What is the integral of e^{-2x^2}*(-4)x^3 ?

The integral of e^{-2x^2}*(-4)x^3 is e^{-2x^2}x^2+1/2 e^{-2x^2}+C

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integral of e^{-2x^2}*(-4)x^3

\int e^{- 2 x^{2}} \cdot (- 4) x^{3} d x

e^{- 2 x^{2}} x^{2} + \frac{1}{2} e^{- 2 x^{2}} + C

Solution steps

\int e^{- 2 x^{2}} (- 4) x^{3} d x

Take the constant out:

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

= - 4 \cdot \int e^{- 2 x^{2}} x^{3} d x

Apply u-substitution

: \int \frac{e ^{- 2 u} u}{2} d u

= - 4 \cdot \int \frac{e ^{- 2 u} u}{2} d u

Take the constant out:

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

= - 4 \cdot \frac{1}{2} \cdot \int e^{- 2 u} u d u

Apply u-substitution

: \int \frac{e ^{v} v}{4} d v

= - 4 \cdot \frac{1}{2} \cdot \int \frac{e ^{v} v}{4} d v

Take the constant out:

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

= - 4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int e^{v} v d v

Apply Integration By Parts

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

= - 4 \cdot \frac{1}{2} \cdot \frac{1}{4} (e^{v} v - \int e^{v} d v)

\int e^{v} d v = e^{v}

= - 4 \cdot \frac{1}{2} \cdot \frac{1}{4} (e^{v} v - e^{v})

Substitute back

= - 4 \cdot \frac{1}{2} \cdot \frac{1}{4} (e^{- 2 x^{2}} (- 2 x^{2}) - e^{- 2 x^{2}})

Simplify

- 4 \cdot \frac{1}{2} \cdot \frac{1}{4} (e^{- 2 x^{2}} (- 2 x^{2}) - e^{- 2 x^{2}}) : e^{- 2 x^{2}} x^{2} + \frac{1}{2} e^{- 2 x^{2}}

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

Add a constant to the solution

= e^{- 2 x^{2}} x^{2} + \frac{1}{2} e^{- 2 x^{2}} + C

n = 1 \sum \infty 8 (0.2)^{n}

n = 1 \sum \infty (1 + \frac{3}{n})^{n}

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

y = 5 x^{3} - 5 x, (1, 0)

n = 1 \sum \infty \frac{2 ^{n} x ^{n}}{n}

What is the integral of e^{-2x^2}*(-4)x^3 ?
The integral of e^{-2x^2}*(-4)x^3 is e^{-2x^2}x^2+1/2 e^{-2x^2}+C