What is the integral of (4-x)8^{(4-x)^2} ?

The integral of (4-x)8^{(4-x)^2} is -32(4-x)^2+C

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integral of (4-x)8^{(4-x)^2}

\int (4 - x) 8^{(4 - x)^{2}} d x

- 32 (4 - x)^{2} + C

Solution steps

\int (4 - x) \cdot 8^{(4 - x)^{2}} d x

Apply u-substitution

= \int - 32 d u

Integral of a constant:

\int a d x = a x

= (- 32) u

Substitute back

u = (4 - x)^{2}

= (- 32) (4 - x)^{2}

Simplify

= - 32 (4 - x)^{2}

Add a constant to the solution

= - 32 (4 - x)^{2} + C

What is the integral of (4-x)8^{(4-x)^2} ?
The integral of (4-x)8^{(4-x)^2} is -32(4-x)^2+C