Calculus Examples

$y = \frac{1}{x}$ ; $x = 1$ ; $y = \frac{1}{2}$ ; $x = 0$

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.

$\frac{1}{x} = \frac{1}{2}$

Step 1.2

Solve $\frac{1}{x} = \frac{1}{2}$ for $x$ .

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Step 1.2.1

Find the LCD of the terms in the equation.

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Step 1.2.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x, 2 + y = \frac{1}{2}$

Step 1.2.1.2

Since $x, 2$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 2$ then find LCM for the variable part $x^{1}$ .

Step 1.2.1.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

$1$ List the prime factors of each number.

Step 1.2.1.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 1.2.1.5

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 1.2.1.6

The LCM of $1, 2$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2 + y = \frac{1}{2}$

Step 1.2.1.7

The factor for $x^{1}$ is $x$ itself.

$x = x$

Step 1.2.1.8

The LCM of $x^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x + y = \frac{1}{2}$

Step 1.2.1.9

The LCM for $x, 2$ is the numeric part $2$ multiplied by the variable part.

$2 x + y = \frac{1}{2}$

Step 1.2.2

Multiply each term in $\frac{1}{x} = \frac{1}{2}$ by $2 x$ to eliminate the fractions.

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Step 1.2.2.1

Multiply each term in $\frac{1}{x} = \frac{1}{2}$ by $2 x$ .

$\frac{1}{x} (2 x) = \frac{1}{2} (2 x)$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Rewrite using the commutative property of multiplication.

$2 \frac{1}{x} x = \frac{1}{2} (2 x)$

Step 1.2.2.2.2

Combine $2$ and $\frac{1}{x}$ .

$\frac{2}{x} x = \frac{1}{2} (2 x)$

Step 1.2.2.2.3

Cancel the common factor of $x$ .

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Step 1.2.2.2.3.1

Cancel the common factor.

$\frac{2}{x} x = \frac{1}{2} (2 x)$

Step 1.2.2.2.3.2

Rewrite the expression.

$2 = \frac{1}{2} (2 x)$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Cancel the common factor of $2$ .

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Step 1.2.2.3.1.1

Factor $2$ out of $2 x$ .

$2 = \frac{1}{2} (2 (x))$

Step 1.2.2.3.1.2

Cancel the common factor.

$2 = \frac{1}{2} (2 x)$

Step 1.2.2.3.1.3

Rewrite the expression.

$2 = x$

Step 1.2.3

Rewrite the equation as $x = 2$ .

$x = 2$

Step 1.3

Substitute $2$ for $x$ .

$y = \frac{1}{2}$

Step 1.4

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(2, \frac{1}{2})$

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.

$A r e a = \int_{0}^{1} \frac{1}{x} d x - \int_{0}^{1} \frac{1}{2} d x$

Step 3

Integrate to find the area between $0$ and $1$ .

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Step 3.1

Combine the integrals into a single integral.

$\int_{0}^{1} \frac{1}{x} - (\frac{1}{2}) d x$

Step 3.2

Multiply $- 1$ by $\frac{1}{2}$ .

$\int_{0}^{1} \frac{1}{x} - \frac{1}{2} d x$

Step 3.3

Split the single integral into multiple integrals.

$\int_{0}^{1} \frac{1}{x} d x + \int_{0}^{1} - \frac{1}{2} d x$

Step 3.4

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

$\ln (| x |)]_{0}^{1} + \int_{0}^{1} - \frac{1}{2} d x$

Step 3.5

Apply the constant rule.

$\ln (| x |)]_{0}^{1} + - \frac{1}{2} x]_{0}^{1}$

Step 3.6

Simplify the answer.

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Step 3.6.1

Combine $\ln (| x |)]_{0}^{1}$ and $- \frac{1}{2} x]_{0}^{1}$ .

$\ln (| x |) - \frac{1}{2} x]_{0}^{1}$

Step 3.6.2

Substitute and simplify.

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Step 3.6.2.1

Evaluate $\ln (| x |) - \frac{1}{2} x$ at $1$ and at $0$ .

$(\ln (| 1 |) - \frac{1}{2} \cdot 1) - (\ln (| 0 |) - \frac{1}{2} \cdot 0)$

Step 3.6.2.2

Simplify.

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Step 3.6.2.2.1

Multiply $- 1$ by $1$ .

$\ln (| 1 |) - \frac{1}{2} - (\ln (| 0 |) - \frac{1}{2} \cdot 0)$

Step 3.6.2.2.2

Multiply $0$ by $- 1$ .

$\ln (| 1 |) - \frac{1}{2} - (\ln (| 0 |) + 0 (\frac{1}{2}))$

Step 3.6.2.2.3

Multiply $0$ by $\frac{1}{2}$ .

$\ln (| 1 |) - \frac{1}{2} - (\ln (| 0 |) + 0)$

Step 3.6.2.2.4

Add $\ln (| 0 |)$ and $0$ .

$\ln (| 1 |) - \frac{1}{2} - \ln (| 0 |)$

Step 3.6.3

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

$\ln (\frac{| 1 |}{| 0 |}) - \frac{1}{2}$

Step 3.6.4

Simplify.

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Step 3.6.4.1

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

$\ln (\frac{| 1 |}{0}) - \frac{1}{2}$

Step 3.6.4.2

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\ln (\frac{| 1 |}{0}) - \frac{1}{2}$

Step 3.6.5

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\ln (\frac{| 1 |}{0}) - \frac{1}{2}$

Step 3.7

The expression contains a division by $0$ . The expression is undefined.

Undefined

Step 4