Calculus Examples

$\int_{- \infty}^{0} \frac{1}{3 - 4 x} d x$

Step 1

Write the integral as a limit as $t$ approaches $- \infty$ .

$lim_{t \to - \infty} \int_{t}^{0} \frac{1}{3 - 4 x} d x$

Step 2

Let $u = 3 - 4 x$ . Then $d u = - 4 d x$ , so $- \frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = 3 - 4 x$ . Find $\frac{d u}{d x}$ .

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Step 2.1.1

Differentiate $3 - 4 x$ .

$\frac{d}{d x} [3 - 4 x]$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

By the Sum Rule, the derivative of $3 - 4 x$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [- 4 x]$ .

$\frac{d}{d x} [3] + \frac{d}{d x} [- 4 x]$

Step 2.1.2.2

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

$0 + \frac{d}{d x} [- 4 x]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 2.1.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$0 - 4 \frac{d}{d x} [x]$

Step 2.1.3.2

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

$0 - 4 \cdot 1$

Step 2.1.3.3

Multiply $- 4$ by $1$ .

$0 - 4$

Step 2.1.4

Subtract $4$ from $0$ .

$- 4$

Step 2.2

Substitute the lower limit in for $x$ in $u = 3 - 4 x$ .

$u_{lower} = 3 - 4 t$

Step 2.3

Substitute the upper limit in for $x$ in $u = 3 - 4 x$ .

$u_{upper} = 3 - 4 \cdot 0$

Step 2.4

Simplify.

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Step 2.4.1

Multiply $- 4$ by $0$ .

$u_{upper} = 3 + 0$

Step 2.4.2

Add $3$ and $0$ .

$u_{upper} = 3$

Step 2.5

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

$u_{lower} = 3 - 4 t$

$u_{upper} = 3$

Step 2.6

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

$lim_{t \to - \infty} \int_{3 - 4 t}^{3} \frac{1}{u} \cdot \frac{1}{- 4} d u$

Step 3

Simplify.

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Step 3.1

Move the negative in front of the fraction.

$lim_{t \to - \infty} \int_{3 - 4 t}^{3} \frac{1}{u} (- \frac{1}{4}) d u$

Step 3.2

Multiply $\frac{1}{u}$ by $\frac{1}{4}$ .

$lim_{t \to - \infty} \int_{3 - 4 t}^{3} - \frac{1}{u \cdot 4} d u$

Step 3.3

Move $4$ to the left of $u$ .

$lim_{t \to - \infty} \int_{3 - 4 t}^{3} - \frac{1}{4 u} d u$

Step 4

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

$lim_{t \to - \infty} - \int_{3 - 4 t}^{3} \frac{1}{4 u} d u$

Step 5

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

$lim_{t \to - \infty} - (\frac{1}{4} \int_{3 - 4 t}^{3} \frac{1}{u} d u)$

Step 6

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

$lim_{t \to - \infty} - \frac{1}{4} \ln (| u |)]_{3 - 4 t}^{3}$

Step 7

Combine $\ln (| u |)]_{3 - 4 t}^{3}$ and $\frac{1}{4}$ .

$lim_{t \to - \infty} - \frac{\ln (| u |)]_{3 - 4 t}^{3}}{4}$

Step 8

Evaluate $\ln (| u |)$ at $3$ and at $3 - 4 t$ .

$lim_{t \to - \infty} - \frac{\ln (| 3 |) - \ln (| 3 - 4 t |)}{4}$

Step 9

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

$lim_{t \to - \infty} - \frac{\ln (\frac{| 3 |}{| 3 - 4 t |})}{4}$

Step 10

Since the function $\frac{\ln (\frac{| 3 |}{| 3 - 4 t |})}{4}$ approaches $- \infty$ , the negative constant $- 1$ times the function approaches $\infty$ .

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Step 10.1

Consider the limit with the constant multiple $- 1$ removed.

$lim_{t \to - \infty} \frac{\ln (\frac{| 3 |}{| 3 - 4 t |})}{4}$

Step 10.2

Split the limit using the Limits Quotient Rule on the limit as $t$ approaches $- \infty$ .

$\frac{lim_{t \to - \infty} \ln (\frac{| 3 |}{| 3 - 4 t |})}{lim_{t \to - \infty} 4}$

Step 10.3

As $t$ approaches $- \infty$ from either side, $\ln (\frac{| 3 |}{| 3 - 4 t |})$ decreases without bound.

$\frac{- \infty}{lim_{t \to - \infty} 4}$

Step 10.4

Evaluate the limit of $4$ which is constant as $t$ approaches $- \infty$ .

$\frac{- \infty}{4}$

Step 10.5

Infinity divided by anything that is finite and non-zero is infinity.

$- \infty$

Step 10.6

Since the function $\frac{\ln (\frac{| 3 |}{| 3 - 4 t |})}{4}$ approaches $- \infty$ , the negative constant $- 1$ times the function approaches $\infty$ .

$\infty$