Solve ∫ (from 2 to 3) of (-x^2+4x-3) wrt x

$$\int _ { 2 } ^ { 3 } ( - x ^ { 2 } + 4 x - 3 ) d x$$

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 -1 times \frac{x^{3}}{3}.

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

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

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

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

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