Calculus Examples

$\frac{d s}{d t} = 12 t {(2 t^{2} - 1)}^{3}$

Step 1

Rewrite the equation.

$d s = 12 t {(2 t^{2} - 1)}^{3} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d s = \int 12 t {(2 t^{2} - 1)}^{3} d t$

Step 2.2

Apply the constant rule.

$s + C_{1} = \int 12 t {(2 t^{2} - 1)}^{3} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $12$ is constant with respect to $t$ , move $12$ out of the integral.

$s + C_{1} = 12 \int t {(2 t^{2} - 1)}^{3} d t$

Step 2.3.2

Let $u = 2 t^{2} - 1$ . Then $d u = 4 t d t$ , so $\frac{1}{4} d u = t d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

Let $u = 2 t^{2} - 1$ . Find $\frac{d u}{d t}$ .

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Step 2.3.2.1.1

Differentiate $2 t^{2} - 1$ .

$\frac{d}{d t} [2 t^{2} - 1]$

Step 2.3.2.1.2

By the Sum Rule, the derivative of $2 t^{2} - 1$ with respect to $t$ is $\frac{d}{d t} [2 t^{2}] + \frac{d}{d t} [- 1]$ .

$\frac{d}{d t} [2 t^{2}] + \frac{d}{d t} [- 1]$

Step 2.3.2.1.3

Evaluate $\frac{d}{d t} [2 t^{2}]$ .

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Step 2.3.2.1.3.1

Since $2$ is constant with respect to $t$ , the derivative of $2 t^{2}$ with respect to $t$ is $2 \frac{d}{d t} [t^{2}]$ .

$2 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [- 1]$

Step 2.3.2.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$2 (2 t) + \frac{d}{d t} [- 1]$

Step 2.3.2.1.3.3

Multiply $2$ by $2$ .

$4 t + \frac{d}{d t} [- 1]$

Step 2.3.2.1.4

Differentiate using the Constant Rule.

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Step 2.3.2.1.4.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- 1$ with respect to $t$ is $0$ .

$4 t + 0$

Step 2.3.2.1.4.2

Add $4 t$ and $0$ .

$4 t$

Step 2.3.2.2

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

$s + C_{1} = 12 \int u^{3} \frac{1}{4} d u$

Step 2.3.3

Combine $u^{3}$ and $\frac{1}{4}$ .

$s + C_{1} = 12 \int \frac{u^{3}}{4} d u$

Step 2.3.4

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

$s + C_{1} = 12 (\frac{1}{4} \int u^{3} d u)$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Combine $\frac{1}{4}$ and $12$ .

$s + C_{1} = \frac{12}{4} \int u^{3} d u$

Step 2.3.5.2

Cancel the common factor of $12$ and $4$ .

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Step 2.3.5.2.1

Factor $4$ out of $12$ .

$s + C_{1} = \frac{4 \cdot 3}{4} \int u^{3} d u$

Step 2.3.5.2.2

Cancel the common factors.

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Step 2.3.5.2.2.1

Factor $4$ out of $4$ .

$s + C_{1} = \frac{4 \cdot 3}{4 (1)} \int u^{3} d u$

Step 2.3.5.2.2.2

Cancel the common factor.

$s + C_{1} = \frac{4 \cdot 3}{4 \cdot 1} \int u^{3} d u$

Step 2.3.5.2.2.3

Rewrite the expression.

$s + C_{1} = \frac{3}{1} \int u^{3} d u$

Step 2.3.5.2.2.4

Divide $3$ by $1$ .

$s + C_{1} = 3 \int u^{3} d u$

Step 2.3.6

By the Power Rule, the integral of $u^{3}$ with respect to $u$ is $\frac{1}{4} u^{4}$ .

$s + C_{1} = 3 (\frac{1}{4} u^{4} + C_{2})$

Step 2.3.7

Simplify.

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Step 2.3.7.1

Rewrite $3 (\frac{1}{4} u^{4} + C_{2})$ as $3 (\frac{1}{4}) u^{4} + C_{2}$ .

$s + C_{1} = 3 (\frac{1}{4}) u^{4} + C_{2}$

Step 2.3.7.2

Combine $3$ and $\frac{1}{4}$ .

$s + C_{1} = \frac{3}{4} u^{4} + C_{2}$

Step 2.3.8

Replace all occurrences of $u$ with $2 t^{2} - 1$ .

$s + C_{1} = \frac{3}{4} {(2 t^{2} - 1)}^{4} + C_{2}$

Step 2.4

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

$s = \frac{3}{4} {(2 t^{2} - 1)}^{4} + K$