Calculus Examples

$\int_{1}^{2} \frac{v^{4} + 4 v^{8}}{v^{5}} d v$

Step 1

Simplify.

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Step 1.1

Factor $v^{4}$ out of $v^{4} + 4 v^{8}$ .

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Step 1.1.1

Multiply by $1$ .

$\int_{1}^{2} \frac{v^{4} \cdot 1 + 4 v^{8}}{v^{5}} d v$

Step 1.1.2

Factor $v^{4}$ out of $4 v^{8}$ .

$\int_{1}^{2} \frac{v^{4} \cdot 1 + v^{4} (4 v^{4})}{v^{5}} d v$

Step 1.1.3

Factor $v^{4}$ out of $v^{4} \cdot 1 + v^{4} (4 v^{4})$ .

$\int_{1}^{2} \frac{v^{4} (1 + 4 v^{4})}{v^{5}} d v$

Step 1.2

Cancel the common factors.

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Step 1.2.1

Factor $v^{4}$ out of $v^{5}$ .

$\int_{1}^{2} \frac{v^{4} (1 + 4 v^{4})}{v^{4} v} d v$

Step 1.2.2

Cancel the common factor.

$\int_{1}^{2} \frac{v^{4} (1 + 4 v^{4})}{v^{4} v} d v$

Step 1.2.3

Rewrite the expression.

$\int_{1}^{2} \frac{1 + 4 v^{4}}{v} d v$

Step 2

Reorder $1$ and $4 v^{4}$ .

$\int_{1}^{2} \frac{4 v^{4} + 1}{v} d v$

Step 3

Divide $4 v^{4} + 1$ by $v$ .

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Step 3.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$v$

$0$

$4 v^{4}$

$0 v^{3}$

$0 v^{2}$

$0 v$

$1$

Step 3.2

Divide the highest order term in the dividend $4 v^{4}$ by the highest order term in divisor $v$ .

$4 v^{3}$

$v$

$0$

$4 v^{4}$

$0 v^{3}$

$0 v^{2}$

$0 v$

$1$

Step 3.3

Multiply the new quotient term by the divisor.

$4 v^{3}$

$v$

$0$

$4 v^{4}$

$0 v^{3}$

$0 v^{2}$

$0 v$

$1$

$4 v^{4}$

$0$

Step 3.4

The expression needs to be subtracted from the dividend, so change all the signs in $4 v^{4} + 0$

$4 v^{3}$

$v$

$0$

$4 v^{4}$

$0 v^{3}$

$0 v^{2}$

$0 v$

$1$

$4 v^{4}$

$0$

Step 3.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$4 v^{3}$

$v$

$0$

$4 v^{4}$

$0 v^{3}$

$0 v^{2}$

$0 v$

$1$

$4 v^{4}$

$0$

Step 3.6

Pull the next term from the original dividend down into the current dividend.

$4 v^{3}$

$v$

$0$

$4 v^{4}$

$0 v^{3}$

$0 v^{2}$

$0 v$

$1$

$4 v^{4}$

$0$

$0 v^{2}$

$0 v$

Step 3.7

The final answer is the quotient plus the remainder over the divisor.

$\int_{1}^{2} 4 v^{3} + \frac{1}{v} d v$

Step 4

Split the single integral into multiple integrals.

$\int_{1}^{2} 4 v^{3} d v + \int_{1}^{2} \frac{1}{v} d v$

Step 5

Since $4$ is constant with respect to $v$ , move $4$ out of the integral.

$4 \int_{1}^{2} v^{3} d v + \int_{1}^{2} \frac{1}{v} d v$

Step 6

By the Power Rule, the integral of $v^{3}$ with respect to $v$ is $\frac{1}{4} v^{4}$ .

$4 (\frac{1}{4} v^{4}]_{1}^{2}) + \int_{1}^{2} \frac{1}{v} d v$

Step 7

Combine $\frac{1}{4}$ and $v^{4}$ .

$4 (\frac{v^{4}}{4}]_{1}^{2}) + \int_{1}^{2} \frac{1}{v} d v$

Step 8

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

$4 (\frac{v^{4}}{4}]_{1}^{2}) + \ln (| v |)]_{1}^{2}$

Step 9

Simplify the answer.

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Step 9.1

Substitute and simplify.

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Step 9.1.1

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

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

Step 9.1.2

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

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

Step 9.1.3

Simplify.

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Step 9.1.3.1

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

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

Step 9.1.3.2

Cancel the common factor of $16$ and $4$ .

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Step 9.1.3.2.1

Factor $4$ out of $16$ .

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

Step 9.1.3.2.2

Cancel the common factors.

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Step 9.1.3.2.2.1

Factor $4$ out of $4$ .

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

Step 9.1.3.2.2.2

Cancel the common factor.

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

Step 9.1.3.2.2.3

Rewrite the expression.

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

Step 9.1.3.2.2.4

Divide $4$ by $1$ .

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

Step 9.1.3.3

One to any power is one.

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

Step 9.1.3.4

To write $4$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 9.1.3.5

Combine $4$ and $\frac{4}{4}$ .

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

Step 9.1.3.6

Combine the numerators over the common denominator.

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

Step 9.1.3.7

Simplify the numerator.

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Step 9.1.3.7.1

Multiply $4$ by $4$ .

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

Step 9.1.3.7.2

Subtract $1$ from $16$ .

$4 (\frac{15}{4}) + \ln (| 2 |) - \ln (| 1 |)$

Step 9.1.3.8

Combine $4$ and $\frac{15}{4}$ .

$\frac{4 \cdot 15}{4} + \ln (| 2 |) - \ln (| 1 |)$

Step 9.1.3.9

Multiply $4$ by $15$ .

$\frac{60}{4} + \ln (| 2 |) - \ln (| 1 |)$

Step 9.1.3.10

Cancel the common factor of $60$ and $4$ .

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Step 9.1.3.10.1

Factor $4$ out of $60$ .

$\frac{4 \cdot 15}{4} + \ln (| 2 |) - \ln (| 1 |)$

Step 9.1.3.10.2

Cancel the common factors.

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Step 9.1.3.10.2.1

Factor $4$ out of $4$ .

$\frac{4 \cdot 15}{4 (1)} + \ln (| 2 |) - \ln (| 1 |)$

Step 9.1.3.10.2.2

Cancel the common factor.

$\frac{4 \cdot 15}{4 \cdot 1} + \ln (| 2 |) - \ln (| 1 |)$

Step 9.1.3.10.2.3

Rewrite the expression.

$\frac{15}{1} + \ln (| 2 |) - \ln (| 1 |)$

Step 9.1.3.10.2.4

Divide $15$ by $1$ .

$15 + \ln (| 2 |) - \ln (| 1 |)$

Step 9.2

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

$15 + \ln (\frac{| 2 |}{| 1 |})$

Step 9.3

Simplify.

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Step 9.3.1

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

$15 + \ln (\frac{2}{| 1 |})$

Step 9.3.2

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

$15 + \ln (\frac{2}{1})$

Step 9.3.3

Divide $2$ by $1$ .

$15 + \ln (2)$

Step 10

The result can be shown in multiple forms.

Exact Form:

$15 + \ln (2)$

Decimal Form:

$15.69314718 \dots$

Step 11