Calculus Examples

$y = x^{2} - 3 x$ , $y = 3 x + 7$

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.

$x^{2} - 3 x = 3 x + 7$

Step 1.2

Solve $x^{2} - 3 x = 3 x + 7$ 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 $3 x$ from both sides of the equation.

$x^{2} - 3 x - 3 x = 7$

Step 1.2.1.2

Subtract $3 x$ from $- 3 x$ .

$x^{2} - 6 x = 7$

Step 1.2.2

Subtract $7$ from both sides of the equation.

$x^{2} - 6 x - 7 = 0$

Step 1.2.3

Factor $x^{2} - 6 x - 7$ using the AC method.

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Step 1.2.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 7$ and whose sum is $- 6$ .

$- 7, 1 + y = 3 x + 7$

Step 1.2.3.2

Write the factored form using these integers.

$(x - 7) (x + 1) = 0$

Step 1.2.4

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

$x - 7 = 0$

$x + 1 = 0 + y = 3 x + 7$

Step 1.2.5

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 1.2.5.2

Add $7$ to both sides of the equation.

$x = 7$

Step 1.2.6

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

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Step 1.2.6.1

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

$x + 1 = 0$

Step 1.2.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.2.7

The final solution is all the values that make $(x - 7) (x + 1) = 0$ true.

$x = 7, - 1$

Step 1.3

Evaluate $y$ when $x = 7$ .

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Step 1.3.1

Substitute $7$ for $x$ .

$y = 3 (7) + 7$

Step 1.3.2

Simplify $3 (7) + 7$ .

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Step 1.3.2.1

Multiply $3$ by $7$ .

$y = 21 + 7$

Step 1.3.2.2

Add $21$ and $7$ .

$y = 28$

Step 1.4

Evaluate $y$ when $x = - 1$ .

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Step 1.4.1

Substitute $- 1$ for $x$ .

$y = 3 (- 1) + 7$

Step 1.4.2

Simplify $3 (- 1) + 7$ .

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Step 1.4.2.1

Multiply $3$ by $- 1$ .

$y = - 3 + 7$

Step 1.4.2.2

Add $- 3$ and $7$ .

$y = 4$

Step 1.5

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

$(7, 28)$

$(- 1, 4)$

$(7, 28)$

$(- 1, 4)$

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.

$A r e a = \int_{- 1}^{7} 3 x + 7 d x - \int_{- 1}^{7} x^{2} - 3 x d x$

Step 3

Integrate to find the area between $- 1$ and $7$ .

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Step 3.1

Combine the integrals into a single integral.

$\int_{- 1}^{7} 3 x + 7 - (x^{2} - 3 x) d x$

Step 3.2

Simplify each term.

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Step 3.2.1

Apply the distributive property.

$3 x + 7 - x^{2} - (- 3 x)$

Step 3.2.2

Multiply $- 3$ by $- 1$ .

$3 x + 7 - x^{2} + 3 x$

$\int_{- 1}^{7} 3 x + 7 - x^{2} + 3 x d x$

Step 3.3

Add $3 x$ and $3 x$ .

$\int_{- 1}^{7} 6 x + 7 - x^{2} d x$

Step 3.4

Split the single integral into multiple integrals.

$\int_{- 1}^{7} 6 x d x + \int_{- 1}^{7} 7 d x + \int_{- 1}^{7} - x^{2} d x$

Step 3.5

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$6 \int_{- 1}^{7} x d x + \int_{- 1}^{7} 7 d x + \int_{- 1}^{7} - x^{2} d x$

Step 3.6

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

$6 (\frac{1}{2} x^{2}]_{- 1}^{7}) + \int_{- 1}^{7} 7 d x + \int_{- 1}^{7} - x^{2} d x$

Step 3.7

Combine $\frac{1}{2}$ and $x^{2}$ .

$6 (\frac{x^{2}}{2}]_{- 1}^{7}) + \int_{- 1}^{7} 7 d x + \int_{- 1}^{7} - x^{2} d x$

Step 3.8

Apply the constant rule.

$6 (\frac{x^{2}}{2}]_{- 1}^{7}) + 7 x]_{- 1}^{7} + \int_{- 1}^{7} - x^{2} d x$

Step 3.9

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

$6 (\frac{x^{2}}{2}]_{- 1}^{7}) + 7 x]_{- 1}^{7} - \int_{- 1}^{7} x^{2} d x$

Step 3.10

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

$6 (\frac{x^{2}}{2}]_{- 1}^{7}) + 7 x]_{- 1}^{7} - (\frac{1}{3} x^{3}]_{- 1}^{7})$

Step 3.11

Simplify the answer.

