Algebra Examples

$y = \frac{1}{x}$ , $[5, 12]$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Eliminate the equal sides of each equation and combine.

$\frac{1}{x} = 0$

Step 1.2

Solve $\frac{1}{x} = 0$ for $x$ .

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Step 1.2.1

Set the numerator equal to zero.

$1 = 0$

Step 1.2.2

Since $1 \neq 0$ , there are no solutions.

No solution

Step 2

The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region. The regions are determined by the intersection points of the curves. This can be done algebraically or graphically.

$A r e a = \int_{5}^{12} \frac{1}{x} d x - \int_{5}^{12} 0 d x$

Step 3

Integrate to find the area between $5$ and $12$ .

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Step 3.1

Combine the integrals into a single integral.

$\int_{5}^{12} \frac{1}{x} - (0) d x$

Step 3.2

Subtract $0$ from $\frac{1}{x}$ .

$\int_{5}^{12} \frac{1}{x} d x$

Step 3.3

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$\ln (| x |)]_{5}^{12}$

Step 3.4

Simplify the answer.

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Step 3.4.1

Evaluate $\ln (| x |)$ at $12$ and at $5$ .

$\ln (| 12 |) - \ln (| 5 |)$

Step 3.4.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| 12 |}{| 5 |})$

Step 3.4.3

Simplify.

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Step 3.4.3.1

The absolute value is the distance between a number and zero. The distance between $0$ and $12$ is $12$ .

$\ln (\frac{12}{| 5 |})$

Step 3.4.3.2

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$\ln (\frac{12}{5})$

Step 4