Solve ∫ (from 1 to 3) of x^3+1/x^2-2x

$$\int _ { 1 } ^ { 3 } x ^ { 3 } + \frac { 1 } { x ^ { 2 } } - 2 x$$

Evaluate

$$\frac{38}{3}\approx 12.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}+\frac{1}{x^{2}}-2x\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x+\int -2x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x-2\int x\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+\int \frac{1}{x^{2}}\mathrm{d}x-2\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{2}}\mathrm{d}x with -\frac{1}{x}.

$$\frac{x^{4}}{4}-\frac{1}{x}-2\int x\mathrm{d}x$$

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

$$\frac{x^{4}}{4}-\frac{1}{x}-x^{2}$$

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.

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

Simplify.

$$\frac{38}{3}$$