Solve - 2 / p ∫ ( 8 - 4 x) wrt x=18

$$\frac { - 2 } { p } \int ( 8 - 4 x ) d x = 18$$

Solve for p

$$p=\frac{2x^{2}-8x-С}{9}$$

$$С\neq 2x^{2}-8x\text{ and }С_{1}\neq \frac{8x-2x^{2}}{9}$$

Steps for Solving Linear Equation

Variable p cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by p.

$$-2\int 8-4x\mathrm{d}x=18p$$

Swap sides so that all variable terms are on the left hand side.

$$18p=-2\int 8-4x\mathrm{d}x$$

The equation is in standard form.

$$18p=4x^{2}-16x-2С$$

Divide both sides by 18.

$$\frac{18p}{18}=\frac{4x^{2}-16x-2С}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$p=\frac{4x^{2}-16x-2С}{18}$$

Divide -16x+4x^{2}-2С by 18.

$$p=\frac{2x^{2}-8x-С}{9}$$

Variable p cannot be equal to 0.

$$p=\frac{2x^{2}-8x-С}{9}\text{, }p\neq 0$$