Calculus Examples

$\int_{0}^{3} \frac{1}{9 - y^{2}} d y$

Step 1

Write the fraction using partial fraction decomposition.

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Step 1.1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1.1

Factor the fraction.

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Step 1.1.1.1

Rewrite $9$ as $3^{2}$ .

$\frac{1}{3^{2} - y^{2}}$

Step 1.1.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = y$ .

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

Step 1.1.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{3 + y}$

Step 1.1.3

$\frac{A}{3 + y} + \frac{B}{3 - y}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(3 + y) (3 - y)$ .

$\frac{1 (3 + y) (3 - y)}{(3 + y) (3 - y)} = \frac{(A) (3 + y) (3 - y)}{3 + y} + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.5

Cancel the common factor of $3 + y$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{1 (3 + y) (3 - y)}{(3 + y) (3 - y)} = \frac{(A) (3 + y) (3 - y)}{3 + y} + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.5.2

Rewrite the expression.

$\frac{1 (3 - y)}{3 - y} = \frac{(A) (3 + y) (3 - y)}{3 + y} + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.6

Cancel the common factor of $3 - y$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{1 (3 - y)}{3 - y} = \frac{(A) (3 + y) (3 - y)}{3 + y} + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.6.2

Rewrite the expression.

$1 = \frac{(A) (3 + y) (3 - y)}{3 + y} + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $3 + y$ .

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Step 1.1.7.1.1

Cancel the common factor.

$1 = \frac{A (3 + y) (3 - y)}{3 + y} + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.7.1.2

Divide $(A) (3 - y)$ by $1$ .

$1 = (A) (3 - y) + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.7.2

Apply the distributive property.

$1 = A \cdot 3 + A (- y) + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.7.3

Move $3$ to the left of $A$ .

$1 = 3 \cdot A + A (- y) + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.7.4

Rewrite using the commutative property of multiplication.

$1 = 3 \cdot A - A y + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.7.5

Cancel the common factor of $3 - y$ .

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Step 1.1.7.5.1

Cancel the common factor.

$1 = 3 A - A y + \frac{(B) (3 + y) (3 - y)}{3 - y}$

Step 1.1.7.5.2

Divide $(B) (3 + y)$ by $1$ .

$1 = 3 A - A y + (B) (3 + y)$

Step 1.1.7.6

Apply the distributive property.

$1 = 3 A - A y + B \cdot 3 + B y$

Step 1.1.7.7

Move $3$ to the left of $B$ .

$1 = 3 A - A y + 3 B + B y$

Step 1.1.8

Simplify the expression.

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Step 1.1.8.1

Move $A$ .

$1 = 3 A - y A + 3 B + B y$

Step 1.1.8.2

Reorder $B$ and $y$ .

$1 = 3 A - y A + 3 B + B y$

Step 1.1.8.3

Move $3 B$ .

$1 = 3 A - y A + B y + 3 B$

Step 1.1.8.4

Move $3 A$ .

$1 = - y A + B y + 3 A + 3 B$

Step 1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $y$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = - 1 A + B$

Step 1.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $y$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = 3 A + 3 B$

Step 1.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = - 1 A + B$

$1 = 3 A + 3 B$

$0 = - 1 A + B$

$1 = 3 A + 3 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $B$ in $0 = - 1 A + B$ .

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Step 1.3.1.1

Rewrite the equation as $- 1 A + B = 0$ .

$- 1 A + B = 0$

$1 = 3 A + 3 B$

Step 1.3.1.2

Rewrite $- 1 A$ as $- A$ .

$- A + B = 0$

$1 = 3 A + 3 B$

Step 1.3.1.3

Add $A$ to both sides of the equation.

$B = A$

$1 = 3 A + 3 B$

$B = A$

$1 = 3 A + 3 B$

Step 1.3.2

Replace all occurrences of $B$ with $A$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $B$ in $1 = 3 A + 3 B$ with $A$ .

$1 = 3 A + 3 (A)$

$B = A$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Add $3 A$ and $3 A$ .

