Calculus Examples

$x = \frac{y^{4}}{8} + \frac{1}{4 y^{2}}$ , $1 \leq y \leq 3$

Step 1

Check if $f (y) = \frac{y^{4}}{8} + \frac{1}{4 y^{2}}$ is continuous.

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Step 1.1

To find whether the function is continuous on $[1, 3]$ or not, find the domain of $f (y) = \frac{y^{4}}{8} + \frac{1}{4 y^{2}}$ .

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Step 1.1.1

Set the denominator in $\frac{1}{4 y^{2}}$ equal to $0$ to find where the expression is undefined.

$4 y^{2} = 0$

Step 1.1.2

Solve for $y$ .

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Step 1.1.2.1

Divide each term in $4 y^{2} = 0$ by $4$ and simplify.

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Step 1.1.2.1.1

Divide each term in $4 y^{2} = 0$ by $4$ .

$\frac{4 y^{2}}{4} = \frac{0}{4}$

Step 1.1.2.1.2

Simplify the left side.

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Step 1.1.2.1.2.1

Cancel the common factor of $4$ .

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Step 1.1.2.1.2.1.1

Cancel the common factor.

$\frac{4 y^{2}}{4} = \frac{0}{4}$

Step 1.1.2.1.2.1.2

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{0}{4}$

Step 1.1.2.1.3

Simplify the right side.

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Step 1.1.2.1.3.1

Divide $0$ by $4$ .

$y^{2} = 0$

Step 1.1.2.2

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

$y = \pm \sqrt{0}$

Step 1.1.2.3

Simplify $\pm \sqrt{0}$ .

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Step 1.1.2.3.1

Rewrite $0$ as $0^{2}$ .

$y = \pm \sqrt{0^{2}}$

Step 1.1.2.3.2

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

$y = \pm 0$

Step 1.1.2.3.3

Plus or minus $0$ is $0$ .

$y = 0$

Step 1.1.3

The domain is all values of $y$ that make the expression defined.

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${y | y \neq 0}$

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${y | y \neq 0}$

Step 1.2

$f (y)$ is continuous on $[1, 3]$ .

The function is continuous.

Step 2

Check if $f (y) = \frac{y^{4}}{8} + \frac{1}{4 y^{2}}$ is differentiable.

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Step 2.1

Find the derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

By the Sum Rule, the derivative of $\frac{y^{4}}{8} + \frac{1}{4 y^{2}}$ with respect to $y$ is $\frac{d}{d y} [\frac{y^{4}}{8}] + \frac{d}{d y} [\frac{1}{4 y^{2}}]$ .

$\frac{d}{d y} [\frac{y^{4}}{8}] + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.1.1.2

Evaluate $\frac{d}{d y} [\frac{y^{4}}{8}]$ .

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Step 2.1.1.2.1

Since $\frac{1}{8}$ is constant with respect to $y$ , the derivative of $\frac{y^{4}}{8}$ with respect to $y$ is $\frac{1}{8} \frac{d}{d y} [y^{4}]$ .

$\frac{1}{8} \frac{d}{d y} [y^{4}] + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.1.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 4$ .

$\frac{1}{8} (4 y^{3}) + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.1.1.2.3

Combine $4$ and $\frac{1}{8}$ .

$\frac{4}{8} y^{3} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.1.1.2.4

Combine $\frac{4}{8}$ and $y^{3}$ .

$\frac{4 y^{3}}{8} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.1.1.2.5

Cancel the common factor of $4$ and $8$ .

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Step 2.1.1.2.5.1

Factor $4$ out of $4 y^{3}$ .

$\frac{4 (y^{3})}{8} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.1.1.2.5.2

Cancel the common factors.

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Step 2.1.1.2.5.2.1

Factor $4$ out of $8$ .

$\frac{4 y^{3}}{4 \cdot 2} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.1.1.2.5.2.2

Cancel the common factor.

$\frac{4 y^{3}}{4 \cdot 2} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.1.1.2.5.2.3

Rewrite the expression.

$\frac{y^{3}}{2} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.1.1.3

Evaluate $\frac{d}{d y} [\frac{1}{4 y^{2}}]$ .

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Step 2.1.1.3.1

Since $\frac{1}{4}$ is constant with respect to $y$ , the derivative of $\frac{1}{4 y^{2}}$ with respect to $y$ is $\frac{1}{4} \frac{d}{d y} [\frac{1}{y^{2}}]$ .

