Calculus Examples

$\int_{0}^{\frac{π}{2}} \frac{r}{4 + r^{2}} d r$

Step 1

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

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Step 1.1

Let $u = 4 + r^{2}$ . Find $\frac{d u}{d r}$ .

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Step 1.1.1

Differentiate $4 + r^{2}$ .

$\frac{d}{d r} [4 + r^{2}]$

Step 1.1.2

By the Sum Rule, the derivative of $4 + r^{2}$ with respect to $r$ is $\frac{d}{d r} [4] + \frac{d}{d r} [r^{2}]$ .

$\frac{d}{d r} [4] + \frac{d}{d r} [r^{2}]$

Step 1.1.3

Since $4$ is constant with respect to $r$ , the derivative of $4$ with respect to $r$ is $0$ .

$0 + \frac{d}{d r} [r^{2}]$

Step 1.1.4

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

$0 + 2 r$

Step 1.1.5

Add $0$ and $2 r$ .

$2 r$

Step 1.2

Substitute the lower limit in for $r$ in $u = 4 + r^{2}$ .

$u_{lower} = 4 + 0^{2}$

Step 1.3

Simplify.

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Step 1.3.1

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

$u_{lower} = 4 + 0$

Step 1.3.2

Add $4$ and $0$ .

$u_{lower} = 4$

Step 1.4

Substitute the upper limit in for $r$ in $u = 4 + r^{2}$ .

$u_{upper} = 4 + {(\frac{π}{2})}^{2}$

Step 1.5

Simplify each term.

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Step 1.5.1

Apply the product rule to $\frac{π}{2}$ .

$u_{upper} = 4 + \frac{π^{2}}{2^{2}}$

Step 1.5.2

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

$u_{upper} = 4 + \frac{π^{2}}{4}$

Step 1.6

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

$u_{lower} = 4$

$u_{upper} = 4 + \frac{π^{2}}{4}$

Step 1.7

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

$\int_{4}^{4 + \frac{π^{2}}{4}} \frac{1}{u} \cdot \frac{1}{2} d u$

Step 2

Simplify.

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Step 2.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$\int_{4}^{4 + \frac{π^{2}}{4}} \frac{1}{u \cdot 2} d u$

Step 2.2

Move $2$ to the left of $u$ .

$\int_{4}^{4 + \frac{π^{2}}{4}} \frac{1}{2 u} d u$

Step 3

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int_{4}^{4 + \frac{π^{2}}{4}} \frac{1}{u} d u$

Step 4

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\frac{1}{2} \ln (| u |)]_{4}^{4 + \frac{π^{2}}{4}}$

Step 5

Evaluate $\ln (| u |)$ at $4 + \frac{π^{2}}{4}$ and at $4$ .

$\frac{1}{2} (\ln (| 4 + \frac{π^{2}}{4} |) - \ln (| 4 |))$

Step 6

Simplify.

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Step 6.1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{1}{2} \ln (\frac{| 4 + \frac{π^{2}}{4} |}{| 4 |})$

Step 6.2

Combine $\frac{1}{2}$ and $\ln (\frac{| 4 + \frac{π^{2}}{4} |}{| 4 |})$ .

$\frac{\ln (\frac{| 4 + \frac{π^{2}}{4} |}{| 4 |})}{2}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

$4 + \frac{π^{2}}{4}$ is approximately $6.4674011$ which is positive so remove the absolute value

$\frac{\ln (\frac{4 + \frac{π^{2}}{4}}{| 4 |})}{2}$

Step 7.1.2

To write $4$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{\ln (\frac{4 \cdot \frac{4}{4} + \frac{π^{2}}{4}}{| 4 |})}{2}$

Step 7.1.3

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

$\frac{\ln (\frac{\frac{4 \cdot 4}{4} + \frac{π^{2}}{4}}{| 4 |})}{2}$

Step 7.1.4

Combine the numerators over the common denominator.

$\frac{\ln (\frac{\frac{4 \cdot 4 + π^{2}}{4}}{| 4 |})}{2}$

Step 7.1.5

Multiply $4$ by $4$ .

$\frac{\ln (\frac{\frac{16 + π^{2}}{4}}{| 4 |})}{2}$

Step 7.2

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$\frac{\ln (\frac{\frac{16 + π^{2}}{4}}{4})}{2}$

Step 7.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{\ln (\frac{16 + π^{2}}{4} \cdot \frac{1}{4})}{2}$

Step 7.4

Multiply $\frac{16 + π^{2}}{4} \cdot \frac{1}{4}$ .

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Step 7.4.1

Multiply $\frac{16 + π^{2}}{4}$ by $\frac{1}{4}$ .

$\frac{\ln (\frac{16 + π^{2}}{4 \cdot 4})}{2}$

Step 7.4.2

Multiply $4$ by $4$ .

$\frac{\ln (\frac{16 + π^{2}}{16})}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{\ln (\frac{16 + π^{2}}{16})}{2}$

Decimal Form:

$0.24023999 \dots$

Step 9