Solve ∫ (from 1 to 3) of (x^3-3x^2-x+3) wrt x

$$\int_{ 1 }^{ 3 } ( { x }^{ 3 } -3 { x }^{ 2 } -x+3) d x$$

Evaluate

$$-4$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

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 -1 times \frac{x^{2}}{2}.

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

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

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

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.

$$\frac{3^{4}}{4}-3^{3}-\frac{3^{2}}{2}+3\times 3-\left(\frac{1^{4}}{4}-1^{3}-\frac{1^{2}}{2}+3\times 1\right)$$

Simplify.

$$-4$$