Solve ∫ (from 5 to 6) of 1/x^2-1/x^3 wrt x

$$\int _ { 5 } ^ { 6 } \frac { 1 } { x ^ { 2 } } - \frac { 1 } { x ^ { 3 } } d x$$

Evaluate

$$\frac{49}{1800}\approx 0.027222222$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{3}}\mathrm{d}x with -\frac{1}{2x^{2}}. Multiply -1 times -\frac{1}{2x^{2}}.

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

Simplify.

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

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.

$$\left(\frac{1}{2}-6\right)\times 6^{-2}-\left(\frac{1}{2}-5\right)\times 5^{-2}$$

Simplify.

$$\frac{49}{1800}$$