Calculus Examples

$f (x) = 5 \csc (x)$ , $0 < x < 2 π$

Step 1

To find the average value of a function, the function should be continuous on the closed interval $[a, b]$ . To find whether $f (x)$ is continuous on $[0, 2 π]$ or not, find the domain of $f (x) = 5 \csc (x)$ .

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Step 1.1

Set the argument in $\csc (x)$ equal to $π n$ to find where the expression is undefined.

$x = π n$ , for any integer $n$

Step 1.2

The domain is all values of $x$ that make the expression defined.

Set-Builder Notation:

${x | x \neq π n}$ , for any integer $n$

Set-Builder Notation:

${x | x \neq π n}$ , for any integer $n$

Step 2

$f (x)$ is continuous on $[0, 2 π]$ .

$f (x)$ is continuous

Step 3

The average value of function $f$ over the interval $[a, b]$ is defined as $A (x) = \frac{1}{b - a} \int_{a}^{b} f (x) d x$ .

$A (x) = \frac{1}{b - a} \int_{a}^{b} f (x) d x$

Step 4

Substitute the actual values into the formula for the average value of a function.

$A (x) = \frac{1}{2 π - 0} (\int_{0}^{2 π} 5 \csc (x) d x)$

Step 5

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$A (x) = \frac{1}{2 π - 0} (5 \int_{0}^{2 π} \csc (x) d x)$

Step 6

The integral of $\csc (x)$ with respect to $x$ is $\ln (| \csc (x) - \cot (x) |)$ .

$A (x) = \frac{1}{2 π - 0} (5 (\ln (| \csc (x) - \cot (x) |)]_{0}^{2 π}))$

Step 7

Simplify the answer.

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Step 7.1

Evaluate $\ln (| \csc (x) - \cot (x) |)$ at $2 π$ and at $0$ .

$A (x) = \frac{1}{2 π - 0} (5 (\ln (| \csc (2 π) - \cot (2 π) |) - \ln (| \csc (0) - \cot (0) |)))$

Step 7.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \csc (2 π) - \cot (2 π) |}{| \csc (0) - \cot (0) |}))$

Step 7.3

Simplify.

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Step 7.3.1

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \csc (2 π) - \cot (0) |}{| \csc (0) - \cot (0) |}))$

Step 7.3.2

Simplify the numerator.

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Step 7.3.2.1

Rewrite $\csc (2 π)$ in terms of sines and cosines.

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \frac{1}{\sin (2 π)} - \cot (0) |}{| \csc (0) - \cot (0) |}))$

Step 7.3.2.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \frac{1}{\sin (0)} - \cot (0) |}{| \csc (0) - \cot (0) |}))$

Step 7.3.2.3

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

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \frac{1}{0} - \cot (0) |}{| \csc (0) - \cot (0) |}))$

Step 7.3.2.4

The expression contains a division by $0$ . The expression is undefined.

$A (x) = \frac{1}{2 π - 0} (Undefined)$

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \frac{1}{0} - \cot (0) |}{| \csc (0) - \cot (0) |}))$

Step 7.3.3

Simplify the denominator.

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Step 7.3.3.1

Rewrite $\csc (0)$ in terms of sines and cosines.

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \frac{1}{0} - \cot (0) |}{| \frac{1}{\sin (0)} - \cot (0) |}))$

Step 7.3.3.2

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

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \frac{1}{0} - \cot (0) |}{| \frac{1}{0} - \cot (0) |}))$

Step 7.3.3.3

The expression contains a division by $0$ . The expression is undefined.

$A (x) = \frac{1}{2 π - 0} (Undefined)$

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \frac{1}{0} - \cot (0) |}{| \frac{1}{0} - \cot (0) |}))$

Step 7.3.4

Cancel the common factor of $| \frac{1}{0} - \cot (0) |$ .

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Step 7.3.4.1

Cancel the common factor.

$A (x) = \frac{1}{2 π - 0} (5 \ln (\frac{| \frac{1}{0} - \cot (0) |}{| \frac{1}{0} - \cot (0) |}))$

Step 7.3.4.2

Rewrite the expression.

$A (x) = \frac{1}{2 π - 0} (5 \ln (1))$

Step 7.3.5

The natural logarithm of $1$ is $0$ .

$A (x) = \frac{1}{2 π - 0} (5 \cdot 0)$

Step 7.3.6

Multiply $5$ by $0$ .

$A (x) = \frac{1}{2 π - 0} (0)$

Step 8

Simplify the denominator.

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Step 8.1

Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{2 π + 0} \cdot 0$

Step 8.2

Add $2 π$ and $0$ .

$A (x) = \frac{1}{2 π} \cdot 0$

Step 9

Multiply $\frac{1}{2 π}$ by $0$ .

$A (x) = 0$

Step 10