Calculus Examples

$\int_{2}^{4} \frac{1}{5 - 3 x} d x$

Step 1

Let $u = 5 - 3 x$ . Then $d u = - 3 d x$ , so $- \frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 5 - 3 x$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $5 - 3 x$ .

$\frac{d}{d x} [5 - 3 x]$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $5 - 3 x$ with respect to $x$ is $\frac{d}{d x} [5] + \frac{d}{d x} [- 3 x]$ .

$\frac{d}{d x} [5] + \frac{d}{d x} [- 3 x]$

Step 1.1.2.2

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- 3 x]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 3 x]$ .

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Step 1.1.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$0 - 3 \frac{d}{d x} [x]$

Step 1.1.3.2

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

$0 - 3 \cdot 1$

Step 1.1.3.3

Multiply $- 3$ by $1$ .

$0 - 3$

Step 1.1.4

Subtract $3$ from $0$ .

$- 3$

Step 1.2

Substitute the lower limit in for $x$ in $u = 5 - 3 x$ .

$u_{lower} = 5 - 3 \cdot 2$

Step 1.3

Simplify.

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Step 1.3.1

Multiply $- 3$ by $2$ .

$u_{lower} = 5 - 6$

Step 1.3.2

Subtract $6$ from $5$ .

$u_{lower} = - 1$

Step 1.4

Substitute the upper limit in for $x$ in $u = 5 - 3 x$ .

$u_{upper} = 5 - 3 \cdot 4$

Step 1.5

Simplify.

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Step 1.5.1

Multiply $- 3$ by $4$ .

$u_{upper} = 5 - 12$

Step 1.5.2

Subtract $12$ from $5$ .

$u_{upper} = - 7$

Step 1.6

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

$u_{lower} = - 1$

$u_{upper} = - 7$

Step 1.7

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

$\int_{- 1}^{- 7} \frac{1}{u} \cdot \frac{1}{- 3} d u$

Step 2

Simplify.

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Step 2.1

Move the negative in front of the fraction.

$\int_{- 1}^{- 7} \frac{1}{u} (- \frac{1}{3}) d u$

Step 2.2

Multiply $\frac{1}{u}$ by $\frac{1}{3}$ .

$\int_{- 1}^{- 7} - \frac{1}{u \cdot 3} d u$

Step 2.3

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

$\int_{- 1}^{- 7} - \frac{1}{3 u} d u$

Step 3

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

$- \int_{- 1}^{- 7} \frac{1}{3 u} d u$

Step 4

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$- (\frac{1}{3} \int_{- 1}^{- 7} \frac{1}{u} d u)$

Step 5

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

$- \frac{1}{3} \ln (| u |)]_{- 1}^{- 7}$

Step 6

Simplify.

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Step 6.1

Evaluate $\ln (| u |)$ at $- 7$ and at $- 1$ .

$- \frac{1}{3} ((\ln (| - 7 |)) - \ln (| - 1 |))$

Step 6.2

Simplify.

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Step 6.2.1

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

$- \frac{1}{3} \ln (\frac{| - 7 |}{| - 1 |})$

Step 6.2.2

Combine $\ln (\frac{| - 7 |}{| - 1 |})$ and $\frac{1}{3}$ .

$- \frac{\ln (\frac{| - 7 |}{| - 1 |})}{3}$

Step 6.3

Simplify.

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Step 6.3.1

The absolute value is the distance between a number and zero. The distance between $- 7$ and $0$ is $7$ .

$- \frac{\ln (\frac{7}{| - 1 |})}{3}$

Step 6.3.2

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

$- \frac{\ln (\frac{7}{1})}{3}$

Step 6.3.3

Divide $7$ by $1$ .

$- \frac{\ln (7)}{3}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$- \frac{\ln (7)}{3}$

Decimal Form:

$- 0.64863671 \dots$