Calculus Examples

$f (x) = \frac{6}{8 x - 1}$ , $[4, 8]$

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.

$\frac{6}{8 x - 1} = 0$

Step 1.2

Solve $\frac{6}{8 x - 1} = 0$ for $x$ .

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Step 1.2.1

Set the numerator equal to zero.

$6 = 0$

Step 1.2.2

Since $6 \neq 0$ , 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.

$A r e a = \int_{4}^{8} \frac{6}{8 x - 1} d x - \int_{4}^{8} 0 d x$

Step 3

Integrate to find the area between $4$ and $8$ .

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Step 3.1

Combine the integrals into a single integral.

$\int_{4}^{8} \frac{6}{8 x - 1} - (0) d x$

Step 3.2

Subtract $0$ from $\frac{6}{8 x - 1}$ .

$\int_{4}^{8} \frac{6}{8 x - 1} d x$

Step 3.3

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$6 \int_{4}^{8} \frac{1}{8 x - 1} d x$

Step 3.4

Let $u = 8 x - 1$ . Then $d u = 8 d x$ , so $\frac{1}{8} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.4.1

Let $u = 8 x - 1$ . Find $\frac{d u}{d x}$ .

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Step 3.4.1.1

Differentiate $8 x - 1$ .

$\frac{d}{d x} [8 x - 1]$

Step 3.4.1.2

By the Sum Rule, the derivative of $8 x - 1$ with respect to $x$ is $\frac{d}{d x} [8 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [8 x] + \frac{d}{d x} [- 1]$

Step 3.4.1.3

Evaluate $\frac{d}{d x} [8 x]$ .

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Step 3.4.1.3.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$8 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 3.4.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$8 \cdot 1 + \frac{d}{d x} [- 1]$

Step 3.4.1.3.3

Multiply $8$ by $1$ .

$8 + \frac{d}{d x} [- 1]$

Step 3.4.1.4

Differentiate using the Constant Rule.

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Step 3.4.1.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$8 + 0$

Step 3.4.1.4.2

Add $8$ and $0$ .

$8$

Step 3.4.2

Substitute the lower limit in for $x$ in $u = 8 x - 1$ .

$u_{lower} = 8 \cdot 4 - 1$

Step 3.4.3

Simplify.

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Step 3.4.3.1

Multiply $8$ by $4$ .

$u_{lower} = 32 - 1$

Step 3.4.3.2

Subtract $1$ from $32$ .

$u_{lower} = 31$

Step 3.4.4

Substitute the upper limit in for $x$ in $u = 8 x - 1$ .

$u_{upper} = 8 \cdot 8 - 1$

Step 3.4.5

Simplify.

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Step 3.4.5.1

Multiply $8$ by $8$ .

$u_{upper} = 64 - 1$

Step 3.4.5.2

Subtract $1$ from $64$ .

$u_{upper} = 63$

Step 3.4.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 31$

$u_{upper} = 63$

Step 3.4.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$6 \int_{31}^{63} \frac{1}{u} \cdot \frac{1}{8} d u$

Step 3.5

Simplify.

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Step 3.5.1

Multiply $\frac{1}{u}$ by $\frac{1}{8}$ .

$6 \int_{31}^{63} \frac{1}{u \cdot 8} d u$

Step 3.5.2

Move $8$ to the left of $u$ .

$6 \int_{31}^{63} \frac{1}{8 u} d u$

Step 3.6

Since $\frac{1}{8}$ is constant with respect to $u$ , move $\frac{1}{8}$ out of the integral.

$6 (\frac{1}{8} \int_{31}^{63} \frac{1}{u} d u)$

Step 3.7

Simplify.

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Step 3.7.1

Combine $\frac{1}{8}$ and $6$ .

$\frac{6}{8} \int_{31}^{63} \frac{1}{u} d u$

Step 3.7.2

Cancel the common factor of $6$ and $8$ .

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Step 3.7.2.1

Factor $2$ out of $6$ .

$\frac{2 (3)}{8} \int_{31}^{63} \frac{1}{u} d u$

Step 3.7.2.2

Cancel the common factors.

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Step 3.7.2.2.1

Factor $2$ out of $8$ .

$\frac{2 \cdot 3}{2 \cdot 4} \int_{31}^{63} \frac{1}{u} d u$

Step 3.7.2.2.2

Cancel the common factor.

$\frac{2 \cdot 3}{2 \cdot 4} \int_{31}^{63} \frac{1}{u} d u$

Step 3.7.2.2.3

Rewrite the expression.

$\frac{3}{4} \int_{31}^{63} \frac{1}{u} d u$

Step 3.8

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

$\frac{3}{4} \ln (| u |)]_{31}^{63}$

Step 3.9

Evaluate $\ln (| u |)$ at $63$ and at $31$ .

$\frac{3}{4} (\ln (| 63 |) - \ln (| 31 |))$

Step 3.10

Simplify.

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Step 3.10.1

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

$\frac{3}{4} \ln (\frac{| 63 |}{| 31 |})$

Step 3.10.2

Combine $\frac{3}{4}$ and $\ln (\frac{| 63 |}{| 31 |})$ .

$\frac{3 \ln (\frac{| 63 |}{| 31 |})}{4}$

Step 3.11

Simplify.

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Step 3.11.1

The absolute value is the distance between a number and zero. The distance between $0$ and $63$ is $63$ .

$\frac{3 \ln (\frac{63}{| 31 |})}{4}$

Step 3.11.2

The absolute value is the distance between a number and zero. The distance between $0$ and $31$ is $31$ .

$\frac{3 \ln (\frac{63}{31})}{4}$

Step 4

Add the areas $\frac{3 \ln (\frac{63}{31})}{4}$ .

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Step 4.1

Simplify $3 \ln (\frac{63}{31})$ by moving $3$ inside the logarithm.

$\frac{\ln ({(\frac{63}{31})}^{3})}{4}$

Step 4.2

Simplify the numerator.

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Step 4.2.1

Apply the product rule to $\frac{63}{31}$ .

$\frac{\ln (\frac{63^{3}}{31^{3}})}{4}$

Step 4.2.2

Raise $63$ to the power of $3$ .

$\frac{\ln (\frac{250047}{31^{3}})}{4}$

Step 4.2.3

Raise $31$ to the power of $3$ .

$\frac{\ln (\frac{250047}{29791})}{4}$

Step 4.3

Rewrite $\frac{\ln (\frac{250047}{29791})}{4}$ as $\frac{1}{4} \ln (\frac{250047}{29791})$ .

$\frac{1}{4} \ln (\frac{250047}{29791})$

Step 4.4

Simplify $\frac{1}{4} \ln (\frac{250047}{29791})$ by moving $\frac{1}{4}$ inside the logarithm.

$\ln ({(\frac{250047}{29791})}^{\frac{1}{4}})$

Step 4.5

Apply the product rule to $\frac{250047}{29791}$ .

$\ln (\frac{250047^{\frac{1}{4}}}{29791^{\frac{1}{4}}})$

Step 5