Calculus Examples Applications of Integration

Find the Area Between the Curves

$y = x^{2} + x$ , $y = x + 9$

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.

$x^{2} + x = x + 9$

Step 1.2

Solve $x^{2} + x = x + 9$ for $x$ .

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Step 1.2.1

Move all terms containing $x$ to the left side of the equation.

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Step 1.2.1.1

Subtract $x$ from both sides of the equation.

$x^{2} + x - x = 9$

Step 1.2.1.2

Combine the opposite terms in $x^{2} + x - x$ .

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Step 1.2.1.2.1

Subtract $x$ from $x$ .

$x^{2} + 0 = 9$

Step 1.2.1.2.2

Add $x^{2}$ and $0$ .

$x^{2} = 9$

Step 1.2.2

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

$x = \pm \sqrt{9}$

Step 1.2.3

Simplify $\pm \sqrt{9}$ .

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Step 1.2.3.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

Step 1.2.3.2

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

$x = \pm 3$

Step 1.2.4

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

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Step 1.2.4.1

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

$x = 3$

Step 1.2.4.2

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

$x = - 3$

Step 1.2.4.3

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

$x = 3, - 3$

Step 1.3

Evaluate $y$ when $x = 3$ .

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Step 1.3.1

Substitute $3$ for $x$ .

$y = (3) + 9$

Step 1.3.2

Substitute $3$ for $x$ in $y = (3) + 9$ and solve for $y$ .

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Step 1.3.2.1

Remove parentheses.

$y = 3 + 9$

Step 1.3.2.2

Remove parentheses.

$y = (3) + 9$

Step 1.3.2.3

Add $3$ and $9$ .

$y = 12$

Step 1.4

Evaluate $y$ when $x = - 3$ .

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Step 1.4.1

Substitute $- 3$ for $x$ .

$y = (- 3) + 9$

Step 1.4.2

Substitute $- 3$ for $x$ in $y = (- 3) + 9$ and solve for $y$ .

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Step 1.4.2.1

Remove parentheses.

$y = - 3 + 9$

Step 1.4.2.2

Remove parentheses.

$y = (- 3) + 9$

Step 1.4.2.3

Add $- 3$ and $9$ .

$y = 6$

Step 1.5

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

$(3, 12)$

$(- 3, 6)$

$(3, 12)$

$(- 3, 6)$

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.

$A r e a = \int_{- 3}^{3} x + 9 d x - \int_{- 3}^{3} x^{2} + x d x$

Step 3

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

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Step 3.1

Combine the integrals into a single integral.

$\int_{- 3}^{3} x + 9 - (x^{2} + x) d x$

Step 3.2

Apply the distributive property.

$\int_{- 3}^{3} x + 9 - x^{2} - x d x$

Step 3.3

Combine the opposite terms in $x + 9 - x^{2} - x$ .

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Step 3.3.1

Subtract $x$ from $x$ .

$9 - x^{2} + 0$

Step 3.3.2

Add $9 - x^{2}$ and $0$ .

$9 - x^{2}$

$\int_{- 3}^{3} 9 - x^{2} d x$

Step 3.4

Split the single integral into multiple integrals.

$\int_{- 3}^{3} 9 d x + \int_{- 3}^{3} - x^{2} d x$

Step 3.5

Apply the constant rule.

$9 x]_{- 3}^{3} + \int_{- 3}^{3} - x^{2} d x$

Step 3.6

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

$9 x]_{- 3}^{3} - \int_{- 3}^{3} x^{2} d x$

Step 3.7

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

$9 x]_{- 3}^{3} - (\frac{1}{3} x^{3}]_{- 3}^{3})$

Step 3.8

Simplify the answer.

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Step 3.8.1

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

$9 x]_{- 3}^{3} - (\frac{x^{3}}{3}]_{- 3}^{3})$

Step 3.8.2

Substitute and simplify.

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Step 3.8.2.1

Evaluate $9 x$ at $3$ and at $- 3$ .

$(9 \cdot 3) - 9 \cdot - 3 - (\frac{x^{3}}{3}]_{- 3}^{3})$

Step 3.8.2.2

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

$9 \cdot 3 - 9 \cdot - 3 - (\frac{3^{3}}{3} - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3

Simplify.

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Step 3.8.2.3.1

Multiply $9$ by $3$ .

$27 - 9 \cdot - 3 - (\frac{3^{3}}{3} - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3.2

Multiply $- 9$ by $- 3$ .

$27 + 27 - (\frac{3^{3}}{3} - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3.3

Add $27$ and $27$ .

$54 - (\frac{3^{3}}{3} - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3.4

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

$54 - (\frac{27}{3} - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3.5

Cancel the common factor of $27$ and $3$ .

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Step 3.8.2.3.5.1

Factor $3$ out of $27$ .

$54 - (\frac{3 \cdot 9}{3} - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3.5.2

Cancel the common factors.

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Step 3.8.2.3.5.2.1

Factor $3$ out of $3$ .

$54 - (\frac{3 \cdot 9}{3 (1)} - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3.5.2.2

Cancel the common factor.

$54 - (\frac{3 \cdot 9}{3 \cdot 1} - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3.5.2.3

Rewrite the expression.

$54 - (\frac{9}{1} - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3.5.2.4

Divide $9$ by $1$ .

$54 - (9 - \frac{{(- 3)}^{3}}{3})$

Step 3.8.2.3.6

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

$54 - (9 - \frac{- 27}{3})$

Step 3.8.2.3.7

Cancel the common factor of $- 27$ and $3$ .

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Step 3.8.2.3.7.1

Factor $3$ out of $- 27$ .

$54 - (9 - \frac{3 \cdot - 9}{3})$

Step 3.8.2.3.7.2

Cancel the common factors.

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Step 3.8.2.3.7.2.1

Factor $3$ out of $3$ .

$54 - (9 - \frac{3 \cdot - 9}{3 (1)})$

Step 3.8.2.3.7.2.2

Cancel the common factor.

$54 - (9 - \frac{3 \cdot - 9}{3 \cdot 1})$

Step 3.8.2.3.7.2.3

Rewrite the expression.

$54 - (9 - \frac{- 9}{1})$

Step 3.8.2.3.7.2.4

Divide $- 9$ by $1$ .

$54 - (9 - - 9)$

Step 3.8.2.3.8

Multiply $- 1$ by $- 9$ .

$54 - (9 + 9)$

Step 3.8.2.3.9

Add $9$ and $9$ .

$54 - 1 \cdot 18$

Step 3.8.2.3.10

Multiply $- 1$ by $18$ .

$54 - 18$

Step 3.8.2.3.11

Subtract $18$ from $54$ .

$36$

Step 4

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Find the Area Between the Curves

Applications of Integration

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