Solve ∫ (from 0 to 2) of x^2-5x+6

$$\int _ { 0 } ^ { 2 } x ^ { 2 } - 5 x + 6$$

Evaluate

$$\frac{14}{3}\approx 4.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}-5x+6\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -5x\mathrm{d}x+\int 6\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-5\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-5\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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 -5 times \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}-\frac{5x^{2}}{2}+\int 6\mathrm{d}x$$

Find the integral of 6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-\frac{5x^{2}}{2}+6x$$

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.

$$\frac{2^{3}}{3}-\frac{5}{2}\times 2^{2}+6\times 2-\left(\frac{0^{3}}{3}-\frac{5}{2}\times 0^{2}+6\times 0\right)$$

Simplify.

$$\frac{14}{3}$$