Solve ∫ (from 2 to 3) of (x^3-5x^2+6x)

$$\int _ { 2 } ^ { 3 } ( x ^ { 3 } - 5 x ^ { 2 } + 6 x )$$

Evaluate

$$-\frac{5}{12}\approx -0.416666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$\frac{x^{4}}{4}-5\int x^{2}\mathrm{d}x+6\int x\mathrm{d}x$$

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 -5 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}-\frac{5x^{3}}{3}+6\int x\mathrm{d}x$$

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

$$\frac{x^{4}}{4}-\frac{5x^{3}}{3}+3x^{2}$$

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.

$$\frac{3^{4}}{4}-\frac{5}{3}\times 3^{3}+3\times 3^{2}-\left(\frac{2^{4}}{4}-\frac{5}{3}\times 2^{3}+3\times 2^{2}\right)$$

Simplify.

$$-\frac{5}{12}$$