Calculus Examples

$y = x^{2}$ , $y = 2 x - x^{2}$

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.

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

Step 1.2

Solve $x^{2} = 2 x - x^{2}$ for $x$ .

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Step 1.2.1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 1.2.2

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

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Step 1.2.2.1

Subtract $x^{2}$ from both sides of the equation.

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

Step 1.2.2.2

Subtract $x^{2}$ from $- x^{2}$ .

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

Step 1.2.3

Factor the left side of the equation.

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Step 1.2.3.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$2 u - 2 u^{2} = 0$

Step 1.2.3.2

Factor $2 u$ out of $2 u - 2 u^{2}$ .

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Step 1.2.3.2.1

Factor $2 u$ out of $2 u$ .

$2 u (1) - 2 u^{2} = 0$

Step 1.2.3.2.2

Factor $2 u$ out of $- 2 u^{2}$ .

$2 u (1) + 2 u (- u) = 0$

Step 1.2.3.2.3

Factor $2 u$ out of $2 u (1) + 2 u (- u)$ .

$2 u (1 - u) = 0$

Step 1.2.3.3

Replace all occurrences of $u$ with $x$ .

$2 x (1 - x) = 0$

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 $0$ .

$x = 0$

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

Step 1.2.5

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.6

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

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Step 1.2.6.1

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

$1 - x = 0$

Step 1.2.6.2

Solve $1 - x = 0$ for $x$ .

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Step 1.2.6.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 1.2.6.2.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 1.2.6.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 1.2.6.2.2.2

Simplify the left side.

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Step 1.2.6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 1.2.6.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 1.2.6.2.2.3

Simplify the right side.

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Step 1.2.6.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 1.2.7

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

$x = 0, 1$

Step 1.3

Evaluate $y$ when $x = 0$ .

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Step 1.3.1

Substitute $0$ for $x$ .

$y = 2 (0) - {(0)}^{2}$

Step 1.3.2

Substitute $0$ for $x$ in $y = 2 (0) - {(0)}^{2}$ and solve for $y$ .

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Step 1.3.2.1

Remove parentheses.

$y = 2 (0) - 0^{2}$

Step 1.3.2.2

Simplify $2 (0) - 0^{2}$ .

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Step 1.3.2.2.1

Simplify each term.

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Step 1.3.2.2.1.1

Multiply $2$ by $0$ .

$y = 0 - 0^{2}$

Step 1.3.2.2.1.2

Raising $0$ to any positive power yields $0$ .

$y = 0 - 0$

Step 1.3.2.2.1.3

Multiply $- 1$ by $0$ .

$y = 0 + 0$

Step 1.3.2.2.2

Add $0$ and $0$ .

$y = 0$

Step 1.4

Evaluate $y$ when $x = 1$ .

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Step 1.4.1

Substitute $1$ for $x$ .

$y = 2 (1) - {(1)}^{2}$

Step 1.4.2

Substitute $1$ for $x$ in $y = 2 (1) - {(1)}^{2}$ and solve for $y$ .

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Step 1.4.2.1

Remove parentheses.

$y = 2 (1) - 1^{2}$

Step 1.4.2.2

Simplify $2 (1) - 1^{2}$ .

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Step 1.4.2.2.1

Simplify each term.

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Step 1.4.2.2.1.1

Multiply $2$ by $1$ .

$y = 2 - 1^{2}$

Step 1.4.2.2.1.2

One to any power is one.

$y = 2 - 1 \cdot 1$

Step 1.4.2.2.1.3

Multiply $- 1$ by $1$ .

$y = 2 - 1$

Step 1.4.2.2.2

Subtract $1$ from $2$ .

$y = 1$

Step 1.5

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

$(0, 0)$

$(1, 1)$

$(0, 0)$

$(1, 1)$

Step 2

Reorder $2 x$ and $- x^{2}$ .

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

Step 3

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.

$A r e a = \int_{0}^{1} - x^{2} + 2 x d x - \int_{0}^{1} x^{2} d x$

Step 4

Integrate to find the area between $0$ and $1$ .

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Step 4.1

Combine the integrals into a single integral.

$\int_{0}^{1} - x^{2} + 2 x - (x^{2}) d x$

Step 4.2

Subtract $x^{2}$ from $- x^{2}$ .

$\int_{0}^{1} - 2 x^{2} + 2 x d x$

Step 4.3

Split the single integral into multiple integrals.

$\int_{0}^{1} - 2 x^{2} d x + \int_{0}^{1} 2 x d x$

Step 4.4

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

$- 2 \int_{0}^{1} x^{2} d x + \int_{0}^{1} 2 x d x$

Step 4.5

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

$- 2 (\frac{1}{3} x^{3}]_{0}^{1}) + \int_{0}^{1} 2 x d x$

Step 4.6

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

$- 2 (\frac{x^{3}}{3}]_{0}^{1}) + \int_{0}^{1} 2 x d x$

Step 4.7

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$- 2 (\frac{x^{3}}{3}]_{0}^{1}) + 2 \int_{0}^{1} x d x$

Step 4.8

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

$- 2 (\frac{x^{3}}{3}]_{0}^{1}) + 2 (\frac{1}{2} x^{2}]_{0}^{1})$

Step 4.9

Simplify the answer.

