Calculus Examples

$x = 1$ , $x = 3$ , $y = x^{3} - 8$ , $y = 0$

Step 1

Solve by substitution to find the intersection between the curves.

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Eliminate the equal sides of each equation and combine.

$x^{3} - 8 = 0$

Solve $x^{3} - 8 = 0$ for $x$ .

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Add $8$ to both sides of the equation.

$x^{3} = 8$

Subtract $8$ from both sides of the equation.

$x^{3} - 8 = 0$

Factor the left side of the equation.

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

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

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 $b = 2$ .

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

Simplify.

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

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

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

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

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

$x - 2 = 0$

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

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

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

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

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Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a} + y = 0$

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

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1} + y = 0$

Simplify.

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Simplify the numerator.

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

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

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

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Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

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

Subtract $16$ from $4$ .

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

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

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

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

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

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

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

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

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Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

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

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

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

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Simplify the numerator.

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

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

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

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Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

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

Subtract $16$ from $4$ .

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

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

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

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

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

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

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

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

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Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

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

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

Change the $\pm$ to $+$ .

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

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

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Simplify the numerator.

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

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

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

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Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

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

Subtract $16$ from $4$ .

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

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

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

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

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

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

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

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

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Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

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

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

Change the $\pm$ to $-$ .

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

The final answer is the combination of both solutions.

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

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

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

Substitute $2$ for $x$ .

$y = 0$

List all of the solutions.

$y = 0, x = 2$

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

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

$y = 0, x = 2$

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

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

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.

$A r e a = \int_{1}^{3} x^{3} - 8 d x - \int_{1}^{3} 0 d x$

Step 3

Integrate to find the area between $1$ and $3$ .

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Integrate to find the area between $1$ and $3$ .

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Combine the integrals into a single integral.

$\int_{1}^{3} 0 - (x^{3} - 8) d x$

Subtract $x^{3} - 8$ from $0$ .

$\int_{1}^{3} - (x^{3} - 8) d x$

Apply the distributive property.

$\int_{1}^{3} - x^{3} + 8 d x$

Split the single integral into multiple integrals.

$\int_{1}^{3} - x^{3} d x + \int_{1}^{3} 8 d x$

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$- \int_{1}^{3} x^{3} d x + \int_{1}^{3} 8 d x$

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

$- (\frac{1}{4} x^{4}]_{1}^{3}) + \int_{1}^{3} 8 d x$

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

$- (\frac{x^{4}}{4}]_{1}^{3}) + \int_{1}^{3} 8 d x$

Apply the constant rule.

$- (\frac{x^{4}}{4}]_{1}^{3}) + 8 x]_{1}^{3}$

Substitute and simplify.

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Evaluate $\frac{x^{4}}{4}$ at $3$ and at $1$ .

$- ((\frac{3^{4}}{4}) - \frac{1^{4}}{4}) + 8 x]_{1}^{3}$

Evaluate $8 x$ at $3$ and at $1$ .

$- (\frac{3^{4}}{4} - \frac{1^{4}}{4}) + 8 \cdot 3 - 8 \cdot 1$

Simplify.

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Raise $3$ to the power of $4$ .

$- (\frac{81}{4} - \frac{1^{4}}{4}) + 8 \cdot 3 - 8 \cdot 1$

One to any power is one.

$- (\frac{81}{4} - \frac{1}{4}) + 8 \cdot 3 - 8 \cdot 1$

Combine the numerators over the common denominator.

$- \frac{81 - 1}{4} + 8 \cdot 3 - 8 \cdot 1$

Subtract $1$ from $81$ .

$- \frac{80}{4} + 8 \cdot 3 - 8 \cdot 1$

Cancel the common factor of $80$ and $4$ .

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Factor $4$ out of $80$ .

$- \frac{4 \cdot 20}{4} + 8 \cdot 3 - 8 \cdot 1$

Cancel the common factors.

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Factor $4$ out of $4$ .

$- \frac{4 \cdot 20}{4 (1)} + 8 \cdot 3 - 8 \cdot 1$

Cancel the common factor.

$- \frac{4 \cdot 20}{4 \cdot 1} + 8 \cdot 3 - 8 \cdot 1$

Rewrite the expression.

$- \frac{20}{1} + 8 \cdot 3 - 8 \cdot 1$

Divide $20$ by $1$ .

