Solve ∫ (from 0 to 2) of (4x^2-2x) wrt x

$$\int _ { 0 } ^ { 2 } ( 4 x ^ { 2 } - 2 x ) d x$$

Evaluate

$$\frac{20}{3}\approx 6.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4x^{2}-2x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int -2x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x-2\int x\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}-2\int x\mathrm{d}x$$

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 -2 times \frac{x^{2}}{2}.

$$\frac{4x^{3}}{3}-x^{2}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{4}{3}\times 2^{3}-2^{2}-\left(\frac{4}{3}\times 0^{3}-0^{2}\right)$$

Simplify.

$$\frac{20}{3}$$