Calculus Examples

$y = \frac{16}{x}$ , $y = x$ , $y = \frac{x}{4}$

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.

$\frac{16}{x} = x$

Step 1.2

Solve $\frac{16}{x} = x$ 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.

$x, 1 + y = x$

$y = \frac{x}{4}$

Step 1.2.1.2

The LCM of one and any expression is the expression.

$x + y = x$

$y = \frac{x}{4}$

$x + y = x$

$y = \frac{x}{4}$

Step 1.2.2

Multiply each term in $\frac{16}{x} = x$ by $x$ to eliminate the fractions.

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Step 1.2.2.1

Multiply each term in $\frac{16}{x} = x$ by $x$ .

$\frac{16}{x} \cdot x = x \cdot x + y = x$

$y = \frac{x}{4}$

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

Cancel the common factor.

$\frac{16}{x} \cdot x = x \cdot x + y = x$

$y = \frac{x}{4}$

Step 1.2.2.2.1.2

Rewrite the expression.

$16 = x \cdot x + y = x$

$y = \frac{x}{4}$

$16 = x \cdot x + y = x$

$y = \frac{x}{4}$

$16 = x \cdot x + y = x$

$y = \frac{x}{4}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Multiply $x$ by $x$ .

$16 = x^{2} + y = x$

$y = \frac{x}{4}$

$16 = x^{2} + y = x$

$y = \frac{x}{4}$

$16 = x^{2} + y = x$

$y = \frac{x}{4}$

Step 1.2.3

Solve the equation.

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Step 1.2.3.1

Rewrite the equation as $x^{2} = 16$ .

$x^{2} = 16 + y = x$

$y = \frac{x}{4}$

Step 1.2.3.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{16} + y = x$

$y = \frac{x}{4}$

Step 1.2.3.3

Simplify $\pm \sqrt{16}$ .

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Step 1.2.3.3.1

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}} + y = x$

$y = \frac{x}{4}$

Step 1.2.3.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4 + y = x$

$y = \frac{x}{4}$

$x = \pm 4 + y = x$

$y = \frac{x}{4}$

Step 1.2.3.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.2.3.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 4 + y = x$

$y = \frac{x}{4}$

Step 1.2.3.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 4 + y = x$

$y = \frac{x}{4}$

Step 1.2.3.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 4, - 4 + y = x$

$y = \frac{x}{4}$

$x = 4, - 4 + y = x$

$y = \frac{x}{4}$

$x = 4, - 4 + y = x$

$y = \frac{x}{4}$

$x = 4, - 4 + y = x$

$y = \frac{x}{4}$

Step 1.3

Evaluate $y$ when $x = 4$ .

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Step 1.3.1

Substitute $4$ for $x$ .

$y = 4 y = \frac{4}{4}$

Step 1.3.2

Remove parentheses.

$y = 4$

Step 1.4

Substitute $4$ for $x$ in $y = \frac{4}{4}$ and solve for $y$ .

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Step 1.4.1

Remove parentheses.

$y = \frac{4}{4}$

Step 1.4.2

Divide $4$ by $4$ .

$y = 1$

Step 1.5

Evaluate $y$ when $x = - 4$ .

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Step 1.5.1

Substitute $- 4$ for $x$ .

$y = - 4 y = \frac{- 4}{4}$

Step 1.5.2

Remove parentheses.

$y = - 4$

Step 1.6

Divide $- 4$ by $4$ .

$y = - 1$

Step 1.7

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

$(4, 1)$

$(- 4, - 1)$

$(4, 1)$

$(- 4, - 1)$

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.

$A r e a = \int_{- 4}^{4} x d x - \int_{- 4}^{4} \frac{16}{x} d x$

Step 3

Integrate to find the area between $- 4$ and $4$ .

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Step 3.1

Combine the integrals into a single integral.

$\int_{- 4}^{4} x - (\frac{16}{x}) d x$

Step 3.2

Multiply $- 1$ by $\frac{16}{x}$ .

$\int_{- 4}^{4} x - \frac{16}{x} d x$

Step 3.3

Split the single integral into multiple integrals.

$\int_{- 4}^{4} x d x + \int_{- 4}^{4} - \frac{16}{x} d x$

Step 3.4

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

$\frac{1}{2} x^{2}]_{- 4}^{4} + \int_{- 4}^{4} - \frac{16}{x} d x$

Step 3.5

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

$\frac{1}{2} x^{2}]_{- 4}^{4} - \int_{- 4}^{4} \frac{16}{x} d x$

Step 3.6

Since $16$ is constant with respect to $x$ , move $16$ out of the integral.