$\frac{y^{3}}{2} + \frac{1}{4} \frac{d}{d y} [\frac{1}{y^{2}}]$

Step 2.1.1.3.2

Rewrite $\frac{1}{y^{2}}$ as ${(y^{2})}^{- 1}$ .

$\frac{y^{3}}{2} + \frac{1}{4} \frac{d}{d y} [{(y^{2})}^{- 1}]$

Step 2.1.1.3.3

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{- 1}$ and $g (y) = y^{2}$ .

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Step 2.1.1.3.3.1

To apply the Chain Rule, set $u$ as $y^{2}$ .

$\frac{y^{3}}{2} + \frac{1}{4} (\frac{d}{d u} [u^{- 1}] \frac{d}{d y} [y^{2}])$

Step 2.1.1.3.3.2

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

$\frac{y^{3}}{2} + \frac{1}{4} (- u^{- 2} \frac{d}{d y} [y^{2}])$

Step 2.1.1.3.3.3

Replace all occurrences of $u$ with $y^{2}$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- {(y^{2})}^{- 2} \frac{d}{d y} [y^{2}])$

Step 2.1.1.3.4

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 2$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- {(y^{2})}^{- 2} (2 y))$

Step 2.1.1.3.5

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

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Step 2.1.1.3.5.1

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

$\frac{y^{3}}{2} + \frac{1}{4} (- y^{2 \cdot - 2} (2 y))$

Step 2.1.1.3.5.2

Multiply $2$ by $- 2$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- y^{- 4} (2 y))$

Step 2.1.1.3.6

Multiply $2$ by $- 1$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- 2 y^{- 4} y)$

Step 2.1.1.3.7

Raise $y$ to the power of $1$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- 2 (y^{1} y^{- 4}))$

Step 2.1.1.3.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{y^{3}}{2} + \frac{1}{4} (- 2 y^{1 - 4})$

Step 2.1.1.3.9

Subtract $4$ from $1$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- 2 y^{- 3})$

Step 2.1.1.3.10

Combine $- 2$ and $\frac{1}{4}$ .

$\frac{y^{3}}{2} + \frac{- 2}{4} y^{- 3}$

Step 2.1.1.3.11

Combine $\frac{- 2}{4}$ and $y^{- 3}$ .

$\frac{y^{3}}{2} + \frac{- 2 y^{- 3}}{4}$

Step 2.1.1.3.12

Move $y^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{y^{3}}{2} + \frac{- 2}{4 y^{3}}$

Step 2.1.1.3.13

Cancel the common factor of $- 2$ and $4$ .

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Step 2.1.1.3.13.1

Factor $2$ out of $- 2$ .

$\frac{y^{3}}{2} + \frac{2 (- 1)}{4 y^{3}}$

Step 2.1.1.3.13.2

Cancel the common factors.

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Step 2.1.1.3.13.2.1

Factor $2$ out of $4 y^{3}$ .

$\frac{y^{3}}{2} + \frac{2 (- 1)}{2 (2 y^{3})}$

Step 2.1.1.3.13.2.2

Cancel the common factor.

$\frac{y^{3}}{2} + \frac{2 \cdot - 1}{2 (2 y^{3})}$

Step 2.1.1.3.13.2.3

Rewrite the expression.

$\frac{y^{3}}{2} + \frac{- 1}{2 y^{3}}$

Step 2.1.1.3.14

Move the negative in front of the fraction.

$f' (y) = \frac{y^{3}}{2} - \frac{1}{2 y^{3}}$

Step 2.1.2

The first derivative of $f (y)$ with respect to $y$ is $\frac{y^{3}}{2} - \frac{1}{2 y^{3}}$ .

$\frac{y^{3}}{2} - \frac{1}{2 y^{3}}$

Step 2.2

Find if the derivative is continuous on $[1, 3]$ .

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Step 2.2.1

To find whether the function is continuous on $[1, 3]$ or not, find the domain of $f' (y) = \frac{y^{3}}{2} - \frac{1}{2 y^{3}}$ .

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Step 2.2.1.1

Set the denominator in $\frac{1}{2 y^{3}}$ equal to $0$ to find where the expression is undefined.

$2 y^{3} = 0$

Step 2.2.1.2

Solve for $y$ .

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Step 2.2.1.2.1

Divide each term in $2 y^{3} = 0$ by $2$ and simplify.

