Calculus Examples

$y = - 4 \sin (x)$ , $y = \sin (2 x)$

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.

$- 4 \sin (x) = \sin (2 x)$

Step 1.2

Solve $- 4 \sin (x) = \sin (2 x)$ for $x$ .

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Step 1.2.1

Subtract $\sin (2 x)$ from both sides of the equation.

$- 4 \sin (x) - \sin (2 x) = 0$

Step 1.2.2

Simplify each term.

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Step 1.2.2.1

Apply the sine double-angle identity.

$- 4 \sin (x) - (2 \sin (x) \cos (x)) = 0$

Step 1.2.2.2

Multiply $2$ by $- 1$ .

$- 4 \sin (x) - 2 \sin (x) \cos (x) = 0$

Step 1.2.3

Factor $- 4 \sin (x) - 2 \sin (x) \cos (x)$ .

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Step 1.2.3.1

Factor $2 \sin (x)$ out of $- 4 \sin (x) - 2 \sin (x) \cos (x)$ .

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Step 1.2.3.1.1

Factor $2 \sin (x)$ out of $- 4 \sin (x)$ .

$2 \sin (x) \cdot - 2 - 2 \sin (x) \cos (x) = 0$

Step 1.2.3.1.2

Factor $2 \sin (x)$ out of $- 2 \sin (x) \cos (x)$ .

$2 \sin (x) \cdot - 2 + 2 \sin (x) (- 1 \cos (x)) = 0$

Step 1.2.3.1.3

Factor $2 \sin (x)$ out of $2 \sin (x) \cdot - 2 + 2 \sin (x) (- 1 \cos (x))$ .

$2 \sin (x) (- 2 - 1 \cos (x)) = 0$

Step 1.2.3.2

Rewrite $- 1 \cos (x)$ as $- \cos (x)$ .

$2 \sin (x) (- 2 - \cos (x)) = 0$

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 $0$ .

$\sin (x) = 0$

$- 2 - \cos (x) = 0 + y = \sin (2 x)$

Step 1.2.5

Set $\sin (x)$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $\sin (x)$ equal to $0$ .

$\sin (x) = 0$

Step 1.2.5.2

Solve $\sin (x) = 0$ for $x$ .

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Step 1.2.5.2.1

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (0)$

Step 1.2.5.2.2

Simplify the right side.

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Step 1.2.5.2.2.1

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

Step 1.2.5.2.3

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - 0$

Step 1.2.5.2.4

Subtract $0$ from $π$ .

$x = π$

Step 1.2.5.2.5

Find the period of $\sin (x)$ .

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Step 1.2.5.2.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.2.5.2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 1.2.5.2.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 1.2.5.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.5.2.6

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 1.2.6

Set $- 2 - \cos (x)$ equal to $0$ and solve for $x$ .

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Step 1.2.6.1

Set $- 2 - \cos (x)$ equal to $0$ .

$- 2 - \cos (x) = 0$

Step 1.2.6.2

Solve $- 2 - \cos (x) = 0$ for $x$ .

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Step 1.2.6.2.1

Add $2$ to both sides of the equation.

$- \cos (x) = 2$

Step 1.2.6.2.2

Divide each term in $- \cos (x) = 2$ by $- 1$ and simplify.

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Step 1.2.6.2.2.1

Divide each term in $- \cos (x) = 2$ by $- 1$ .

$\frac{- \cos (x)}{- 1} = \frac{2}{- 1}$

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.

$\frac{\cos (x)}{1} = \frac{2}{- 1}$

Step 1.2.6.2.2.2.2

Divide $\cos (x)$ by $1$ .

$\cos (x) = \frac{2}{- 1}$

Step 1.2.6.2.2.3

Simplify the right side.

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Step 1.2.6.2.2.3.1

Divide $2$ by $- 1$ .

$\cos (x) = - 2$

Step 1.2.6.2.3

The range of cosine is $- 1 \leq y \leq 1$ . Since $- 2$ does not fall in this range, there is no solution.

No $s o l u t i o n + y = \sin (2 x)$

Step 1.2.7

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

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 1.2.8

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 1.3

Evaluate $y$ when $x = π n$ .

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Step 1.3.1

Substitute $π n$ for $x$ .

$y = \sin (2 (π n))$

Step 1.3.2

Remove parentheses.

$y = \sin (2 π n)$

Step 1.4

List all of the solutions.

$y = \sin (2 π n), x = π n$

Step 2

The area between the given curves is unbounded.

Unbounded area

Step 3