Calculus Examples

$y = | x^{2} - 4 |$ $y = 0$ $y = 5$

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.

$| x^{2} - 4 | = 0$

Step 1.2

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

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Step 1.2.1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x^{2} - 4 = \pm 0 + y = 0$

$y = 5$

Step 1.2.2

Plus or minus $0$ is $0$ .

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

$y = 5$

Step 1.2.3

Add $4$ to both sides of the equation.

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

$y = 5$

Step 1.2.4

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

$x = \pm \sqrt{4} + y = 0$

$y = 5$

Step 1.2.5

Simplify $\pm \sqrt{4}$ .

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Step 1.2.5.1

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

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

$y = 5$

Step 1.2.5.2

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

$x = \pm 2 + y = 0$

$y = 5$

$x = \pm 2 + y = 0$

$y = 5$

Step 1.2.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.2.6.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 2 + y = 0$

$y = 5$

Step 1.2.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2 + y = 0$

$y = 5$

Step 1.2.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2 + y = 0$

$y = 5$

$x = 2, - 2 + y = 0$

$y = 5$

$x = 2, - 2 + y = 0$

$y = 5$

Step 1.3

Substitute $2$ for $x$ .

$y = 0 y = 5$

Step 1.4

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(2, 5)$

$(- 2, 5)$

$(2, 5)$

$(- 2, 5)$

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.

$A r e a = \int_{- 2}^{2} | x^{2} - 4 | d x - \int_{- 2}^{2} 0 d x$

Step 3

Integrate to find the area between $- 2$ and $2$ .

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Step 3.1

Combine the integrals into a single integral.

$\int_{- 2}^{2} | x^{2} - 4 | - (0) d x$

Step 3.2

Subtract $0$ from $| x^{2} - 4 |$ .

$\int_{- 2}^{2} | x^{2} - 4 | d x$

Step 3.3

Split up the integral depending on where $x^{2} - 4$ is positive and negative.

$\int_{- 2}^{2} - x^{2} + 4 d x$

Step 3.4

Split the single integral into multiple integrals.

$\int_{- 2}^{2} - x^{2} d x + \int_{- 2}^{2} 4 d x$

Step 3.5

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

$- \int_{- 2}^{2} x^{2} d x + \int_{- 2}^{2} 4 d x$

Step 3.6

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

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

Step 3.7

Combine $\frac{1}{3}$ and $x^{3}$ .

$- (\frac{x^{3}}{3}]_{- 2}^{2}) + \int_{- 2}^{2} 4 d x$

Step 3.8

Apply the constant rule.

$- (\frac{x^{3}}{3}]_{- 2}^{2}) + 4 x]_{- 2}^{2}$

Step 3.9

Substitute and simplify.

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Step 3.9.1

Evaluate $\frac{x^{3}}{3}$ at $2$ and at $- 2$ .

$- ((\frac{2^{3}}{3}) - \frac{{(- 2)}^{3}}{3}) + 4 x]_{- 2}^{2}$

Step 3.9.2

Evaluate $4 x$ at $2$ and at $- 2$ .

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

Step 3.9.3

Simplify.

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Step 3.9.3.1

Raise $2$ to the power of $3$ .

$- (\frac{8}{3} - \frac{{(- 2)}^{3}}{3}) + 4 \cdot 2 - 4 \cdot - 2$

Step 3.9.3.2

Raise $- 2$ to the power of $3$ .

$- (\frac{8}{3} - \frac{- 8}{3}) + 4 \cdot 2 - 4 \cdot - 2$

Step 3.9.3.3

Move the negative in front of the fraction.

$- (\frac{8}{3} - - \frac{8}{3}) + 4 \cdot 2 - 4 \cdot - 2$

Step 3.9.3.4

Multiply $- 1$ by $- 1$ .

$- (\frac{8}{3} + 1 (\frac{8}{3})) + 4 \cdot 2 - 4 \cdot - 2$

Step 3.9.3.5

Multiply $\frac{8}{3}$ by $1$ .

$- (\frac{8}{3} + \frac{8}{3}) + 4 \cdot 2 - 4 \cdot - 2$

Step 3.9.3.6

Combine the numerators over the common denominator.

$- \frac{8 + 8}{3} + 4 \cdot 2 - 4 \cdot - 2$

Step 3.9.3.7

Add $8$ and $8$ .

$- \frac{16}{3} + 4 \cdot 2 - 4 \cdot - 2$

Step 3.9.3.8

Multiply $4$ by $2$ .

$- \frac{16}{3} + 8 - 4 \cdot - 2$

Step 3.9.3.9

Multiply $- 4$ by $- 2$ .

$- \frac{16}{3} + 8 + 8$

Step 3.9.3.10

Add $8$ and $8$ .

$- \frac{16}{3} + 16$

Step 3.9.3.11

To write $16$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{16}{3} + 16 \cdot \frac{3}{3}$

Step 3.9.3.12

Combine $16$ and $\frac{3}{3}$ .

$- \frac{16}{3} + \frac{16 \cdot 3}{3}$

Step 3.9.3.13

Combine the numerators over the common denominator.

$\frac{- 16 + 16 \cdot 3}{3}$

Step 3.9.3.14

Simplify the numerator.

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Step 3.9.3.14.1

Multiply $16$ by $3$ .

$\frac{- 16 + 48}{3}$

Step 3.9.3.14.2

Add $- 16$ and $48$ .

$\frac{32}{3}$

Step 4