What is the integral of te^{-2} ?

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

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integral of te^{-2}

\int t e^{- 2} d t

\frac{t ^{2}}{2 e ^{2}} + C

Solution steps

\int t e^{- 2} d t

Take the constant out:

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

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

Apply exponent rule:

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

= \frac{1}{e ^{2}} \cdot \int t d t

Apply the Power Rule

: \frac{t ^{2}}{2}

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

Simplify

\frac{1}{e ^{2}} \cdot \frac{t ^{2}}{2} : \frac{t ^{2}}{2 e ^{2}}

= \frac{t ^{2}}{2 e ^{2}}

Add a constant to the solution

= \frac{t ^{2}}{2 e ^{2}} + C

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