Calculus Examples

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

Step 1

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

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

Step 2

Apply basic rules of exponents.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Multiply the exponents in ${(x^{\frac{1}{2}})}^{- 1}$ .

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Move the negative in front of the fraction.

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

Step 3

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

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Step 3.1

Let $u = - x^{\frac{1}{2}}$ . Find $\frac{d u}{d x}$ .

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Step 3.1.1

Differentiate $- x^{\frac{1}{2}}$ .

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

Step 3.1.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{\frac{1}{2}}$ with respect to $x$ is $- \frac{d}{d x} [x^{\frac{1}{2}}]$ .

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

Step 3.1.3

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

$- (\frac{1}{2} x^{\frac{1}{2} - 1})$

Step 3.1.4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.1.5

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

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

Step 3.1.6

Combine the numerators over the common denominator.

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

Step 3.1.7

Simplify the numerator.

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Step 3.1.7.1

Multiply $- 1$ by $2$ .

$- (\frac{1}{2} x^{\frac{1 - 2}{2}})$

Step 3.1.7.2

Subtract $2$ from $1$ .

$- (\frac{1}{2} x^{\frac{- 1}{2}})$

Step 3.1.8

Move the negative in front of the fraction.

$- (\frac{1}{2} x^{- \frac{1}{2}})$

Step 3.1.9

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

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

Step 3.1.10

Move $x^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.2

Substitute the lower limit in for $x$ in $u = - x^{\frac{1}{2}}$ .

$u_{lower} = - 1^{\frac{1}{2}}$

Step 3.3

Simplify.

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Step 3.3.1

One to any power is one.

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

Step 3.3.2

Multiply $- 1$ by $1$ .

$u_{lower} = - 1$

Step 3.4

Substitute the upper limit in for $x$ in $u = - x^{\frac{1}{2}}$ .

$u_{upper} = - t^{\frac{1}{2}}$

Step 3.5

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

$u_{lower} = - 1$

$u_{upper} = - t^{\frac{1}{2}}$

Step 3.6

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

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

Step 4

Multiply $- 1$ by $2$ .

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

Step 5

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

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

Step 6

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

$lim_{t \to \infty} - 2 (e^{u}]_{- 1}^{- t^{\frac{1}{2}}})$

Step 7

Evaluate $e^{u}$ at $- t^{\frac{1}{2}}$ and at $- 1$ .

$lim_{t \to \infty} - 2 (e^{- t^{\frac{1}{2}}} - e^{- 1})$

Step 8

Evaluate the limit.

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Step 8.1

Evaluate the limit.

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Step 8.1.1

Move the term $- 2$ outside of the limit because it is constant with respect to $t$ .

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

Step 8.1.2

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

$- 2 (lim_{t \to \infty} e^{- t^{\frac{1}{2}}} - lim_{t \to \infty} e^{- 1})$

Step 8.2

Since the exponent $- t^{\frac{1}{2}}$ approaches $- \infty$ , the quantity $e^{- t^{\frac{1}{2}}}$ approaches $0$ .

$- 2 (0 - lim_{t \to \infty} e^{- 1})$

Step 8.3

Evaluate the limit.

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Step 8.3.1

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

$- 2 (0 - e^{- 1})$

Step 8.3.2

Simplify the answer.

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Step 8.3.2.1

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

$- 2 (0 - \frac{1}{e})$

Step 8.3.2.2

Subtract $\frac{1}{e}$ from $0$ .

$- 2 (- \frac{1}{e})$

Step 8.3.2.3

Multiply $- 2 (- \frac{1}{e})$ .

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Step 8.3.2.3.1

Multiply $- 1$ by $- 2$ .

$2 \frac{1}{e}$

Step 8.3.2.3.2

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

$\frac{2}{e}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$\frac{2}{e}$

Decimal Form:

$0.73575888 \dots$