Calculus Examples

$y = \sqrt{2 x + 3}$ , $y = 0$ , $x = 0$ , $x = 1$

Step 1

To find the volume of the solid, first define the area of each slice then integrate across the range. The area of each slice is the area of a circle with radius $f (x)$ and $A = π r^{2}$ .

$V = π \int_{0}^{1} {(f (x))}^{2} d x$ where $f (x) = \sqrt{2 x + 3}$

Step 2

Rewrite ${\sqrt{2 x + 3}}^{2}$ as $2 x + 3$ .

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Step 2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x + 3}$ as ${(2 x + 3)}^{\frac{1}{2}}$ .

$V = {({(2 x + 3)}^{\frac{1}{2}})}^{2}$

Step 2.2

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

$V = {(2 x + 3)}^{\frac{1}{2} \cdot 2}$

Step 2.3

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

$V = {(2 x + 3)}^{\frac{2}{2}}$

Step 2.4

Cancel the common factor of $2$ .

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Step 2.4.1

Cancel the common factor.

$V = {(2 x + 3)}^{\frac{2}{2}}$

Step 2.4.2

Rewrite the expression.

$V = 2 x + 3$

Step 2.5

Simplify.

$V = 2 x + 3$

Step 3

Split the single integral into multiple integrals.

$V = π (\int_{0}^{1} 2 x d x + \int_{0}^{1} 3 d x)$

Step 4

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$V = π (2 \int_{0}^{1} x d x + \int_{0}^{1} 3 d x)$

Step 5

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

$V = π (2 (\frac{1}{2} x^{2}]_{0}^{1}) + \int_{0}^{1} 3 d x)$

Step 6

Combine $\frac{1}{2}$ and $x^{2}$ .

$V = π (2 (\frac{x^{2}}{2}]_{0}^{1}) + \int_{0}^{1} 3 d x)$

Step 7

Apply the constant rule.

$V = π (2 (\frac{x^{2}}{2}]_{0}^{1}) + 3 x]_{0}^{1})$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $\frac{x^{2}}{2}$ at $1$ and at $0$ .

$V = π (2 ((\frac{1^{2}}{2}) - \frac{0^{2}}{2}) + 3 x]_{0}^{1})$

Step 8.2

Evaluate $3 x$ at $1$ and at $0$ .

$V = π (2 (\frac{1^{2}}{2} - \frac{0^{2}}{2}) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3

Simplify.

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Step 8.3.1

One to any power is one.

$V = π (2 (\frac{1}{2} - \frac{0^{2}}{2}) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.2

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

$V = π (2 (\frac{1}{2} - \frac{0}{2}) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.3

Cancel the common factor of $0$ and $2$ .

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Step 8.3.3.1

Factor $2$ out of $0$ .

$V = π (2 (\frac{1}{2} - \frac{2 (0)}{2}) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.3.2

Cancel the common factors.

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Step 8.3.3.2.1

Factor $2$ out of $2$ .

$V = π (2 (\frac{1}{2} - \frac{2 \cdot 0}{2 \cdot 1}) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.3.2.2

Cancel the common factor.

$V = π (2 (\frac{1}{2} - \frac{2 \cdot 0}{2 \cdot 1}) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.3.2.3

Rewrite the expression.

$V = π (2 (\frac{1}{2} - \frac{0}{1}) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.3.2.4

Divide $0$ by $1$ .

$V = π (2 (\frac{1}{2} - 0) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.4

Multiply $- 1$ by $0$ .

$V = π (2 (\frac{1}{2} + 0) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.5

Add $\frac{1}{2}$ and $0$ .

$V = π (2 (\frac{1}{2}) + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.6

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

$V = π (\frac{2}{2} + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.7

Cancel the common factor of $2$ .

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Step 8.3.7.1

Cancel the common factor.

$V = π (\frac{2}{2} + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.7.2

Rewrite the expression.

$V = π (1 + 3 \cdot 1 - 3 \cdot 0)$

Step 8.3.8

Multiply $3$ by $1$ .

$V = π (1 + 3 - 3 \cdot 0)$

Step 8.3.9

Multiply $- 3$ by $0$ .

$V = π (1 + 3 + 0)$

Step 8.3.10

Add $3$ and $0$ .

$V = π (1 + 3)$

Step 8.3.11

Add $1$ and $3$ .

$V = π \cdot 4$

Step 8.3.12

Move $4$ to the left of $π$ .

$V = 4 π$

Step 9

The result can be shown in multiple forms.

Exact Form:

$V = 4 π$

Decimal Form:

$V = 12.56637061 \dots$

Step 10