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Step 2.2.1.2.1.1

Divide each term in $2 y^{3} = 0$ by $2$ .

$\frac{2 y^{3}}{2} = \frac{0}{2}$

Step 2.2.1.2.1.2

Simplify the left side.

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Step 2.2.1.2.1.2.1

Cancel the common factor of $2$ .

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Step 2.2.1.2.1.2.1.1

Cancel the common factor.

$\frac{2 y^{3}}{2} = \frac{0}{2}$

Step 2.2.1.2.1.2.1.2

Divide $y^{3}$ by $1$ .

$y^{3} = \frac{0}{2}$

Step 2.2.1.2.1.3

Simplify the right side.

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Step 2.2.1.2.1.3.1

Divide $0$ by $2$ .

$y^{3} = 0$

Step 2.2.1.2.2

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

$y = \sqrt[3]{0}$

Step 2.2.1.2.3

Simplify $\sqrt[3]{0}$ .

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Step 2.2.1.2.3.1

Rewrite $0$ as $0^{3}$ .

$y = \sqrt[3]{0^{3}}$

Step 2.2.1.2.3.2

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

$y = 0$

Step 2.2.1.3

The domain is all values of $y$ that make the expression defined.

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${y | y \neq 0}$

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${y | y \neq 0}$

Step 2.2.2

$f' (y)$ is continuous on $[1, 3]$ .

The function is continuous.

Step 2.3

The function is differentiable on $[1, 3]$ because the derivative is continuous on $[1, 3]$ .

The function is differentiable.

Step 3

For arc length to be guaranteed, the function and its derivative must both be continuous on the closed interval $[1, 3]$ .

The function and its derivative are continuous on the closed interval $[1, 3]$ .

Step 4

Find the derivative of $f (y) = \frac{y^{4}}{8} + \frac{1}{4 y^{2}}$ .

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Step 4.1

By the Sum Rule, the derivative of $\frac{y^{4}}{8} + \frac{1}{4 y^{2}}$ with respect to $y$ is $\frac{d}{d y} [\frac{y^{4}}{8}] + \frac{d}{d y} [\frac{1}{4 y^{2}}]$ .

$\frac{d}{d y} [\frac{y^{4}}{8}] + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 4.2

Evaluate $\frac{d}{d y} [\frac{y^{4}}{8}]$ .

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Step 4.2.1

Since $\frac{1}{8}$ is constant with respect to $y$ , the derivative of $\frac{y^{4}}{8}$ with respect to $y$ is $\frac{1}{8} \frac{d}{d y} [y^{4}]$ .

$\frac{1}{8} \frac{d}{d y} [y^{4}] + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 4.2.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 4$ .

$\frac{1}{8} (4 y^{3}) + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 4.2.3

Combine $4$ and $\frac{1}{8}$ .

$\frac{4}{8} y^{3} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 4.2.4

Combine $\frac{4}{8}$ and $y^{3}$ .

$\frac{4 y^{3}}{8} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 4.2.5

Cancel the common factor of $4$ and $8$ .

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Step 4.2.5.1

Factor $4$ out of $4 y^{3}$ .

$\frac{4 (y^{3})}{8} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 4.2.5.2

Cancel the common factors.

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Step 4.2.5.2.1

Factor $4$ out of $8$ .

$\frac{4 y^{3}}{4 \cdot 2} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 4.2.5.2.2

Cancel the common factor.

$\frac{4 y^{3}}{4 \cdot 2} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 4.2.5.2.3

Rewrite the expression.

$\frac{y^{3}}{2} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 4.3

Evaluate $\frac{d}{d y} [\frac{1}{4 y^{2}}]$ .

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Step 4.3.1

Since $\frac{1}{4}$ is constant with respect to $y$ , the derivative of $\frac{1}{4 y^{2}}$ with respect to $y$ is $\frac{1}{4} \frac{d}{d y} [\frac{1}{y^{2}}]$ .

$\frac{y^{3}}{2} + \frac{1}{4} \frac{d}{d y} [\frac{1}{y^{2}}]$

Step 4.3.2

Rewrite $\frac{1}{y^{2}}$ as ${(y^{2})}^{- 1}$ .

$\frac{y^{3}}{2} + \frac{1}{4} \frac{d}{d y} [{(y^{2})}^{- 1}]$

Step 4.3.3

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{- 1}$ and $g (y) = y^{2}$ .