$1 = 6 A$

$B = A$

$1 = 6 A$

$B = A$

$1 = 6 A$

$B = A$

Step 1.3.3

Solve for $A$ in $1 = 6 A$ .

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Step 1.3.3.1

Rewrite the equation as $6 A = 1$ .

$6 A = 1$

$B = A$

Step 1.3.3.2

Divide each term in $6 A = 1$ by $6$ and simplify.

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Step 1.3.3.2.1

Divide each term in $6 A = 1$ by $6$ .

$\frac{6 A}{6} = \frac{1}{6}$

$B = A$

Step 1.3.3.2.2

Simplify the left side.

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Step 1.3.3.2.2.1

Cancel the common factor of $6$ .

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Step 1.3.3.2.2.1.1

Cancel the common factor.

$\frac{6 A}{6} = \frac{1}{6}$

$B = A$

Step 1.3.3.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{6}$

$B = A$

$A = \frac{1}{6}$

$B = A$

$A = \frac{1}{6}$

$B = A$

$A = \frac{1}{6}$

$B = A$

$A = \frac{1}{6}$

$B = A$

Step 1.3.4

Replace all occurrences of $A$ with $\frac{1}{6}$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $A$ in $B = A$ with $\frac{1}{6}$ .

$B = \frac{1}{6}$

$A = \frac{1}{6}$

Step 1.3.4.2

Simplify the left side.

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Step 1.3.4.2.1

Remove parentheses.

$B = \frac{1}{6}$

$A = \frac{1}{6}$

$B = \frac{1}{6}$

$A = \frac{1}{6}$

$B = \frac{1}{6}$

$A = \frac{1}{6}$

Step 1.3.5

List all of the solutions.

$B = \frac{1}{6}, A = \frac{1}{6}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{3 + y} + \frac{B}{3 - y}$ with the values found for $A$ and $B$ .

$\frac{\frac{1}{6}}{3 + y} + \frac{\frac{1}{6}}{3 - y}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{6} \cdot \frac{1}{3 + y} + \frac{\frac{1}{6}}{3 - y}$

Step 1.5.2

Multiply $\frac{1}{6}$ by $\frac{1}{3 + y}$ .

$\frac{1}{6 (3 + y)} + \frac{\frac{1}{6}}{3 - y}$

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{6 (3 + y)} + \frac{1}{6} \cdot \frac{1}{3 - y}$

Step 1.5.4

Multiply $\frac{1}{6}$ by $\frac{1}{3 - y}$ .

$\int_{0}^{3} \frac{1}{6 (3 + y)} + \frac{1}{6 (3 - y)} d y$

Step 2

Split the single integral into multiple integrals.

$\int_{0}^{3} \frac{1}{6 (3 + y)} d y + \int_{0}^{3} \frac{1}{6 (3 - y)} d y$

Step 3

Since $\frac{1}{6}$ is constant with respect to $y$ , move $\frac{1}{6}$ out of the integral.

$\frac{1}{6} \int_{0}^{3} \frac{1}{3 + y} d y + \int_{0}^{3} \frac{1}{6 (3 - y)} d y$

Step 4

Let $u_{1} = 3 + y$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.1

Let $u_{1} = 3 + y$ . Find $\frac{d u_{1}}{d y}$ .

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Step 4.1.1

Differentiate $3 + y$ .

$\frac{d}{d y} [3 + y]$

Step 4.1.2

By the Sum Rule, the derivative of $3 + y$ with respect to $y$ is $\frac{d}{d y} [3] + \frac{d}{d y} [y]$ .

$\frac{d}{d y} [3] + \frac{d}{d y} [y]$

Step 4.1.3

Since $3$ is constant with respect to $y$ , the derivative of $3$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [y]$

Step 4.1.4

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$0 + 1$

Step 4.1.5

Add $0$ and $1$ .

$1$

Step 4.2

Substitute the lower limit in for $y$ in $u_{1} = 3 + y$ .

$u_{lower} = 3 + 0$

Step 4.3

Add $3$ and $0$ .

$u_{lower} = 3$

Step 4.4

Substitute the upper limit in for $y$ in $u_{1} = 3 + y$ .

