What is the integral of xe^{4-x^2} ?

The integral of xe^{4-x^2} is -1/2 e^{4-x^2}+C

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integral of xe^{4-x^2}

\int x e^{4 - x^{2}} d x

- \frac{1}{2} e^{4 - x^{2}} + C

Solution steps

\int x e^{4 - x^{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{2} e^{u}

Substitute back

u = 4 - x^{2}

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

Add a constant to the solution

= - \frac{1}{2} e^{4 - x^{2}} + C

x \to \frac{π}{6} lim (\frac{sin ( x )}{x})

y^{^{''}} + 64 y = \frac{32}{3} cos (7 t)

3 - x^{2}, 7 - 5 x

f (x) = 2 x^{2} + 3 x - 2, at x = - 2

\frac{x}{1 + x ^{2}}, at x = 2

What is the integral of xe^{4-x^2} ?
The integral of xe^{4-x^2} is -1/2 e^{4-x^2}+C