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Step 4.3.3.1

To apply the Chain Rule, set $u$ as $y^{2}$ .

$\frac{y^{3}}{2} + \frac{1}{4} (\frac{d}{d u} [u^{- 1}] \frac{d}{d y} [y^{2}])$

Step 4.3.3.2

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

$\frac{y^{3}}{2} + \frac{1}{4} (- u^{- 2} \frac{d}{d y} [y^{2}])$

Step 4.3.3.3

Replace all occurrences of $u$ with $y^{2}$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- {(y^{2})}^{- 2} \frac{d}{d y} [y^{2}])$

Step 4.3.4

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 2$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- {(y^{2})}^{- 2} (2 y))$

Step 4.3.5

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

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Step 4.3.5.1

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

$\frac{y^{3}}{2} + \frac{1}{4} (- y^{2 \cdot - 2} (2 y))$

Step 4.3.5.2

Multiply $2$ by $- 2$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- y^{- 4} (2 y))$

Step 4.3.6

Multiply $2$ by $- 1$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- 2 y^{- 4} y)$

Step 4.3.7

Raise $y$ to the power of $1$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- 2 (y^{1} y^{- 4}))$

Step 4.3.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{y^{3}}{2} + \frac{1}{4} (- 2 y^{1 - 4})$

Step 4.3.9

Subtract $4$ from $1$ .

$\frac{y^{3}}{2} + \frac{1}{4} (- 2 y^{- 3})$

Step 4.3.10

Combine $- 2$ and $\frac{1}{4}$ .

$\frac{y^{3}}{2} + \frac{- 2}{4} y^{- 3}$

Step 4.3.11

Combine $\frac{- 2}{4}$ and $y^{- 3}$ .

$\frac{y^{3}}{2} + \frac{- 2 y^{- 3}}{4}$

Step 4.3.12

Move $y^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{y^{3}}{2} + \frac{- 2}{4 y^{3}}$

Step 4.3.13

Cancel the common factor of $- 2$ and $4$ .

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Step 4.3.13.1

Factor $2$ out of $- 2$ .

$\frac{y^{3}}{2} + \frac{2 (- 1)}{4 y^{3}}$

Step 4.3.13.2

Cancel the common factors.

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Step 4.3.13.2.1

Factor $2$ out of $4 y^{3}$ .

$\frac{y^{3}}{2} + \frac{2 (- 1)}{2 (2 y^{3})}$

Step 4.3.13.2.2

Cancel the common factor.

$\frac{y^{3}}{2} + \frac{2 \cdot - 1}{2 (2 y^{3})}$

Step 4.3.13.2.3

Rewrite the expression.

$\frac{y^{3}}{2} + \frac{- 1}{2 y^{3}}$

Step 4.3.14

Move the negative in front of the fraction.

$\frac{y^{3}}{2} - \frac{1}{2 y^{3}}$

Step 5

To find the arc length of a function, use the formula $L = \int_{a}^{b} \sqrt{1 + {(f' (y))}^{2}} d y$ .

$\int_{1}^{3} \sqrt{1 + {(\frac{y^{3}}{2} - \frac{1}{2 y^{3}})}^{2}} d y$

Step 6

Evaluate the integral.

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Step 6.1

Since $\frac{1}{2}$ is constant with respect to $y$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int_{1}^{3} \frac{(y^{2} + 1) (y^{4} - y^{2} + 1)}{y^{3}} d y$

Step 6.2

Apply basic rules of exponents.

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Step 6.2.1

Move $y^{3}$ out of the denominator by raising it to the $- 1$ power.

$\frac{1}{2} \int_{1}^{3} (y^{2} + 1) (y^{4} - y^{2} + 1) {(y^{3})}^{- 1} d y$

Step 6.2.2

Multiply the exponents in ${(y^{3})}^{- 1}$ .

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Step 6.2.2.1

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

$\frac{1}{2} \int_{1}^{3} (y^{2} + 1) (y^{4} - y^{2} + 1) y^{3 \cdot - 1} d y$

Step 6.2.2.2

Multiply $3$ by $- 1$ .

$\frac{1}{2} \int_{1}^{3} (y^{2} + 1) (y^{4} - y^{2} + 1) y^{- 3} d y$

Step 6.3

Expand $(y^{2} + 1) (y^{4} - y^{2} + 1) y^{- 3}$ .

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Step 6.3.1