integral of (x-2)/x^3 from 1 to 2 — step-by-step solution

Let's compute the integral:

\int_{1}^{2} \frac{x - 2}{x^{3}} d x

For

x

≠ 0, divide each numerator term by the power in the denominator:

\frac{x - 2}{x^{3}} = \frac{1}{x^{2}} - \frac{2}{x^{3}}

ⅰ

\int \frac{x - 2}{x^{3}} d x

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

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

ⓘ An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

\int \frac{1}{x^{2}} d x

\int x^{-2} d x

- x^{-1}

- \frac{1}{x}

■A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

- \int \frac{2}{x^{3}} d x

- \int 2 x^{-3} d x

x^{-2}

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

▷

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

Let's evaluate for the interval

[1, 2]

(\frac{1}{2^{2}} - \frac{1}{2}) - (\frac{1}{1^{2}} - \frac{1}{1})

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

2^{2}

2 \cdot 2

4

■A minus sign before a parenthesis changes the sign of each of its terms:

- (\frac{1}{1^{2}} - \frac{1}{1})

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

\frac{1}{4}

- \frac{1}{2}

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

+ \frac{1}{1}

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{1}{4} - \frac{1}{2}

\frac{1 + 2 \cdot (-1)}{4}

■Start from

-1

and move forward

1

on the number line, up to zero:

- 1 + 1

0

\frac{1 + 2 \cdot (-1)}{4} + 0

ⓘ Adding 0 changes nothing:

\frac{1 + 2 \cdot (-1)}{4} + 0

\frac{1 + 2 \cdot (-1)}{4}

■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-1)

-2

\frac{1 - 2}{4}

ⓘ The numerator and the denominator are computed before the division:

1 - 2

-1

▶Integral:

- \frac{1}{4}

The numeric value of:

- \frac{1}{4}

equals:

▷

-0.25