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Step 3.11.1

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

$6 (\frac{x^{2}}{2}]_{- 1}^{7}) + 7 x]_{- 1}^{7} - (\frac{x^{3}}{3}]_{- 1}^{7})$

Step 3.11.2

Substitute and simplify.

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Step 3.11.2.1

Evaluate $\frac{x^{2}}{2}$ at $7$ and at $- 1$ .

$6 ((\frac{7^{2}}{2}) - \frac{{(- 1)}^{2}}{2}) + 7 x]_{- 1}^{7} - (\frac{x^{3}}{3}]_{- 1}^{7})$

Step 3.11.2.2

Evaluate $7 x$ at $7$ and at $- 1$ .

$6 (\frac{7^{2}}{2} - \frac{{(- 1)}^{2}}{2}) + 7 \cdot 7 - 7 \cdot - 1 - (\frac{x^{3}}{3}]_{- 1}^{7})$

Step 3.11.2.3

Evaluate $\frac{x^{3}}{3}$ at $7$ and at $- 1$ .

$6 (\frac{7^{2}}{2} - \frac{{(- 1)}^{2}}{2}) + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4

Simplify.

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Step 3.11.2.4.1

Raise $7$ to the power of $2$ .

$6 (\frac{49}{2} - \frac{{(- 1)}^{2}}{2}) + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.2

Raise $- 1$ to the power of $2$ .

$6 (\frac{49}{2} - \frac{1}{2}) + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.3

Combine the numerators over the common denominator.

$6 \frac{49 - 1}{2} + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.4

Subtract $1$ from $49$ .

$6 (\frac{48}{2}) + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.5

Cancel the common factor of $48$ and $2$ .

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Step 3.11.2.4.5.1

Factor $2$ out of $48$ .

$6 \frac{2 \cdot 24}{2} + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.5.2

Cancel the common factors.

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Step 3.11.2.4.5.2.1

Factor $2$ out of $2$ .

$6 \frac{2 \cdot 24}{2 (1)} + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.5.2.2

Cancel the common factor.

$6 \frac{2 \cdot 24}{2 \cdot 1} + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.5.2.3

Rewrite the expression.

$6 (\frac{24}{1}) + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.5.2.4

Divide $24$ by $1$ .

$6 \cdot 24 + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.6

Multiply $6$ by $24$ .

$144 + 7 \cdot 7 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.7

Multiply $7$ by $7$ .

$144 + 49 - 7 \cdot - 1 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.8

Multiply $- 7$ by $- 1$ .

$144 + 49 + 7 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.9

Add $49$ and $7$ .

$144 + 56 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.10

Add $144$ and $56$ .

$200 - ((\frac{7^{3}}{3}) - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.11

Raise $7$ to the power of $3$ .

$200 - (\frac{343}{3} - \frac{{(- 1)}^{3}}{3})$

Step 3.11.2.4.12

Raise $- 1$ to the power of $3$ .

$200 - (\frac{343}{3} - \frac{- 1}{3})$

Step 3.11.2.4.13

Move the negative in front of the fraction.

$200 - (\frac{343}{3} - - \frac{1}{3})$

Step 3.11.2.4.14

Multiply $- 1$ by $- 1$ .

$200 - (\frac{343}{3} + 1 (\frac{1}{3}))$

Step 3.11.2.4.15

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

$200 - (\frac{343}{3} + \frac{1}{3})$

Step 3.11.2.4.16

Combine the numerators over the common denominator.

$200 - \frac{343 + 1}{3}$

Step 3.11.2.4.17

Add $343$ and $1$ .

$200 - \frac{344}{3}$

Step 3.11.2.4.18

To write $200$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$200 \cdot \frac{3}{3} - \frac{344}{3}$

Step 3.11.2.4.19

Combine $200$ and $\frac{3}{3}$ .

$\frac{200 \cdot 3}{3} - \frac{344}{3}$

Step 3.11.2.4.20

Combine the numerators over the common denominator.

$\frac{200 \cdot 3 - 344}{3}$

Step 3.11.2.4.21

Simplify the numerator.

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Step 3.11.2.4.21.1

Multiply $200$ by $3$ .

$\frac{600 - 344}{3}$

Step 3.11.2.4.21.2

Subtract $344$ from $600$ .

$\frac{256}{3}$

Step 4