$\frac{1}{2} x^{2}]_{- 4}^{4} - (16 \int_{- 4}^{4} \frac{1}{x} d x)$

Step 3.7

Multiply $16$ by $- 1$ .

$\frac{1}{2} x^{2}]_{- 4}^{4} - 16 \int_{- 4}^{4} \frac{1}{x} d x$

Step 3.8

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

$\frac{1}{2} x^{2}]_{- 4}^{4} - 16 (\ln (| x |)]_{- 4}^{4})$

Step 3.9

Simplify the answer.

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Step 3.9.1

Substitute and simplify.

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Step 3.9.1.1

Evaluate $\frac{1}{2} x^{2}$ at $4$ and at $- 4$ .

$(\frac{1}{2} \cdot 4^{2}) - \frac{1}{2} {(- 4)}^{2} - 16 (\ln (| x |)]_{- 4}^{4})$

Step 3.9.1.2

Evaluate $\ln (| x |)$ at $4$ and at $- 4$ .

$\frac{1}{2} \cdot 4^{2} - \frac{1}{2} {(- 4)}^{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3

Simplify.

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Step 3.9.1.3.1

Raise $4$ to the power of $2$ .

$\frac{1}{2} \cdot 16 - \frac{1}{2} {(- 4)}^{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.2

Combine $\frac{1}{2}$ and $16$ .

$\frac{16}{2} - \frac{1}{2} {(- 4)}^{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.3

Cancel the common factor of $16$ and $2$ .

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Step 3.9.1.3.3.1

Factor $2$ out of $16$ .

$\frac{2 \cdot 8}{2} - \frac{1}{2} {(- 4)}^{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.3.2

Cancel the common factors.

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Step 3.9.1.3.3.2.1

Factor $2$ out of $2$ .

$\frac{2 \cdot 8}{2 (1)} - \frac{1}{2} {(- 4)}^{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.3.2.2

Cancel the common factor.

$\frac{2 \cdot 8}{2 \cdot 1} - \frac{1}{2} {(- 4)}^{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.3.2.3

Rewrite the expression.

$\frac{8}{1} - \frac{1}{2} {(- 4)}^{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.3.2.4

Divide $8$ by $1$ .

$8 - \frac{1}{2} {(- 4)}^{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.4

Raise $- 4$ to the power of $2$ .

$8 - \frac{1}{2} \cdot 16 - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.5

Multiply $16$ by $- 1$ .

$8 - 16 (\frac{1}{2}) - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.6

Combine $- 16$ and $\frac{1}{2}$ .

$8 + \frac{- 16}{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.7

Cancel the common factor of $- 16$ and $2$ .

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Step 3.9.1.3.7.1

Factor $2$ out of $- 16$ .

$8 + \frac{2 \cdot - 8}{2} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.7.2

Cancel the common factors.

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Step 3.9.1.3.7.2.1

Factor $2$ out of $2$ .

$8 + \frac{2 \cdot - 8}{2 (1)} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.7.2.2

Cancel the common factor.

$8 + \frac{2 \cdot - 8}{2 \cdot 1} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.7.2.3

Rewrite the expression.

$8 + \frac{- 8}{1} - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.7.2.4

Divide $- 8$ by $1$ .

$8 - 8 - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.8

Subtract $8$ from $8$ .

$0 - 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.1.3.9

Subtract $16 (\ln (| 4 |) - \ln (| - 4 |))$ from $0$ .

$- 16 (\ln (| 4 |) - \ln (| - 4 |))$

Step 3.9.2

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

$- 16 \ln (\frac{| 4 |}{| - 4 |})$

Step 3.9.3

Simplify.

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Step 3.9.3.1

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$- 16 \ln (\frac{4}{| - 4 |})$

Step 3.9.3.2

The absolute value is the distance between a number and zero. The distance between $- 4$ and $0$ is $4$ .

$- 16 \ln (\frac{4}{4})$

Step 3.9.3.3

Divide $4$ by $4$ .

$- 16 \ln (1)$

Step 3.9.3.4

The natural logarithm of $1$ is $0$ .

$- 16 \cdot 0$

Step 3.9.3.5

Multiply $- 16$ by $0$ .

$0$

Step 4