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Step 4.9.1

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

$- 2 (\frac{x^{3}}{3}]_{0}^{1}) + 2 (\frac{x^{2}}{2}]_{0}^{1})$

Step 4.9.2

Substitute and simplify.

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Step 4.9.2.1

Evaluate $\frac{x^{3}}{3}$ at $1$ and at $0$ .

$- 2 ((\frac{1^{3}}{3}) - \frac{0^{3}}{3}) + 2 (\frac{x^{2}}{2}]_{0}^{1})$

Step 4.9.2.2

Evaluate $\frac{x^{2}}{2}$ at $1$ and at $0$ .

$- 2 (\frac{1^{3}}{3} - \frac{0^{3}}{3}) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3

Simplify.

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Step 4.9.2.3.1

One to any power is one.

$- 2 (\frac{1}{3} - \frac{0^{3}}{3}) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.2

Raising $0$ to any positive power yields $0$ .

$- 2 (\frac{1}{3} - \frac{0}{3}) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.3

Cancel the common factor of $0$ and $3$ .

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Step 4.9.2.3.3.1

Factor $3$ out of $0$ .

$- 2 (\frac{1}{3} - \frac{3 (0)}{3}) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.3.2

Cancel the common factors.

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Step 4.9.2.3.3.2.1

Factor $3$ out of $3$ .

$- 2 (\frac{1}{3} - \frac{3 \cdot 0}{3 \cdot 1}) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.3.2.2

Cancel the common factor.

$- 2 (\frac{1}{3} - \frac{3 \cdot 0}{3 \cdot 1}) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.3.2.3

Rewrite the expression.

$- 2 (\frac{1}{3} - \frac{0}{1}) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.3.2.4

Divide $0$ by $1$ .

$- 2 (\frac{1}{3} - 0) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.4

Multiply $- 1$ by $0$ .

$- 2 (\frac{1}{3} + 0) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.5

Add $\frac{1}{3}$ and $0$ .

$- 2 (\frac{1}{3}) + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.6

Combine $- 2$ and $\frac{1}{3}$ .

$\frac{- 2}{3} + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.7

Move the negative in front of the fraction.

$- \frac{2}{3} + 2 (\frac{1^{2}}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.8

One to any power is one.

$- \frac{2}{3} + 2 (\frac{1}{2} - \frac{0^{2}}{2})$

Step 4.9.2.3.9

Raising $0$ to any positive power yields $0$ .

$- \frac{2}{3} + 2 (\frac{1}{2} - \frac{0}{2})$

Step 4.9.2.3.10

Cancel the common factor of $0$ and $2$ .

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Step 4.9.2.3.10.1

Factor $2$ out of $0$ .

$- \frac{2}{3} + 2 (\frac{1}{2} - \frac{2 (0)}{2})$

Step 4.9.2.3.10.2

Cancel the common factors.

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Step 4.9.2.3.10.2.1

Factor $2$ out of $2$ .

$- \frac{2}{3} + 2 (\frac{1}{2} - \frac{2 \cdot 0}{2 \cdot 1})$

Step 4.9.2.3.10.2.2

Cancel the common factor.

$- \frac{2}{3} + 2 (\frac{1}{2} - \frac{2 \cdot 0}{2 \cdot 1})$

Step 4.9.2.3.10.2.3

Rewrite the expression.

$- \frac{2}{3} + 2 (\frac{1}{2} - \frac{0}{1})$

Step 4.9.2.3.10.2.4

Divide $0$ by $1$ .

$- \frac{2}{3} + 2 (\frac{1}{2} - 0)$

Step 4.9.2.3.11

Multiply $- 1$ by $0$ .

$- \frac{2}{3} + 2 (\frac{1}{2} + 0)$

Step 4.9.2.3.12

Add $\frac{1}{2}$ and $0$ .

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

Step 4.9.2.3.13

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

$- \frac{2}{3} + \frac{2}{2}$

Step 4.9.2.3.14

Cancel the common factor of $2$ .

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Step 4.9.2.3.14.1

Cancel the common factor.

$- \frac{2}{3} + \frac{2}{2}$

Step 4.9.2.3.14.2

Rewrite the expression.

$- \frac{2}{3} + 1$

Step 4.9.2.3.15

Write $1$ as a fraction with a common denominator.

$- \frac{2}{3} + \frac{3}{3}$

Step 4.9.2.3.16

Combine the numerators over the common denominator.

$\frac{- 2 + 3}{3}$

Step 4.9.2.3.17

Add $- 2$ and $3$ .

$\frac{1}{3}$

Step 5