$- 1 \cdot 20 + 8 \cdot 3 - 8 \cdot 1$

Multiply $- 1$ by $20$ .

$- 20 + 8 \cdot 3 - 8 \cdot 1$

Multiply $8$ by $3$ .

$- 20 + 24 - 8 \cdot 1$

Multiply $- 8$ by $1$ .

$- 20 + 24 - 8$

Subtract $8$ from $24$ .

$- 20 + 16$

Add $- 20$ and $16$ .

$- 4$

Combine the integrals into a single integral.

$\int_{1}^{3} x^{3} - 8 - (0) d x$

Subtract $0$ from $x^{3} - 8$ .

$\int_{1}^{3} x^{3} - 8 d x$

Split the single integral into multiple integrals.

$\int_{1}^{3} x^{3} d x + \int_{1}^{3} - 8 d x$

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

$\frac{1}{4} x^{4}]_{1}^{3} + \int_{1}^{3} - 8 d x$

Apply the constant rule.

$\frac{1}{4} x^{4}]_{1}^{3} + - 8 x]_{1}^{3}$

Simplify the answer.

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Combine $\frac{1}{4} x^{4}]_{1}^{3}$ and $- 8 x]_{1}^{3}$ .

$\frac{1}{4} x^{4} - 8 x]_{1}^{3}$

Substitute and simplify.

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Evaluate $\frac{1}{4} x^{4} - 8 x$ at $3$ and at $1$ .

$(\frac{1}{4} \cdot 3^{4} - 8 \cdot 3) - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

Simplify.

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Raise $3$ to the power of $4$ .

$\frac{1}{4} \cdot 81 - 8 \cdot 3 - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

Combine $\frac{1}{4}$ and $81$ .

$\frac{81}{4} - 8 \cdot 3 - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

Multiply $- 8$ by $3$ .

$\frac{81}{4} - 24 - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

To write $- 24$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{81}{4} - 24 \cdot \frac{4}{4} - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

Combine $- 24$ and $\frac{4}{4}$ .

$\frac{81}{4} + \frac{- 24 \cdot 4}{4} - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

Combine the numerators over the common denominator.

$\frac{81 - 24 \cdot 4}{4} - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

Simplify the numerator.

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Multiply $- 24$ by $4$ .

$\frac{81 - 96}{4} - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

Subtract $96$ from $81$ .

$\frac{- 15}{4} - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

Move the negative in front of the fraction.

$- \frac{15}{4} - (\frac{1}{4} \cdot 1^{4} - 8 \cdot 1)$

One to any power is one.

$- \frac{15}{4} - (\frac{1}{4} \cdot 1 - 8 \cdot 1)$

Multiply $\frac{1}{4}$ by $1$ .

$- \frac{15}{4} - (\frac{1}{4} - 8 \cdot 1)$

Multiply $- 8$ by $1$ .

$- \frac{15}{4} - (\frac{1}{4} - 8)$

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$- \frac{15}{4} - (\frac{1}{4} - 8 \cdot \frac{4}{4})$

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

$- \frac{15}{4} - (\frac{1}{4} + \frac{- 8 \cdot 4}{4})$

Combine the numerators over the common denominator.

$- \frac{15}{4} - \frac{1 - 8 \cdot 4}{4}$

Simplify the numerator.

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Multiply $- 8$ by $4$ .

$- \frac{15}{4} - \frac{1 - 32}{4}$

Subtract $32$ from $1$ .

$- \frac{15}{4} - \frac{- 31}{4}$

Move the negative in front of the fraction.

$- \frac{15}{4} - - \frac{31}{4}$

Multiply $- 1$ by $- 1$ .

$- \frac{15}{4} + 1 (\frac{31}{4})$

Multiply $\frac{31}{4}$ by $1$ .

$- \frac{15}{4} + \frac{31}{4}$

Combine the numerators over the common denominator.

$\frac{- 15 + 31}{4}$

Add $- 15$ and $31$ .

$\frac{16}{4}$

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

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Factor $4$ out of $16$ .

$\frac{4 \cdot 4}{4}$

Cancel the common factors.

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Factor $4$ out of $4$ .

$\frac{4 \cdot 4}{4 (1)}$

Cancel the common factor.

$\frac{4 \cdot 4}{4 \cdot 1}$

Rewrite the expression.

$\frac{4}{1}$

Divide $4$ by $1$ .

$4$

Step 4