Calculus Examples

$\int_{1}^{\infty} \frac{e^{- \frac{1}{x}}}{x^{2}} d x$

Step 1

Write the integral as a limit as $t$ approaches $\infty$ .

$lim_{t \to \infty} \int_{1}^{t} \frac{e^{- \frac{1}{x}}}{x^{2}} d x$

Step 2

Let $u = - \frac{1}{x}$ . Then $d u = \frac{1}{x^{2}} d x$ , so $x^{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

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

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Step 2.1.1

Differentiate $- \frac{1}{x}$ .

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

Step 2.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{x}$ .

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

Step 2.1.3

Differentiate.

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Step 2.1.3.1

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 2.1.3.2

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

$- - x^{- 2} + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 2.1.3.3

Multiply.

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Step 2.1.3.3.1

Multiply $- 1$ by $- 1$ .

$1 x^{- 2} + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 2.1.3.3.2

Multiply $x^{- 2}$ by $1$ .

$x^{- 2} + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 2.1.3.4

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

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

Step 2.1.3.5

Simplify the expression.

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Step 2.1.3.5.1

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

$x^{- 2} + 0$

Step 2.1.3.5.2

Add $x^{- 2}$ and $0$ .

$x^{- 2}$

Step 2.1.4

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

$\frac{1}{x^{2}}$

Step 2.2

Substitute the lower limit in for $x$ in $u = - \frac{1}{x}$ .

$u_{lower} = - \frac{1}{1}$

Step 2.3

Simplify.

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Step 2.3.1

Cancel the common factor of $1$ .

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Step 2.3.1.1

Cancel the common factor.

$u_{lower} = - \frac{1}{1}$

Step 2.3.1.2

Rewrite the expression.

$u_{lower} = - 1 \cdot 1$

Step 2.3.2

Multiply $- 1$ by $1$ .

$u_{lower} = - 1$

Step 2.4

Substitute the upper limit in for $x$ in $u = - \frac{1}{x}$ .

$u_{upper} = - \frac{1}{t}$

Step 2.5

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

$u_{lower} = - 1$

$u_{upper} = - \frac{1}{t}$

Step 2.6

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

$lim_{t \to \infty} \int_{- 1}^{- \frac{1}{t}} e^{u} d u$

Step 3

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$lim_{t \to \infty} e^{u}]_{- 1}^{- \frac{1}{t}}$

Step 4

Evaluate $e^{u}$ at $- \frac{1}{t}$ and at $- 1$ .

$lim_{t \to \infty} e^{- \frac{1}{t}} - e^{- 1}$

Step 5

Evaluate the limit.

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Step 5.1

Evaluate the limit.

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Step 5.1.1

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $\infty$ .

$lim_{t \to \infty} e^{- \frac{1}{t}} - lim_{t \to \infty} e^{- 1}$

Step 5.1.2

Move the limit into the exponent.

$e^{lim_{t \to \infty} - \frac{1}{t}} - lim_{t \to \infty} e^{- 1}$

Step 5.1.3

Move the term $- 1$ outside of the limit because it is constant with respect to $t$ .

$e^{- lim_{t \to \infty} \frac{1}{t}} - lim_{t \to \infty} e^{- 1}$

Step 5.2

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{t}$ approaches $0$ .

$e^{- 0} - lim_{t \to \infty} e^{- 1}$

Step 5.3

Evaluate the limit.

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Step 5.3.1

Evaluate the limit of $e^{- 1}$ which is constant as $t$ approaches $\infty$ .

$e^{- 0} - e^{- 1}$

Step 5.3.2

Simplify each term.

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Step 5.3.2.1

Multiply $- 1$ by $0$ .

$e^{0} - e^{- 1}$

Step 5.3.2.2

Anything raised to $0$ is $1$ .

$1 - e^{- 1}$

Step 5.3.2.3

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

$1 - \frac{1}{e}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$1 - \frac{1}{e}$

Decimal Form:

$0.63212055 \dots$