Calculus Examples

y = sin (x)

$y = \sin (x)$ ,

x = 0

$x = 0$ ,

x = π

$x = π$

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.

sin (x) = 0

$\sin (x) = 0$

Step 1.2

Solve

sin (x) = 0

$\sin (x) = 0$ for

x

$x$ .

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Step 1.2.1

Take the inverse sine of both sides of the equation to extract

x

$x$ from inside the sine.

x = arcsin (0)

$x = \arcsin (0)$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

The exact value of

arcsin (0)

$\arcsin (0)$ is

0

$0$ .

x = 0

$x = 0$

x = 0

$x = 0$

Step 1.2.3

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

π

$π$ to find the solution in the second quadrant.

x = π - 0

$x = π - 0$

Step 1.2.4

Subtract

0

$0$ from

π

$π$ .

x = π

$x = π$

Step 1.2.5

Find the period of

sin (x)

$\sin (x)$ .

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Step 1.2.5.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 1.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 1.2.5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 1.2.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 1.2.6

The period of the

sin (x)

$\sin (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = 2 π n, π + 2 π n

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

n

$n$

Step 1.2.7

Consolidate the answers.

x = π n

$x = π n$ , for any integer

n

$n$

x = π n

$x = π n$ , for any integer

n

$n$

Step 1.3

Substitute

π n

$π n$ for

x

$x$ .

y = 0

$y = 0$

Step 1.4

List all of the solutions.

y = 0, x = π n

$y = 0, x = π n$

y = 0, x = π n

$y = 0, x = π n$

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_{0}^{π} sin (x) d x - \int_{0}^{π} 0 d x

$A r e a = \int_{0}^{π} \sin (x) d x - \int_{0}^{π} 0 d x$

Step 3

Integrate to find the area between

0

$0$ and

π

$π$ .

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Step 3.1

Combine the integrals into a single integral.

\int_{0}^{π} sin (x) - (0) d x

$\int_{0}^{π} \sin (x) - (0) d x$

Step 3.2

Subtract

0

$0$ from

sin (x)

$\sin (x)$ .

\int_{0}^{π} sin (x) d x

$\int_{0}^{π} \sin (x) d x$

Step 3.3

The integral of

sin (x)

$\sin (x)$ with respect to

x

$x$ is

- cos (x)

$- \cos (x)$ .

- cos (x)]_{0}^{π}

$- \cos (x)]_{0}^{π}$

Step 3.4

Simplify the answer.

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Step 3.4.1

Evaluate

- cos (x)

$- \cos (x)$ at

π

$π$ and at

0

$0$ .

- cos (π) + cos (0)

$- \cos (π) + \cos (0)$

Step 3.4.2

The exact value of

cos (0)

$\cos (0)$ is

1

$1$ .

- cos (π) + 1

$- \cos (π) + 1$

Step 3.4.3

Simplify.

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Step 3.4.3.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

- - cos (0) + 1

$- - \cos (0) + 1$

Step 3.4.3.2

The exact value of

cos (0)

$\cos (0)$ is

1

$1$ .

- (- 1 \cdot 1) + 1

$- (- 1 \cdot 1) + 1$

Step 3.4.3.3

Multiply

- 1

$- 1$ by

1

$1$ .

- - 1 + 1

$- - 1 + 1$

Step 3.4.3.4

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

1 + 1

$1 + 1$

Step 3.4.3.5

Add

1

$1$ and

1

$1$ .

2

$2$

2

$2$

2

$2$

2

$2$

Step 4