What is the integral of (8-x)(2x+8) ?

The integral of (8-x)(2x+8) is 2(2x^2-(x^3)/3+32x)+C

integral of (8-x)(2x+8)

\int (8 - x) (2 x + 8) d x

2 (2 x^{2} - \frac{x ^{3}}{3} + 32 x) + C

Solution steps

\int (8 - x) (2 x + 8) d x

(8 - x) (2 x + 8) = 2 (8 - x) (x + 4)

= \int 2 (8 - x) (x + 4) d x

Take the constant out:

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

= 2 \cdot \int (8 - x) (x + 4) d x

Expand

(8 - x) (x + 4) : 4 x - x^{2} + 32

= 2 \cdot \int 4 x - x^{2} + 32 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int 4 x d x - \int x^{2} d x + \int 32 d x)

\int 4 x d x = 2 x^{2}

\int x^{2} d x = \frac{x ^{3}}{3}

\int 32 d x = 32 x

= 2 (2 x^{2} - \frac{x ^{3}}{3} + 32 x)

Add a constant to the solution

= 2 (2 x^{2} - \frac{x ^{3}}{3} + 32 x) + C

\frac{\partial}{\partial x} (\frac{1}{4} (x + y)^{4})

- 4 x + \frac{2 x d y}{d x} = \frac{2 d y}{d x} + 3 x^{2} + 2

t an g e n t y = \frac{1}{( 1 + x ^{2} ) , at ( 1 , \frac{1}{2} )}

\int_{0}^{1} 1 + 7 x d x

\frac{\partial}{\partial x} (x - y^{2})

What is the integral of (8-x)(2x+8) ?
The integral of (8-x)(2x+8) is 2(2x^2-(x^3)/3+32x)+C