Calculus Examples

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

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.

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

Step 1.2

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

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

Step 1.2.1.2

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

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

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

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

Step 1.2.1.6

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

$x = x$

Step 1.2.1.7

The factors for $x^{2}$ are $x \cdot x$ , which is $x$ multiplied by each other $2$ times.

$x^{2} = x \cdot x$

Step 1.2.1.8

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

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

Step 1.2.1.9

Multiply $x$ by $x$ .

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $x$ .

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Step 1.2.2.2.1.1

Factor $x$ out of $x^{2}$ .

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

Step 1.2.2.2.1.2

Cancel the common factor.

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

Step 1.2.2.2.1.3

Rewrite the expression.

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

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Cancel the common factor of $x^{2}$ .

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Step 1.2.2.3.1.1

Cancel the common factor.

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

Step 1.2.2.3.1.2

Rewrite the expression.

$x = 1$

Step 1.3

Evaluate $y$ when $x = 1$ .

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Step 1.3.1

Substitute $1$ for $x$ .

$y = \frac{1}{{(1)}^{2}}$

Step 1.3.2

Substitute $1$ for $x$ in $y = \frac{1}{{(1)}^{2}}$ and solve for $y$ .

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Step 1.3.2.1

Remove parentheses.

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

Step 1.3.2.2

Simplify $\frac{1}{1^{2}}$ .

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Step 1.3.2.2.1

One to any power is one.

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

Step 1.3.2.2.2

Divide $1$ by $1$ .

$y = 1$

Step 1.4

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

$(1, 1)$

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.

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

Step 3

Integrate to find the area between $1$ and $2$ .

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Step 3.1

Combine the integrals into a single integral.

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

Step 3.2

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

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

Step 3.3

Split the single integral into multiple integrals.

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

Step 3.4

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

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

Step 3.5

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Step 3.6

Apply basic rules of exponents.

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Step 3.6.1

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 3.6.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 3.6.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.6.2.2

Multiply $2$ by $- 1$ .

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

Step 3.7

By the Power Rule, the integral of $x^{- 2}$ with respect to $x$ is $- x^{- 1}$ .

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

Step 3.8

Simplify the answer.

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Step 3.8.1

Substitute and simplify.

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Step 3.8.1.1

Evaluate $\ln (| x |)$ at $2$ and at $1$ .

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

Step 3.8.1.2

Evaluate $- x^{- 1}$ at $2$ and at $1$ .

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

Step 3.8.1.3

Simplify.

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Step 3.8.1.3.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.8.1.3.2

One to any power is one.

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

Step 3.8.1.3.3

Write $1$ as a fraction with a common denominator.

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

Step 3.8.1.3.4

Combine the numerators over the common denominator.

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

Step 3.8.1.3.5

Add $- 1$ and $2$ .

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

Step 3.8.2

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

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

Step 3.8.3

Simplify.

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Step 3.8.3.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

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

Step 3.8.3.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

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

Step 3.8.3.3

Divide $2$ by $1$ .

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

Step 4