Calculus Examples

$y = x^{2} - 2 x + 3$ , $y = 4 - 2 x$

Step 1

Solve by substitution to find the intersection between the curves.

Tap for more steps...

Step 1.1

Eliminate the equal sides of each equation and combine.

$x^{2} - 2 x + 3 = 4 - 2 x$

Step 1.2

Solve $x^{2} - 2 x + 3 = 4 - 2 x$ for $x$ .

Tap for more steps...

Step 1.2.1

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

Tap for more steps...

Step 1.2.1.1

Add $2 x$ to both sides of the equation.

$x^{2} - 2 x + 3 + 2 x = 4$

Step 1.2.1.2

Combine the opposite terms in $x^{2} - 2 x + 3 + 2 x$ .

Tap for more steps...

Step 1.2.1.2.1

Add $- 2 x$ and $2 x$ .

$x^{2} + 0 + 3 = 4$

Step 1.2.1.2.2

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

$x^{2} + 3 = 4$

Step 1.2.2

Move all terms not containing $x$ to the right side of the equation.

Tap for more steps...

Step 1.2.2.1

Subtract $3$ from both sides of the equation.

$x^{2} = 4 - 3$

Step 1.2.2.2

Subtract $3$ from $4$ .

$x^{2} = 1$

Step 1.2.3

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

$x = \pm \sqrt{1}$

Step 1.2.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 1.2.5

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

Tap for more steps...

Step 1.2.5.1

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

$x = 1$

Step 1.2.5.2

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

$x = - 1$

Step 1.2.5.3

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

$x = 1, - 1$

Step 1.3

Evaluate $y$ when $x = 1$ .

Tap for more steps...

Step 1.3.1

Substitute $1$ for $x$ .

$y = 4 - 2 \cdot 1$

Step 1.3.2

Substitute $1$ for $x$ in $y = 4 - 2 \cdot 1$ and solve for $y$ .

Tap for more steps...

Step 1.3.2.1

Remove parentheses.

$y = 4 - 2 \cdot 1$

Step 1.3.2.2

Simplify $4 - 2 \cdot 1$ .

Tap for more steps...

Step 1.3.2.2.1

Multiply $- 2$ by $1$ .

$y = 4 - 2$

Step 1.3.2.2.2

Subtract $2$ from $4$ .

$y = 2$

Step 1.4

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

Tap for more steps...

Step 1.4.1

Substitute $- 1$ for $x$ .

$y = 4 - 2 \cdot - 1$

Step 1.4.2

Substitute $- 1$ for $x$ in $y = 4 - 2 \cdot - 1$ and solve for $y$ .

Tap for more steps...

Step 1.4.2.1

Remove parentheses.

$y = 4 - 2 \cdot - 1$

Step 1.4.2.2

Simplify $4 - 2 \cdot - 1$ .

Tap for more steps...

Step 1.4.2.2.1

Multiply $- 2$ by $- 1$ .

$y = 4 + 2$

Step 1.4.2.2.2

Add $4$ and $2$ .

$y = 6$

Step 1.5

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

$(1, 2)$

$(- 1, 6)$

$(1, 2)$

$(- 1, 6)$

Step 2

Reorder $4$ and $- 2 x$ .

$y = - 2 x + 4$

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.

$A r e a = \int_{- 1}^{1} - 2 x + 4 d x - \int_{- 1}^{1} x^{2} - 2 x + 3 d x$

Step 4

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

Tap for more steps...

Step 4.1

Combine the integrals into a single integral.

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

Step 4.2

Simplify each term.

Tap for more steps...

Step 4.2.1

Apply the distributive property.

$- 2 x + 4 - x^{2} - (- 2 x) - 1 \cdot 3$

Step 4.2.2

Simplify.

Tap for more steps...

Step 4.2.2.1

Multiply $- 2$ by $- 1$ .

$- 2 x + 4 - x^{2} + 2 x - 1 \cdot 3$

Step 4.2.2.2

Multiply $- 1$ by $3$ .

$- 2 x + 4 - x^{2} + 2 x - 3$

$\int_{- 1}^{1} - 2 x + 4 - x^{2} + 2 x - 3 d x$

Step 4.3

Simplify by adding terms.

Tap for more steps...

Step 4.3.1

Combine the opposite terms in $- 2 x + 4 - x^{2} + 2 x - 3$ .

Tap for more steps...

Step 4.3.1.1

Add $- 2 x$ and $2 x$ .

$4 - x^{2} + 0 - 3$

Step 4.3.1.2

Add $4 - x^{2}$ and $0$ .

$4 - x^{2} - 3$

Step 4.3.2

Subtract $3$ from $4$ .

$- x^{2} + 1$

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

Step 4.4

Split the single integral into multiple integrals.

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

Step 4.5

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

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

Step 4.6

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

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

Step 4.7

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

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

Step 4.8

Apply the constant rule.

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

Step 4.9

Substitute and simplify.

Tap for more steps...

Step 4.9.1

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

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

Step 4.9.2

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

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

Step 4.9.3

Simplify.

Tap for more steps...

Step 4.9.3.1

One to any power is one.

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

Step 4.9.3.2

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

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

Step 4.9.3.3

Move the negative in front of the fraction.

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

Step 4.9.3.4

Multiply $- 1$ by $- 1$ .

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

Step 4.9.3.5

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

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

Step 4.9.3.6

Combine the numerators over the common denominator.

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

Step 4.9.3.7

Add $1$ and $1$ .

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

Step 4.9.3.8

Add $1$ and $1$ .

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

Step 4.9.3.9

To write $2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 4.9.3.10

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

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

Step 4.9.3.11

Combine the numerators over the common denominator.

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

Step 4.9.3.12

Simplify the numerator.

Tap for more steps...

Step 4.9.3.12.1

Multiply $2$ by $3$ .

$\frac{- 2 + 6}{3}$

Step 4.9.3.12.2

Add $- 2$ and $6$ .

$\frac{4}{3}$

Step 5