Calculus Examples

$y = x^{4}$ and $y = 8 x$

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.

$x^{4} = 8 x$

Step 1.2

Solve $x^{4} = 8 x$ for $x$ .

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Step 1.2.1

Subtract $8 x$ from both sides of the equation.

$x^{4} - 8 x = 0$

Step 1.2.2

Factor the left side of the equation.

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Step 1.2.2.1

Factor $x$ out of $x^{4} - 8 x$ .

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Step 1.2.2.1.1

Factor $x$ out of $x^{4}$ .

$x \cdot x^{3} - 8 x = 0$

Step 1.2.2.1.2

Factor $x$ out of $- 8 x$ .

$x \cdot x^{3} + x \cdot - 8 = 0$

Step 1.2.2.1.3

Factor $x$ out of $x \cdot x^{3} + x \cdot - 8$ .

$x (x^{3} - 8) = 0$

Step 1.2.2.2

Rewrite $8$ as $2^{3}$ .

$x (x^{3} - 2^{3}) = 0$

Step 1.2.2.3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 2$ .

$x ((x - 2) (x^{2} + x \cdot 2 + 2^{2})) = 0$

Step 1.2.2.4

Factor.

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Step 1.2.2.4.1

Simplify.

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Step 1.2.2.4.1.1

Move $2$ to the left of $x$ .

$x ((x - 2) (x^{2} + 2 x + 2^{2})) = 0$

Step 1.2.2.4.1.2

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

$x ((x - 2) (x^{2} + 2 x + 4)) = 0$

Step 1.2.2.4.2

Remove unnecessary parentheses.

$x (x - 2) (x^{2} + 2 x + 4) = 0$

Step 1.2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 2 = 0$

$x^{2} + 2 x + 4 = 0 + y = 8 x$

Step 1.2.4

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.5

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 1.2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 1.2.6

Set $x^{2} + 2 x + 4$ equal to $0$ and solve for $x$ .

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Step 1.2.6.1

Set $x^{2} + 2 x + 4$ equal to $0$ .

$x^{2} + 2 x + 4 = 0$

Step 1.2.6.2

Solve $x^{2} + 2 x + 4 = 0$ for $x$ .

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Step 1.2.6.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a} + y = 8 x$

Step 1.2.6.2.2

Substitute the values $a = 1$ , $b = 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1} + y = 8 x$

Step 1.2.6.2.3

Simplify.

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Step 1.2.6.2.3.1

Simplify the numerator.

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Step 1.2.6.2.3.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 1.2.6.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 1.2.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Step 1.2.6.2.3.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 1.2.6.2.3.1.3

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 1.2.6.2.3.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 1.2.6.2.3.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 1.2.6.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 1.2.6.2.3.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.2.6.2.3.1.7.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 1.2.6.2.3.1.7.2

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 1.2.6.2.3.1.8

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 1.2.6.2.3.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 1.2.6.2.3.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Step 1.2.6.2.3.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Step 1.2.6.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.2.6.2.4.1

Simplify the numerator.

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Step 1.2.6.2.4.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 1.2.6.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 1.2.6.2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Step 1.2.6.2.4.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 1.2.6.2.4.1.3

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 1.2.6.2.4.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 1.2.6.2.4.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 1.2.6.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 1.2.6.2.4.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.2.6.2.4.1.7.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 1.2.6.2.4.1.7.2

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 1.2.6.2.4.1.8

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 1.2.6.2.4.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 1.2.6.2.4.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Step 1.2.6.2.4.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Step 1.2.6.2.4.4

Change the $\pm$ to $+$ .

$x = - 1 + i \sqrt{3}$

Step 1.2.6.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.2.6.2.5.1

Simplify the numerator.

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Step 1.2.6.2.5.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 1.2.6.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 1.2.6.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Step 1.2.6.2.5.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 1.2.6.2.5.1.3

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 1.2.6.2.5.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 1.2.6.2.5.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 1.2.6.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 1.2.6.2.5.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.2.6.2.5.1.7.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 1.2.6.2.5.1.7.2

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 1.2.6.2.5.1.8

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 1.2.6.2.5.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 1.2.6.2.5.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Step 1.2.6.2.5.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Step 1.2.6.2.5.4

Change the $\pm$ to $-$ .

$x = - 1 - i \sqrt{3}$

Step 1.2.6.2.6

The final answer is the combination of both solutions.

$x = - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Step 1.2.7

The final solution is all the values that make $x (x - 2) (x^{2} + 2 x + 4) = 0$ true.

$x = 0, 2, - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Step 1.3

Evaluate $y$ when $x = 0$ .

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Step 1.3.1

Substitute $0$ for $x$ .

$y = 8 (0)$

Step 1.3.2

Multiply $8$ by $0$ .

$y = 0$

Step 1.4

Evaluate $y$ when $x = 2$ .

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Step 1.4.1

Substitute $2$ for $x$ .

$y = 8 (2)$

Step 1.4.2

Multiply $8$ by $2$ .

$y = 16$

Step 1.5

Evaluate $y$ when $x = - 1 + i \sqrt{3}$ .

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Step 1.5.1

Substitute $- 1 + i \sqrt{3}$ for $x$ .

$y = 8 (- 1 + i \sqrt{3})$

Step 1.5.2

Simplify $8 (- 1 + i \sqrt{3})$ .

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Step 1.5.2.1

Apply the distributive property.

$y = 8 \cdot - 1 + 8 (i \sqrt{3})$

Step 1.5.2.2

Multiply $8$ by $- 1$ .

$y = - 8 + 8 i \sqrt{3}$

Step 1.6

Evaluate $y$ when $x = - 1 - i \sqrt{3}$ .

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Step 1.6.1

Substitute $- 1 - i \sqrt{3}$ for $x$ .

$y = 8 (- 1 - i \sqrt{3})$

Step 1.6.2

Simplify $8 (- 1 - i \sqrt{3})$ .

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Step 1.6.2.1

Apply the distributive property.

$y = 8 \cdot - 1 + 8 (- i \sqrt{3})$

Step 1.6.2.2

Multiply.

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Step 1.6.2.2.1

Multiply $8$ by $- 1$ .

$y = - 8 + 8 (- i \sqrt{3})$

Step 1.6.2.2.2

Multiply $- 1$ by $8$ .

$y = - 8 - 8 i \sqrt{3}$

Step 1.7

List all of the solutions.

$y = 0, x = 0$

$y = 16, x = 2$

$y = - 8 + 8 i \sqrt{3}, x = - 1 + i \sqrt{3}$

$y = - 8 - 8 i \sqrt{3}, x = - 1 - i \sqrt{3}$

$y = 0, x = 0$

$y = 16, x = 2$

$y = - 8 + 8 i \sqrt{3}, x = - 1 + i \sqrt{3}$

$y = - 8 - 8 i \sqrt{3}, x = - 1 - i \sqrt{3}$

Step 2

The area between the given curves is unbounded.

Unbounded area

Step 3