Calculus Examples

$y = \sqrt{x}$ , $y = 0$ , $x = 16$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Eliminate the equal sides of each equation and combine.

$\sqrt{x} = 0$

Step 1.2

Solve $\sqrt{x} = 0$ for $x$ .

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Step 1.2.1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x}}^{2} = 0^{2}$

Step 1.2.2

Simplify each side of the equation.

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Step 1.2.2.1

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

${(x^{\frac{1}{2}})}^{2} = 0^{2}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

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Step 1.2.2.2.1.1

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

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Step 1.2.2.2.1.1.1

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

$x^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 1.2.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 1.2.2.2.1.1.2.1

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 1.2.2.2.1.1.2.2

Rewrite the expression.

$x^{1} = 0^{2}$

Step 1.2.2.2.1.2

Simplify.

$x = 0^{2}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Raising $0$ to any positive power yields $0$ .

$x = 0$

Step 1.3

Substitute $0$ for $x$ .

$y = 0$

Step 1.4

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(0, 0)$

Step 2

The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region. The regions are determined by the intersection points of the curves. This can be done algebraically or graphically.

$A r e a = \int_{0}^{16} \sqrt{x} d x - \int_{0}^{16} 0 d x$

Step 3

Integrate to find the area between $0$ and $16$ .

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Step 3.1

Combine the integrals into a single integral.

$\int_{0}^{16} \sqrt{x} - (0) d x$

Step 3.2

Subtract $0$ from $\sqrt{x}$ .

$\int_{0}^{16} \sqrt{x} d x$

Step 3.3

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

$\int_{0}^{16} x^{\frac{1}{2}} d x$

Step 3.4

By the Power Rule, the integral of $x^{\frac{1}{2}}$ with respect to $x$ is $\frac{2}{3} x^{\frac{3}{2}}$ .

$\frac{2}{3} x^{\frac{3}{2}}]_{0}^{16}$

Step 3.5

Substitute and simplify.

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Step 3.5.1

Evaluate $\frac{2}{3} x^{\frac{3}{2}}$ at $16$ and at $0$ .

$(\frac{2}{3} \cdot 16^{\frac{3}{2}}) - \frac{2}{3} \cdot 0^{\frac{3}{2}}$

Step 3.5.2

Simplify.

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Step 3.5.2.1

Rewrite $16$ as $4^{2}$ .

$\frac{2}{3} \cdot {(4^{2})}^{\frac{3}{2}} - \frac{2}{3} \cdot 0^{\frac{3}{2}}$

Step 3.5.2.2

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

$\frac{2}{3} \cdot 4^{2 (\frac{3}{2})} - \frac{2}{3} \cdot 0^{\frac{3}{2}}$

Step 3.5.2.3

Cancel the common factor of $2$ .

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Step 3.5.2.3.1

Cancel the common factor.

$\frac{2}{3} \cdot 4^{2 (\frac{3}{2})} - \frac{2}{3} \cdot 0^{\frac{3}{2}}$

Step 3.5.2.3.2

Rewrite the expression.

$\frac{2}{3} \cdot 4^{3} - \frac{2}{3} \cdot 0^{\frac{3}{2}}$

Step 3.5.2.4

Raise $4$ to the power of $3$ .

$\frac{2}{3} \cdot 64 - \frac{2}{3} \cdot 0^{\frac{3}{2}}$

Step 3.5.2.5

Combine $\frac{2}{3}$ and $64$ .

$\frac{2 \cdot 64}{3} - \frac{2}{3} \cdot 0^{\frac{3}{2}}$

Step 3.5.2.6

Multiply $2$ by $64$ .

$\frac{128}{3} - \frac{2}{3} \cdot 0^{\frac{3}{2}}$

Step 3.5.2.7

Rewrite $0$ as $0^{2}$ .

$\frac{128}{3} - \frac{2}{3} \cdot {(0^{2})}^{\frac{3}{2}}$

Step 3.5.2.8

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

$\frac{128}{3} - \frac{2}{3} \cdot 0^{2 (\frac{3}{2})}$

Step 3.5.2.9

Cancel the common factor of $2$ .

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Step 3.5.2.9.1

Cancel the common factor.

$\frac{128}{3} - \frac{2}{3} \cdot 0^{2 (\frac{3}{2})}$

Step 3.5.2.9.2

Rewrite the expression.

$\frac{128}{3} - \frac{2}{3} \cdot 0^{3}$

Step 3.5.2.10

Raising $0$ to any positive power yields $0$ .

$\frac{128}{3} - \frac{2}{3} \cdot 0$

Step 3.5.2.11

Multiply $0$ by $- 1$ .

$\frac{128}{3} + 0 (\frac{2}{3})$

Step 3.5.2.12

Multiply $0$ by $\frac{2}{3}$ .

$\frac{128}{3} + 0$

Step 3.5.2.13

Add $\frac{128}{3}$ and $0$ .

$\frac{128}{3}$

Step 4