integral of 2e^{-t}t is 2(-e^{-t}t-e^{-t})+C

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integral of 2e^{-t}t

\int 2 e^{- t} t d t

2 (- e^{- t} t - e^{- t}) + C

Solution steps

\int 2 e^{- t} t d t

Take the constant out:

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

= 2 \cdot \int e^{- t} t d t

Apply Integration By Parts

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

= 2 (- e^{- t} t - \int - e^{- t} d t)

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

= 2 (- e^{- t} t - e^{- t})

Add a constant to the solution

= 2 (- e^{- t} t - e^{- t}) + C

\int e^{- 10 r^{2}} d r

(x - 1) \cdot ln (x - 1), 2

9 - x^{2}, 0

\int \frac{7 x ^{4} - 6 x ^{2} - x + 5}{x + 2} d x

x \to 3 - lim (\frac{1 - 2 x}{x - 3})

What is the integral of 2e^{-t}t ?
The integral of 2e^{-t}t is 2(-e^{-t}t-e^{-t})+C