Calculus Examples

$\frac{d x}{d t} = 4 - 0.2 x$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{4 - 0.2 x}$ .

$\frac{1}{4 - 0.2 x} \frac{d x}{d t} = \frac{1}{4 - 0.2 x} (4 - 0.2 x)$

Step 1.2

Cancel the common factor of $4 - 0.2 x$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{4 - 0.2 x} \frac{d x}{d t} = \frac{1}{4 - 0.2 x} (4 - 0.2 x)$

Step 1.2.2

Rewrite the expression.

$\frac{1}{4 - 0.2 x} \frac{d x}{d t} = 1$

Step 1.3

Rewrite the equation.

$\frac{1}{4 - 0.2 x} d x = 1 d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{4 - 0.2 x} d x = \int d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 4 - 0.2 x$ . Then $d u = - 0.2 d x$ , so $\frac{1}{- 0.2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = 4 - 0.2 x$ . Find $\frac{d u}{d x}$ .

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Step 2.2.1.1.1

Rewrite.

$\frac{- 0.2}{1}$

Step 2.2.1.1.2

Divide $- 0.2$ by $1$ .

$- 0.2$

Step 2.2.1.2

Rewrite the problem using $u$ and $d u$ .

$\int \frac{- 5}{u} d u = \int d t$

Step 2.2.2

Split the fraction into multiple fractions.

$\int - \frac{5}{u} d u = \int d t$

Step 2.2.3

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

$- \int \frac{5}{u} d u = \int d t$

Step 2.2.4

Since $5$ is constant with respect to $u$ , move $5$ out of the integral.

$- (5 \int \frac{1}{u} d u) = \int d t$

Step 2.2.5

Multiply $5$ by $- 1$ .

$- 5 \int \frac{1}{u} d u = \int d t$

Step 2.2.6

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

$- 5 (\ln (| u |) + C_{1}) = \int d t$

Step 2.2.7

Simplify.

$- 5 \ln (| u |) + C_{1} = \int d t$

Step 2.2.8

Replace all occurrences of $u$ with $4 - 0.2 x$ .

$- 5 \ln (| 4 - 0.2 x |) + C_{1} = \int d t$

Step 2.3

Apply the constant rule.

$- 5 \ln (| 4 - 0.2 x |) + C_{1} = t + C_{2}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$- 5 \ln (| 4 - 0.2 x |) = t + C$

Step 3

Solve for $x$ .

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Step 3.1

Divide each term in $- 5 \ln (| 4 - 0.2 x |) = t + C$ by $- 5$ and simplify.

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Step 3.1.1

Divide each term in $- 5 \ln (| 4 - 0.2 x |) = t + C$ by $- 5$ .

$\frac{- 5 \ln (| 4 - 0.2 x |)}{- 5} = \frac{t}{- 5} + \frac{C}{- 5}$

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Cancel the common factor of $- 5$ .

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Step 3.1.2.1.1

Cancel the common factor.

$\frac{- 5 \ln (| 4 - 0.2 x |)}{- 5} = \frac{t}{- 5} + \frac{C}{- 5}$

Step 3.1.2.1.2

Divide $\ln (| 4 - 0.2 x |)$ by $1$ .

$\ln (| 4 - 0.2 x |) = \frac{t}{- 5} + \frac{C}{- 5}$

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Simplify each term.

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Step 3.1.3.1.1

Move the negative in front of the fraction.

$\ln (| 4 - 0.2 x |) = - \frac{t}{5} + \frac{C}{- 5}$

Step 3.1.3.1.2

Move the negative in front of the fraction.

$\ln (| 4 - 0.2 x |) = - \frac{t}{5} - \frac{C}{5}$

Step 3.2

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (| 4 - 0.2 x |)} = e^{- \frac{t}{5} - \frac{C}{5}}$

Step 3.3

Rewrite $\ln (| 4 - 0.2 x |) = - \frac{t}{5} - \frac{C}{5}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- \frac{t}{5} - \frac{C}{5}} = | 4 - 0.2 x |$

Step 3.4

Solve for $x$ .

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Step 3.4.1

Rewrite the equation as $| 4 - 0.2 x | = e^{- \frac{t}{5} - \frac{C}{5}}$ .

$| 4 - 0.2 x | = e^{- \frac{t}{5} - \frac{C}{5}}$

Step 3.4.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$4 - 0.2 x = \pm e^{- \frac{t}{5} - \frac{C}{5}}$

Step 3.4.3

Subtract $4$ from both sides of the equation.

$- 0.2 x = \pm e^{- \frac{t}{5} - \frac{C}{5}} - 4$

Step 3.4.4

Divide each term in $- 0.2 x = \pm e^{- \frac{t}{5} - \frac{C}{5}} - 4$ by $- 0.2$ and simplify.

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Step 3.4.4.1

Divide each term in $- 0.2 x = \pm e^{- \frac{t}{5} - \frac{C}{5}} - 4$ by $- 0.2$ .

$\frac{- 0.2 x}{- 0.2} = \frac{\pm e^{- \frac{t}{5} - \frac{C}{5}}}{- 0.2} + \frac{- 4}{- 0.2}$

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Cancel the common factor of $- 0.2$ .

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Step 3.4.4.2.1.1

Cancel the common factor.

$\frac{- 0.2 x}{- 0.2} = \frac{\pm e^{- \frac{t}{5} - \frac{C}{5}}}{- 0.2} + \frac{- 4}{- 0.2}$

Step 3.4.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\pm e^{- \frac{t}{5} - \frac{C}{5}}}{- 0.2} + \frac{- 4}{- 0.2}$

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

Simplify each term.

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Step 3.4.4.3.1.1

Combine the numerators over the common denominator.

$x = \frac{\pm e^{\frac{- t - C}{5}}}{- 0.2} + \frac{- 4}{- 0.2}$

Step 3.4.4.3.1.2

Simplify $\frac{\pm e^{\frac{- t - C}{5}}}{- 0.2}$ .

$x = \frac{\pm e^{\frac{- t - C}{5}}}{0.2} + \frac{- 4}{- 0.2}$

Step 3.4.4.3.1.3

Multiply by $1$ .

$x = \frac{1 (\pm e^{\frac{- t - C}{5}})}{0.2} + \frac{- 4}{- 0.2}$

Step 3.4.4.3.1.4

Factor $0.2$ out of $0.2$ .

$x = \frac{1 (\pm e^{\frac{- t - C}{5}})}{0.2 (1)} + \frac{- 4}{- 0.2}$

Step 3.4.4.3.1.5

Separate fractions.

$x = \frac{1}{0.2} \cdot \frac{\pm e^{\frac{- t - C}{5}}}{1} + \frac{- 4}{- 0.2}$

Step 3.4.4.3.1.6

Divide $1$ by $0.2$ .

$x = 5 \frac{\pm e^{\frac{- t - C}{5}}}{1} + \frac{- 4}{- 0.2}$

Step 3.4.4.3.1.7

Divide $\pm e^{\frac{- t - C}{5}}$ by $1$ .

$x = 5 (\pm e^{\frac{- t - C}{5}}) + \frac{- 4}{- 0.2}$

Step 3.4.4.3.1.8

Divide $- 4$ by $- 0.2$ .

$x = 5 (\pm e^{\frac{- t - C}{5}}) + 20$

Step 4

Simplify the constant of integration.

$x = 5 (\pm e^{\frac{- t + C}{5}}) + 20$