What is the integral of x^2e^{-1} ?

The integral of x^2e^{-1} is (x^3)/(3e)+C

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integral of x^2e^{-1}

\int x^{2} e^{- 1} d x

\frac{x ^{3}}{3 e} + C

Solution steps

\int x^{2} e^{- 1} d x

Take the constant out:

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

= e^{- 1} \cdot \int x^{2} d x

Apply exponent rule:

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

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

Apply the Power Rule

: \frac{x ^{3}}{3}

= \frac{1}{e} \cdot \frac{x ^{3}}{3}

Simplify

\frac{1}{e} \cdot \frac{x ^{3}}{3} : \frac{x ^{3}}{3 e}

= \frac{x ^{3}}{3 e}

Add a constant to the solution

= \frac{x ^{3}}{3 e} + C

\int \frac{( x ^{2} - 4 x + 9 )}{x ^{2} - 6 x + 9} d x

(2 x + 3 y) d x + (3 x + 2) d y = 0

\frac{( 3 s + 7 )}{s ^{2} + 2 s + 4}

f (x) = - 3 x^{2}

x^{2} + x, at x = 3