Calculus Examples

$\int_{1}^{e} \frac{\ln (x)}{x} d x$

Step 1

Let $u = \ln (x)$ . Then $d u = \frac{1}{x} d x$ , so $x d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = \ln (x)$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $\ln (x)$ .

$\frac{d}{d x} [\ln (x)]$

Step 1.1.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$\frac{1}{x}$

Step 1.2

Substitute the lower limit in for $x$ in $u = \ln (x)$ .

$u_{lower} = \ln (1)$

Step 1.3

The natural logarithm of $1$ is $0$ .

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for $x$ in $u = \ln (x)$ .

$u_{upper} = \ln (e)$

Step 1.5

The natural logarithm of $e$ is $1$ .

$u_{upper} = 1$

Step 1.6

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

$u_{lower} = 0$

$u_{upper} = 1$

Step 1.7

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

$\int_{0}^{1} u d u$

Step 2

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

$\frac{1}{2} u^{2}]_{0}^{1}$

Step 3

Simplify the expression.

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Step 3.1

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

$(\frac{1}{2} \cdot 1^{2}) - \frac{1}{2} \cdot 0^{2}$

Step 3.2

Simplify the expression.

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Step 3.2.1

One to any power is one.

$\frac{1}{2} \cdot 1 - \frac{1}{2} \cdot 0^{2}$

Step 3.2.2

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

$\frac{1}{2} - \frac{1}{2} \cdot 0^{2}$

Step 3.2.3

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

$\frac{1}{2} - \frac{1}{2} \cdot 0$

Step 3.3

Simplify.

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Step 3.3.1

Multiply $0$ by $- 1$ .

$\frac{1}{2} + 0 (\frac{1}{2})$

Step 3.3.2

Multiply $0$ by $\frac{1}{2}$ .

$\frac{1}{2} + 0$

Step 3.4

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

$\frac{1}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{2}$

Decimal Form:

$0.5$