Calculus Examples

$\int_{0}^{2} \frac{x^{2} - 2}{x + 1} d x$

Step 1

Divide $x^{2} - 2$ by $x + 1$ .

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Step 1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{2}$

$0 x$

$2$

Step 1.2

Divide the highest order term in the dividend $x^{2}$ by the highest order term in divisor $x$ .

					$x$
	$x$	+	$1$		$x^{2}$	+	$0 x$	-	$2$

Step 1.3

Multiply the new quotient term by the divisor.

				$x$
$x$	+	$1$		$x^{2}$	+	$0 x$	-	$2$
			+	$x^{2}$	+	$x$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} + x$

				$x$
$x$	+	$1$		$x^{2}$	+	$0 x$	-	$2$
			-	$x^{2}$	-	$x$

Step 1.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x$
$x$	+	$1$		$x^{2}$	+	$0 x$	-	$2$
			-	$x^{2}$	-	$x$
					-	$x$

Step 1.6

Pull the next terms from the original dividend down into the current dividend.

				$x$
$x$	+	$1$		$x^{2}$	+	$0 x$	-	$2$
			-	$x^{2}$	-	$x$
					-	$x$	-	$2$

Step 1.7

Divide the highest order term in the dividend $- x$ by the highest order term in divisor $x$ .

				$x$	-	$1$
$x$	+	$1$		$x^{2}$	+	$0 x$	-	$2$
			-	$x^{2}$	-	$x$
					-	$x$	-	$2$

Step 1.8

Multiply the new quotient term by the divisor.

				$x$	-	$1$
$x$	+	$1$		$x^{2}$	+	$0 x$	-	$2$
			-	$x^{2}$	-	$x$
					-	$x$	-	$2$
					-	$x$	-	$1$

Step 1.9

The expression needs to be subtracted from the dividend, so change all the signs in $- x - 1$

				$x$	-	$1$
$x$	+	$1$		$x^{2}$	+	$0 x$	-	$2$
			-	$x^{2}$	-	$x$
					-	$x$	-	$2$
					+	$x$	+	$1$

Step 1.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x$	-	$1$
$x$	+	$1$		$x^{2}$	+	$0 x$	-	$2$
			-	$x^{2}$	-	$x$
					-	$x$	-	$2$
					+	$x$	+	$1$
							-	$1$

Step 1.11

The final answer is the quotient plus the remainder over the divisor.

$\int_{0}^{2} x - 1 - \frac{1}{x + 1} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{0}^{2} x d x + \int_{0}^{2} - 1 d x + \int_{0}^{2} - \frac{1}{x + 1} d x$

Step 3

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

$\frac{1}{2} x^{2}]_{0}^{2} + \int_{0}^{2} - 1 d x + \int_{0}^{2} - \frac{1}{x + 1} d x$

Step 4

Apply the constant rule.

$\frac{1}{2} x^{2}]_{0}^{2} + - x]_{0}^{2} + \int_{0}^{2} - \frac{1}{x + 1} d x$

Step 5

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

$\frac{1}{2} x^{2}]_{0}^{2} + - x]_{0}^{2} - \int_{0}^{2} \frac{1}{x + 1} d x$

Step 6

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1

Let $u = x + 1$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1

Differentiate $x + 1$ .

$\frac{d}{d x} [x + 1]$

Step 6.1.2

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 6.1.3

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

$1 + \frac{d}{d x} [1]$

Step 6.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

Substitute the lower limit in for $x$ in $u = x + 1$ .

$u_{lower} = 0 + 1$

Step 6.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 6.4

Substitute the upper limit in for $x$ in $u = x + 1$ .

$u_{upper} = 2 + 1$

Step 6.5

Add $2$ and $1$ .

$u_{upper} = 3$

Step 6.6

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

$u_{lower} = 1$

$u_{upper} = 3$

Step 6.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\frac{1}{2} x^{2}]_{0}^{2} + - x]_{0}^{2} - \int_{1}^{3} \frac{1}{u} d u$

Step 7

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

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

Step 8

Combine $\frac{1}{2} x^{2}]_{0}^{2}$ and $- x]_{0}^{2}$ .

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

Step 9

Substitute and simplify.

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Step 9.1

Evaluate $\frac{1}{2} x^{2} - x$ at $2$ and at $0$ .

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

Step 9.2

Evaluate $\ln (| u |)$ at $3$ and at $1$ .

$(\frac{1}{2} \cdot 2^{2} - 1 \cdot 2) - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3

Simplify.

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Step 9.3.1

Raise $2$ to the power of $2$ .

$\frac{1}{2} \cdot 4 - 1 \cdot 2 - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.2

Combine $\frac{1}{2}$ and $4$ .

$\frac{4}{2} - 1 \cdot 2 - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.3

Cancel the common factor of $4$ and $2$ .

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Step 9.3.3.1

Factor $2$ out of $4$ .

$\frac{2 \cdot 2}{2} - 1 \cdot 2 - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.3.2

Cancel the common factors.

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Step 9.3.3.2.1

Factor $2$ out of $2$ .

$\frac{2 \cdot 2}{2 (1)} - 1 \cdot 2 - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.3.2.2

Cancel the common factor.

$\frac{2 \cdot 2}{2 \cdot 1} - 1 \cdot 2 - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.3.2.3

Rewrite the expression.

$\frac{2}{1} - 1 \cdot 2 - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.3.2.4

Divide $2$ by $1$ .

$2 - 1 \cdot 2 - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.4

Multiply $- 1$ by $2$ .

$2 - 2 - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.5

Subtract $2$ from $2$ .

$0 - (\frac{1}{2} \cdot 0^{2} - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.6

Raising $0$ to any positive power yields $0$ .

$0 - (\frac{1}{2} \cdot 0 - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.7

Multiply $\frac{1}{2}$ by $0$ .

$0 - (0 - 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.8

Multiply $- 1$ by $0$ .

$0 - (0 + 0) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.9

Add $0$ and $0$ .

$0 - 0 - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.10

Multiply $- 1$ by $0$ .

$0 + 0 - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.11

Add $0$ and $0$ .

$0 - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 9.3.12

Subtract $\ln (| 3 |) - \ln (| 1 |)$ from $0$ .

$- (\ln (| 3 |) - \ln (| 1 |))$

Step 10

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

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

Step 11

Simplify.

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Step 11.1

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

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

Step 11.2

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

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

Step 11.3

Divide $3$ by $1$ .

$- \ln (3)$

Step 12

The result can be shown in multiple forms.

Exact Form:

$- \ln (3)$

Decimal Form:

$- 1.09861228 \dots$

Step 13