$u_{upper} = 3 + 3$

Step 4.5

Add $3$ and $3$ .

$u_{upper} = 6$

Step 4.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 3$

$u_{upper} = 6$

Step 4.7

Rewrite the problem using $u_{1}$ , $d u_{1}$ , and the new limits of integration.

$\frac{1}{6} \int_{3}^{6} \frac{1}{u_{1}} d u_{1} + \int_{0}^{3} \frac{1}{6 (3 - y)} d y$

Step 5

The integral of $\frac{1}{u_{1}}$ with respect to $u_{1}$ is $\ln (| u_{1} |)$ .

$\frac{1}{6} \ln (| u_{1} |)]_{3}^{6} + \int_{0}^{3} \frac{1}{6 (3 - y)} d y$

Step 6

Since $\frac{1}{6}$ is constant with respect to $y$ , move $\frac{1}{6}$ out of the integral.

$\frac{1}{6} \ln (| u_{1} |)]_{3}^{6} + \frac{1}{6} \int_{0}^{3} \frac{1}{3 - y} d y$

Step 7

Let $u_{2} = 3 - y$ . Then $d u_{2} = - d y$ , so $- d u_{2} = d y$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 7.1

Let $u_{2} = 3 - y$ . Find $\frac{d u_{2}}{d y}$ .

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Step 7.1.1

Rewrite.

$\frac{- 1}{1}$

Step 7.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 7.2

Substitute the lower limit in for $y$ in $u_{2} = 3 - y$ .

$u_{lower} = 3 - 0$

Step 7.3

Subtract $0$ from $3$ .

$u_{lower} = 3$

Step 7.4

Substitute the upper limit in for $y$ in $u_{2} = 3 - y$ .

$u_{upper} = 3 - 1 \cdot 3$

Step 7.5

Simplify.

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Step 7.5.1

Multiply $- 1$ by $3$ .

$u_{upper} = 3 - 3$

Step 7.5.2

Subtract $3$ from $3$ .

$u_{upper} = 0$

Step 7.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 3$

$u_{upper} = 0$

Step 7.7

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$\frac{1}{6} \ln (| u_{1} |)]_{3}^{6} + \frac{1}{6} \int_{3}^{0} \frac{- 1}{u_{2}} d u_{2}$

Step 8

Move the negative in front of the fraction.

$\frac{1}{6} \ln (| u_{1} |)]_{3}^{6} + \frac{1}{6} \int_{3}^{0} - \frac{1}{u_{2}} d u_{2}$

Step 9

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

$\frac{1}{6} \ln (| u_{1} |)]_{3}^{6} + \frac{1}{6} (- \int_{3}^{0} \frac{1}{u_{2}} d u_{2})$

Step 10

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{6} \ln (| u_{1} |)]_{3}^{6} + \frac{1}{6} (- (\ln (| u_{2} |)]_{3}^{0}))$

Step 11

Combine $\frac{1}{6}$ and $\ln (| u_{2} |)]_{3}^{0}$ .

$\frac{1}{6} \ln (| u_{1} |)]_{3}^{6} - \frac{\ln (| u_{2} |)]_{3}^{0}}{6}$

Step 12

Substitute and simplify.

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Step 12.1

Evaluate $\ln (| u_{1} |)$ at $6$ and at $3$ .

$\frac{1}{6} ((\ln (| 6 |)) - \ln (| 3 |)) - \frac{\ln (| u_{2} |)]_{3}^{0}}{6}$

Step 12.2

Evaluate $\ln (| u_{2} |)$ at $0$ and at $3$ .

$\frac{1}{6} ((\ln (| 6 |)) - \ln (| 3 |)) - \frac{(\ln (| 0 |)) - \ln (| 3 |)}{6}$

Step 12.3

Simplify.

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Step 12.3.1

To write $\frac{1}{6} (\ln (| 6 |) - \ln (| 3 |))$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$\frac{1}{6} (\ln (| 6 |) - \ln (| 3 |)) \cdot \frac{6}{6} - \frac{\ln (| 0 |) - \ln (| 3 |)}{6}$

Step 12.3.2

Combine $\frac{1}{6} (\ln (| 6 |) - \ln (| 3 |))$ and $\frac{6}{6}$ .

