Calculus Examples

$\frac{d r}{d q} = \frac{2000}{\sqrt{300 q}}$

Step 1

Rewrite the equation.

$d r = \frac{2000}{\sqrt{300 q}} d q$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d r = \int \frac{2000}{\sqrt{300 q}} d q$

Step 2.2

Apply the constant rule.

$r + C_{1} = \int \frac{2000}{\sqrt{300 q}} d q$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $2000$ is constant with respect to $q$ , move $2000$ out of the integral.

$r + C_{1} = 2000 \int \frac{1}{\sqrt{300 q}} d q$

Step 2.3.2

Let $u = 300 q$ . Then $d u = 300 d q$ , so $\frac{1}{300} d u = d q$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

Let $u = 300 q$ . Find $\frac{d u}{d q}$ .

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Step 2.3.2.1.1

Differentiate $300 q$ .

$\frac{d}{d q} [300 q]$

Step 2.3.2.1.2

Since $300$ is constant with respect to $q$ , the derivative of $300 q$ with respect to $q$ is $300 \frac{d}{d q} [q]$ .

$300 \frac{d}{d q} [q]$

Step 2.3.2.1.3

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

$300 \cdot 1$

Step 2.3.2.1.4

Multiply $300$ by $1$ .

$300$

Step 2.3.2.2

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

$r + C_{1} = 2000 \int \frac{1}{\sqrt{u}} \cdot \frac{1}{300} d u$

Step 2.3.3

Simplify.

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Step 2.3.3.1

Multiply $\frac{1}{\sqrt{u}}$ by $\frac{1}{300}$ .

$r + C_{1} = 2000 \int \frac{1}{\sqrt{u} \cdot 300} d u$

Step 2.3.3.2

Move $300$ to the left of $\sqrt{u}$ .

$r + C_{1} = 2000 \int \frac{1}{300 \sqrt{u}} d u$

Step 2.3.4

Since $\frac{1}{300}$ is constant with respect to $u$ , move $\frac{1}{300}$ out of the integral.

$r + C_{1} = 2000 (\frac{1}{300} \int \frac{1}{\sqrt{u}} d u)$

Step 2.3.5

Simplify the expression.

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Step 2.3.5.1

Simplify.

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Step 2.3.5.1.1

Combine $\frac{1}{300}$ and $2000$ .

$r + C_{1} = \frac{2000}{300} \int \frac{1}{\sqrt{u}} d u$

Step 2.3.5.1.2

Cancel the common factor of $2000$ and $300$ .

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Step 2.3.5.1.2.1

Factor $100$ out of $2000$ .

$r + C_{1} = \frac{100 (20)}{300} \int \frac{1}{\sqrt{u}} d u$

Step 2.3.5.1.2.2

Cancel the common factors.

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Step 2.3.5.1.2.2.1

Factor $100$ out of $300$ .

$r + C_{1} = \frac{100 \cdot 20}{100 \cdot 3} \int \frac{1}{\sqrt{u}} d u$

Step 2.3.5.1.2.2.2

Cancel the common factor.

$r + C_{1} = \frac{100 \cdot 20}{100 \cdot 3} \int \frac{1}{\sqrt{u}} d u$

Step 2.3.5.1.2.2.3

Rewrite the expression.

$r + C_{1} = \frac{20}{3} \int \frac{1}{\sqrt{u}} d u$

Step 2.3.5.2

Apply basic rules of exponents.

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Step 2.3.5.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{u}$ as $u^{\frac{1}{2}}$ .

$r + C_{1} = \frac{20}{3} \int \frac{1}{u^{\frac{1}{2}}} d u$

Step 2.3.5.2.2

Move $u^{\frac{1}{2}}$ out of the denominator by raising it to the $- 1$ power.

$r + C_{1} = \frac{20}{3} \int {(u^{\frac{1}{2}})}^{- 1} d u$

Step 2.3.5.2.3

Multiply the exponents in ${(u^{\frac{1}{2}})}^{- 1}$ .

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Step 2.3.5.2.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$r + C_{1} = \frac{20}{3} \int u^{\frac{1}{2} \cdot - 1} d u$

Step 2.3.5.2.3.2

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

$r + C_{1} = \frac{20}{3} \int u^{\frac{- 1}{2}} d u$

Step 2.3.5.2.3.3

Move the negative in front of the fraction.

$r + C_{1} = \frac{20}{3} \int u^{- \frac{1}{2}} d u$

Step 2.3.6

By the Power Rule, the integral of $u^{- \frac{1}{2}}$ with respect to $u$ is $2 u^{\frac{1}{2}}$ .

$r + C_{1} = \frac{20}{3} (2 u^{\frac{1}{2}} + C_{2})$

Step 2.3.7

Simplify.

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Step 2.3.7.1

Rewrite $\frac{20}{3} (2 u^{\frac{1}{2}} + C_{2})$ as $\frac{20}{3} \cdot 2 u^{\frac{1}{2}} + C_{2}$ .

$r + C_{1} = \frac{20}{3} \cdot 2 u^{\frac{1}{2}} + C_{2}$

Step 2.3.7.2

Simplify.

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Step 2.3.7.2.1

Combine $\frac{20}{3}$ and $2$ .

$r + C_{1} = \frac{20 \cdot 2}{3} u^{\frac{1}{2}} + C_{2}$

Step 2.3.7.2.2

Multiply $20$ by $2$ .

$r + C_{1} = \frac{40}{3} u^{\frac{1}{2}} + C_{2}$

Step 2.3.8

Replace all occurrences of $u$ with $300 q$ .

$r + C_{1} = \frac{40}{3} {(300 q)}^{\frac{1}{2}} + C_{2}$

Step 2.4

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

$r = \frac{40}{3} {(300 q)}^{\frac{1}{2}} + K$