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

$$\int_{ 0 }^{ 2 } { x }^{ 2 } -2+x d x$$

Evaluate

$$\frac{2}{3}\approx 0.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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}.

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

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.

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

Simplify.

$$\frac{2}{3}$$