Calculus Examples

$\frac{d A}{d t} = \frac{1}{2} (12) (- 24) + \frac{1}{2} (9) (18)$

Step 1

Rewrite the equation.

$d A = (\frac{1}{2} (12) (- 24) + \frac{1}{2} (9) (18)) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d A = \int \frac{1}{2} (12) (- 24) + \frac{1}{2} (9) (18) d t$

Step 2.2

Apply the constant rule.

$A + C_{1} = \int \frac{1}{2} (12) (- 24) + \frac{1}{2} (9) (18) d t$

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 $12$ .

$A + C_{1} = \int \frac{12}{2} \cdot - 24 + \frac{1}{2} (9) (18) d t$

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 $12$ .

$A + C_{1} = \int \frac{2 \cdot 6}{2} \cdot - 24 + \frac{1}{2} (9) (18) d t$

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 $2$ .

$A + C_{1} = \int \frac{2 \cdot 6}{2 (1)} \cdot - 24 + \frac{1}{2} (9) (18) d t$

Step 2.3.1.2.2.2

Cancel the common factor.

$A + C_{1} = \int \frac{2 \cdot 6}{2 \cdot 1} \cdot - 24 + \frac{1}{2} (9) (18) d t$

Step 2.3.1.2.2.3

Rewrite the expression.

$A + C_{1} = \int \frac{6}{1} \cdot - 24 + \frac{1}{2} (9) (18) d t$

Step 2.3.1.2.2.4

Divide $6$ by $1$ .

$A + C_{1} = \int 6 \cdot - 24 + \frac{1}{2} (9) (18) d t$

Step 2.3.1.3

Multiply $6$ by $- 24$ .

$A + C_{1} = \int - 144 + \frac{1}{2} (9) (18) d t$

Step 2.3.1.4

Combine $\frac{1}{2}$ and $9$ .

$A + C_{1} = \int - 144 + \frac{9}{2} \cdot 18 d t$

Step 2.3.1.5

Combine $\frac{9}{2}$ and $18$ .

$A + C_{1} = \int - 144 + \frac{9 \cdot 18}{2} d t$

Step 2.3.1.6

Multiply $9$ by $18$ .

$A + C_{1} = \int - 144 + \frac{162}{2} d t$

Step 2.3.1.7

Cancel the common factor of $162$ and $2$ .

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Step 2.3.1.7.1

Factor $2$ out of $162$ .

$A + C_{1} = \int - 144 + \frac{2 \cdot 81}{2} d t$

Step 2.3.1.7.2

Cancel the common factors.

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Step 2.3.1.7.2.1

Factor $2$ out of $2$ .

$A + C_{1} = \int - 144 + \frac{2 \cdot 81}{2 (1)} d t$

Step 2.3.1.7.2.2

Cancel the common factor.

$A + C_{1} = \int - 144 + \frac{2 \cdot 81}{2 \cdot 1} d t$

Step 2.3.1.7.2.3

Rewrite the expression.

$A + C_{1} = \int - 144 + \frac{81}{1} d t$

Step 2.3.1.7.2.4

Divide $81$ by $1$ .

$A + C_{1} = \int - 144 + 81 d t$

Step 2.3.1.8

Add $- 144$ and $81$ .

$A + C_{1} = \int - 63 d t$

Step 2.3.2

Apply the constant rule.

$A + C_{1} = - 63 t + C_{2}$

Step 2.4

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

$A = - 63 t + K$