Solve ∫ (8-4/3x) wrt x

$$\int ( 8 - \frac { 4 } { 3 } x ) d x$$

Evaluate

$$-\frac{2x^{2}}{3}+8x+С$$

Solution Steps

Integrate the sum term by term.

$$\int 8\mathrm{d}x+\int -\frac{4x}{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 8\mathrm{d}x-\frac{4\int x\mathrm{d}x}{3}$$

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$8x-\frac{4\int x\mathrm{d}x}{3}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -\frac{4}{3} times \frac{x^{2}}{2}.

$$8x-\frac{2x^{2}}{3}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$8x-\frac{2x^{2}}{3}+С$$

Differentiate w.r.t. x

$$-\frac{4x}{3}+8$$