Calculus Examples

$\int_{0}^{3} \frac{2}{x - 3^{3}} d x$

Step 1

Simplify.

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Step 1.1

Raise $3$ to the power of $3$ .

$\int_{0}^{3} \frac{2}{x - 1 \cdot 27} d x$

Step 1.2

Multiply $- 1$ by $27$ .

$\int_{0}^{3} \frac{2}{x - 27} d x$

Step 2

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$2 \int_{0}^{3} \frac{1}{x - 27} d x$

Step 3

Let $u = x - 27$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.1

Let $u = x - 27$ . Find $\frac{d u}{d x}$ .

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Step 3.1.1

Differentiate $x - 27$ .

$\frac{d}{d x} [x - 27]$

Step 3.1.2

By the Sum Rule, the derivative of $x - 27$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 27]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 27]$

Step 3.1.3

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

$1 + \frac{d}{d x} [- 27]$

Step 3.1.4

Since $- 27$ is constant with respect to $x$ , the derivative of $- 27$ with respect to $x$ is $0$ .

$1 + 0$

Step 3.1.5

Add $1$ and $0$ .

$1$

Step 3.2

Substitute the lower limit in for $x$ in $u = x - 27$ .

$u_{lower} = 0 - 27$

Step 3.3

Subtract $27$ from $0$ .

$u_{lower} = - 27$

Step 3.4

Substitute the upper limit in for $x$ in $u = x - 27$ .

$u_{upper} = 3 - 27$

Step 3.5

Subtract $27$ from $3$ .

$u_{upper} = - 24$

Step 3.6

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

$u_{lower} = - 27$

$u_{upper} = - 24$

Step 3.7

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

$2 \int_{- 27}^{- 24} \frac{1}{u} d u$

Step 4

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

$2 (\ln (| u |)]_{- 27}^{- 24})$

Step 5

Evaluate $\ln (| u |)$ at $- 24$ and at $- 27$ .

$2 (\ln (| - 24 |) - \ln (| - 27 |))$

Step 6

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

$2 \ln (\frac{| - 24 |}{| - 27 |})$

Step 7

Simplify.

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Step 7.1

The absolute value is the distance between a number and zero. The distance between $- 24$ and $0$ is $24$ .

$2 \ln (\frac{24}{| - 27 |})$

Step 7.2

The absolute value is the distance between a number and zero. The distance between $- 27$ and $0$ is $27$ .

$2 \ln (\frac{24}{27})$

Step 7.3

Cancel the common factor of $24$ and $27$ .

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Step 7.3.1

Factor $3$ out of $24$ .

$2 \ln (\frac{3 (8)}{27})$

Step 7.3.2

Cancel the common factors.

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Step 7.3.2.1

Factor $3$ out of $27$ .

$2 \ln (\frac{3 \cdot 8}{3 \cdot 9})$

Step 7.3.2.2

Cancel the common factor.

$2 \ln (\frac{3 \cdot 8}{3 \cdot 9})$

Step 7.3.2.3

Rewrite the expression.

$2 \ln (\frac{8}{9})$

Step 8

The result can be shown in multiple forms.

Exact Form:

$2 \ln (\frac{8}{9})$

Decimal Form:

$- 0.23556607 \dots$

Step 9