$\frac{\frac{1}{6} (\ln (| 6 |) - \ln (| 3 |)) \cdot 6}{6} - \frac{\ln (| 0 |) - \ln (| 3 |)}{6}$

Step 12.3.3

Combine the numerators over the common denominator.

$\frac{\frac{1}{6} (\ln (| 6 |) - \ln (| 3 |)) \cdot 6 - (\ln (| 0 |) - \ln (| 3 |))}{6}$

Step 12.3.4

Combine $6$ and $\frac{1}{6}$ .

$\frac{\frac{6}{6} (\ln (| 6 |) - \ln (| 3 |)) - (\ln (| 0 |) - \ln (| 3 |))}{6}$

Step 12.3.5

Cancel the common factor of $6$ .

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Step 12.3.5.1

Cancel the common factor.

$\frac{\frac{6}{6} (\ln (| 6 |) - \ln (| 3 |)) - (\ln (| 0 |) - \ln (| 3 |))}{6}$

Step 12.3.5.2

Rewrite the expression.

$\frac{1 (\ln (| 6 |) - \ln (| 3 |)) - (\ln (| 0 |) - \ln (| 3 |))}{6}$

Step 12.3.6

Multiply $\ln (| 6 |) - \ln (| 3 |)$ by $1$ .

$\frac{\ln (| 6 |) - \ln (| 3 |) - (\ln (| 0 |) - \ln (| 3 |))}{6}$

Step 13

Simplify.

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Step 13.1

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

$\frac{\ln (\frac{| 6 |}{| 3 |}) - (\ln (| 0 |) - \ln (| 3 |))}{6}$

Step 13.2

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

$\frac{\ln (\frac{| 6 |}{| 3 |}) - \ln (\frac{| 0 |}{| 3 |})}{6}$

Step 13.3

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

$\frac{\ln (\frac{\frac{| 6 |}{| 3 |}}{\frac{| 0 |}{| 3 |}})}{6}$

Step 13.4

Rewrite $\frac{\frac{| 6 |}{| 3 |}}{\frac{| 0 |}{| 3 |}}$ as a product.

$\frac{\ln (\frac{| 6 |}{| 3 |} \cdot \frac{1}{\frac{| 0 |}{| 3 |}})}{6}$

Step 13.5

Multiply by the reciprocal of the fraction to divide by $\frac{| 0 |}{| 3 |}$ .

$\frac{\ln (\frac{| 6 |}{| 3 |} (1 \frac{| 3 |}{| 0 |}))}{6}$

Step 13.6

Multiply $\frac{| 3 |}{| 0 |}$ by $1$ .

$\frac{\ln (\frac{| 6 |}{| 3 |} \cdot \frac{| 3 |}{| 0 |})}{6}$

Step 13.7

Multiply $\frac{| 6 |}{| 3 |}$ by $\frac{| 3 |}{| 0 |}$ .

$\frac{\ln (\frac{| 6 | | 3 |}{| 3 | | 0 |})}{6}$

Step 13.8

To multiply absolute values, multiply the terms inside each absolute value.

$\frac{\ln (\frac{| 6 \cdot 3 |}{| 3 | | 0 |})}{6}$

Step 13.9

Multiply $6$ by $3$ .

$\frac{\ln (\frac{| 18 |}{| 3 | | 0 |})}{6}$

Step 13.10

To multiply absolute values, multiply the terms inside each absolute value.

$\frac{\ln (\frac{| 18 |}{| 3 \cdot 0 |})}{6}$

Step 13.11

Multiply $3$ by $0$ .

$\frac{\ln (\frac{| 18 |}{| 0 |})}{6}$

Step 14

Simplify.

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Step 14.1

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

$\frac{\ln (\frac{| 18 |}{0})}{6}$

Step 14.2

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

Undefined

$\frac{\ln (\frac{| 18 |}{0})}{6}$

Step 15

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

Undefined