Calculus Examples

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

Step 1

Solve by substitution to find the intersection between the curves.

Tap for more steps...

Step 1.1

Eliminate the equal sides of each equation and combine.

$x^{4} = x$

Step 1.2

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

Tap for more steps...

Step 1.2.1

Subtract $x$ from both sides of the equation.

$x^{4} - x = 0$

Step 1.2.2

Factor the left side of the equation.

Tap for more steps...

Step 1.2.2.1

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

Tap for more steps...

Step 1.2.2.1.1

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

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

Step 1.2.2.1.2

Factor $x$ out of $- x$ .

$x \cdot x^{3} + x \cdot - 1 = 0$

Step 1.2.2.1.3

Factor $x$ out of $x \cdot x^{3} + x \cdot - 1$ .

$x (x^{3} - 1) = 0$

Step 1.2.2.2

Rewrite $1$ as $1^{3}$ .

$x (x^{3} - 1^{3}) = 0$

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 $b = 1$ .

$x ((x - 1) (x^{2} + x \cdot 1 + 1^{2})) = 0$

Step 1.2.2.4

Factor.

Tap for more steps...

Step 1.2.2.4.1

Simplify.

Tap for more steps...

Step 1.2.2.4.1.1

Multiply $x$ by $1$ .

$x ((x - 1) (x^{2} + x + 1^{2})) = 0$

Step 1.2.2.4.1.2

One to any power is one.

$x ((x - 1) (x^{2} + x + 1)) = 0$

Step 1.2.2.4.2

Remove unnecessary parentheses.

$x (x - 1) (x^{2} + x + 1) = 0$

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 $0$ .

$x = 0$

$x - 1 = 0$

$x^{2} + x + 1 = 0 + y = x$

Step 1.2.4

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.5

Set $x - 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 1.2.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 1.2.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.2.6

Set $x^{2} + x + 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 1.2.6.1

Set $x^{2} + x + 1$ equal to $0$ .

$x^{2} + x + 1 = 0$

Step 1.2.6.2

Solve $x^{2} + x + 1 = 0$ for $x$ .

Tap for more steps...

Step 1.2.6.2.1

Use the quadratic formula to find the solutions.

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

Step 1.2.6.2.2

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

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

Step 1.2.6.2.3

Simplify.

Tap for more steps...

Step 1.2.6.2.3.1

Simplify the numerator.

Tap for more steps...

Step 1.2.6.2.3.1.1

One to any power is one.

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

Step 1.2.6.2.3.1.2

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

Tap for more steps...

Step 1.2.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.2.6.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 1.2.6.2.3.1.3

Subtract $4$ from $1$ .

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

Step 1.2.6.2.3.1.4

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 1.2.6.2.3.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Step 1.2.6.2.3.1.6

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

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

Step 1.2.6.2.3.2

Multiply $2$ by $1$ .

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

Step 1.2.6.2.4

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

Tap for more steps...

Step 1.2.6.2.4.1

Simplify the numerator.

Tap for more steps...

Step 1.2.6.2.4.1.1

One to any power is one.

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

Step 1.2.6.2.4.1.2

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

Tap for more steps...

Step 1.2.6.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.2.6.2.4.1.2.2

Multiply $- 4$ by $1$ .

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

Step 1.2.6.2.4.1.3

Subtract $4$ from $1$ .

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

Step 1.2.6.2.4.1.4

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 1.2.6.2.4.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Step 1.2.6.2.4.1.6

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

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

Step 1.2.6.2.4.2

Multiply $2$ by $1$ .

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

Step 1.2.6.2.4.3

Change the $\pm$ to $+$ .

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

Step 1.2.6.2.4.4

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

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

Step 1.2.6.2.4.5

Factor $- 1$ out of $i \sqrt{3}$ .

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

Step 1.2.6.2.4.6

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

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

Step 1.2.6.2.4.7

Move the negative in front of the fraction.

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

Step 1.2.6.2.5

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

Tap for more steps...

Step 1.2.6.2.5.1

Simplify the numerator.

Tap for more steps...

Step 1.2.6.2.5.1.1

One to any power is one.

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

Step 1.2.6.2.5.1.2

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

Tap for more steps...

Step 1.2.6.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.2.6.2.5.1.2.2

Multiply $- 4$ by $1$ .

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

Step 1.2.6.2.5.1.3

Subtract $4$ from $1$ .

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

Step 1.2.6.2.5.1.4

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 1.2.6.2.5.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Step 1.2.6.2.5.1.6

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

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

Step 1.2.6.2.5.2

Multiply $2$ by $1$ .

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

Step 1.2.6.2.5.3

Change the $\pm$ to $-$ .

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

Step 1.2.6.2.5.4

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

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

Step 1.2.6.2.5.5

Factor $- 1$ out of $- i \sqrt{3}$ .

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

Step 1.2.6.2.5.6

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

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

Step 1.2.6.2.5.7

Move the negative in front of the fraction.

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

Step 1.2.6.2.6

The final answer is the combination of both solutions.

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

Step 1.2.7

The final solution is all the values that make $x (x - 1) (x^{2} + x + 1) = 0$ true.

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

Step 1.3

Evaluate $y$ when $x = 0$ .

Tap for more steps...

Step 1.3.1

Substitute $0$ for $x$ .

$y = 0$

Step 1.3.2

Remove parentheses.

$y = 0$

Step 1.4

Evaluate $y$ when $x = 1$ .

Tap for more steps...

Step 1.4.1

Substitute $1$ for $x$ .

$y = 1$

Step 1.4.2

Remove parentheses.

$y = 1$

Step 1.5

Evaluate $y$ when $x = - \frac{1 - i \sqrt{3}}{2}$ .

Tap for more steps...

Step 1.5.1

Substitute $- \frac{1 - i \sqrt{3}}{2}$ for $x$ .

$y = - \frac{1 - i \sqrt{3}}{2}$

Step 1.5.2

Remove parentheses.

$y = - \frac{1 - i \sqrt{3}}{2}$

Step 1.6

Evaluate $y$ when $x = - \frac{1 + i \sqrt{3}}{2}$ .

Tap for more steps...

Step 1.6.1

Substitute $- \frac{1 + i \sqrt{3}}{2}$ for $x$ .

$y = - \frac{1 + i \sqrt{3}}{2}$

Step 1.6.2

Remove parentheses.

$y = - \frac{1 + i \sqrt{3}}{2}$

Step 1.7

List all of the solutions.

$y = 0, x = 0$

$y = 1, x = 1$

$y = - \frac{1 - i \sqrt{3}}{2}, x = - \frac{1 - i \sqrt{3}}{2}$

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

$y = 0, x = 0$

$y = 1, x = 1$

$y = - \frac{1 - i \sqrt{3}}{2}, x = - \frac{1 - i \sqrt{3}}{2}$

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

Step 2

The area between the given curves is unbounded.

Unbounded area

Step 3