Calculus Examples

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

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 $A = π r^{2}$ .

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

Step 2

Simplify the integrand.

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Step 2.1

Apply the product rule to $\frac{3}{1 + x}$ .

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

Step 2.2

Raise $3$ to the power of $2$ .

$V = \frac{9}{{(1 + x)}^{2}}$

Step 3

Since $9$ is constant with respect to $x$ , move $9$ out of the integral.

$V = π (9 \int_{0}^{3} \frac{1}{{(1 + x)}^{2}} d x)$

Step 4

Move $9$ to the left of $π$ .

$V = 9 π \int_{0}^{3} \frac{1}{{(1 + x)}^{2}} d x$

Step 5

Let $u = 1 + x$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = 1 + x$ . Find $\frac{d u}{d x}$ .

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Step 5.1.1

Differentiate $1 + x$ .

$\frac{d}{d x} [1 + x]$

Step 5.1.2

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

$\frac{d}{d x} [1] + \frac{d}{d x} [x]$

Step 5.1.3

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

$0 + \frac{d}{d x} [x]$

Step 5.1.4

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

$0 + 1$

Step 5.1.5

Add $0$ and $1$ .

$1$

Step 5.2

Substitute the lower limit in for $x$ in $u = 1 + x$ .

$u_{lower} = 1 + 0$

Step 5.3

Add $1$ and $0$ .

$u_{lower} = 1$

Step 5.4

Substitute the upper limit in for $x$ in $u = 1 + x$ .

$u_{upper} = 1 + 3$

Step 5.5

Add $1$ and $3$ .

$u_{upper} = 4$

Step 5.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 4$

Step 5.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$V = 9 π \int_{1}^{4} \frac{1}{u^{2}} d u$

Step 6

Apply basic rules of exponents.

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Step 6.1

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

$V = 9 π \int_{1}^{4} {(u^{2})}^{- 1} d u$

Step 6.2

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

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Step 6.2.1

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

$V = 9 π \int_{1}^{4} u^{2 \cdot - 1} d u$

Step 6.2.2

Multiply $2$ by $- 1$ .

$V = 9 π \int_{1}^{4} u^{- 2} d u$

Step 7

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

$V = 9 π - u^{- 1}]_{1}^{4}$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $- u^{- 1}$ at $4$ and at $1$ .

$V = 9 π ((- 4^{- 1}) + 1^{- 1})$

Step 8.2

Simplify.

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Step 8.2.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$V = 9 π (- \frac{1}{4} + 1^{- 1})$

Step 8.2.2

One to any power is one.

$V = 9 π (- \frac{1}{4} + 1)$

Step 8.2.3

Write $1$ as a fraction with a common denominator.

$V = 9 π (- \frac{1}{4} + \frac{4}{4})$

Step 8.2.4

Combine the numerators over the common denominator.

$V = 9 π (\frac{- 1 + 4}{4})$

Step 8.2.5

Add $- 1$ and $4$ .

$V = 9 π (\frac{3}{4})$

Step 8.2.6

Combine $\frac{3}{4}$ and $9$ .

$V = \frac{3 \cdot 9}{4} \cdot π$

Step 8.2.7

Multiply $3$ by $9$ .

$V = \frac{27}{4} \cdot π$

Step 8.2.8

Combine $\frac{27}{4}$ and $π$ .

$V = \frac{27 π}{4}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$V = \frac{27 π}{4}$

Decimal Form:

$V = 21.20575041 \dots$

Step 10