Calculus Examples

$\frac{d A}{d t} = \frac{1}{2} (12) (- 36) + \frac{1}{2} (5) (15)$

Step 1

Rewrite the equation.

$d A = (\frac{1}{2} (12) (- 36) + \frac{1}{2} (5) (15)) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d A = \int \frac{1}{2} (12) (- 36) + \frac{1}{2} (5) (15) d t$

Step 2.2

Apply the constant rule.

$A + C_{1} = \int \frac{1}{2} (12) (- 36) + \frac{1}{2} (5) (15) d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Simplify.

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Step 2.3.1.1

Combine $\frac{1}{2}$ and $12$ .

$A + C_{1} = \int \frac{12}{2} \cdot - 36 + \frac{1}{2} (5) (15) d t$

Step 2.3.1.2

Cancel the common factor of $12$ and $2$ .

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Step 2.3.1.2.1

Factor $2$ out of $12$ .

$A + C_{1} = \int \frac{2 \cdot 6}{2} \cdot - 36 + \frac{1}{2} (5) (15) d t$

Step 2.3.1.2.2

Cancel the common factors.

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Step 2.3.1.2.2.1

Factor $2$ out of $2$ .

$A + C_{1} = \int \frac{2 \cdot 6}{2 (1)} \cdot - 36 + \frac{1}{2} (5) (15) d t$

Step 2.3.1.2.2.2

Cancel the common factor.

$A + C_{1} = \int \frac{2 \cdot 6}{2 \cdot 1} \cdot - 36 + \frac{1}{2} (5) (15) d t$

Step 2.3.1.2.2.3

Rewrite the expression.

$A + C_{1} = \int \frac{6}{1} \cdot - 36 + \frac{1}{2} (5) (15) d t$

Step 2.3.1.2.2.4

Divide $6$ by $1$ .

$A + C_{1} = \int 6 \cdot - 36 + \frac{1}{2} (5) (15) d t$

Step 2.3.1.3

Multiply $6$ by $- 36$ .

$A + C_{1} = \int - 216 + \frac{1}{2} (5) (15) d t$

Step 2.3.1.4

Combine $\frac{1}{2}$ and $5$ .

$A + C_{1} = \int - 216 + \frac{5}{2} \cdot 15 d t$

Step 2.3.1.5

Combine $\frac{5}{2}$ and $15$ .

$A + C_{1} = \int - 216 + \frac{5 \cdot 15}{2} d t$

Step 2.3.1.6

Multiply $5$ by $15$ .

$A + C_{1} = \int - 216 + \frac{75}{2} d t$

Step 2.3.1.7

To write $- 216$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$A + C_{1} = \int - 216 \cdot \frac{2}{2} + \frac{75}{2} d t$

Step 2.3.1.8

Combine $- 216$ and $\frac{2}{2}$ .

$A + C_{1} = \int \frac{- 216 \cdot 2}{2} + \frac{75}{2} d t$

Step 2.3.1.9

Combine the numerators over the common denominator.

$A + C_{1} = \int \frac{- 216 \cdot 2 + 75}{2} d t$

Step 2.3.1.10

Simplify the numerator.

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Step 2.3.1.10.1

Multiply $- 216$ by $2$ .

$A + C_{1} = \int \frac{- 432 + 75}{2} d t$

Step 2.3.1.10.2

Add $- 432$ and $75$ .

$A + C_{1} = \int \frac{- 357}{2} d t$

Step 2.3.1.11

Move the negative in front of the fraction.

$A + C_{1} = \int - \frac{357}{2} d t$

Step 2.3.2

Apply the constant rule.

$A + C_{1} = - \frac{357}{2} t + C_{2}$

Step 2.4

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

$A = - \frac{357}{2} t + K$