Calculus Examples

$\int_{1}^{\infty} 4 x e^{- x^{2}} d x$

Step 1

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

$lim_{t \to \infty} \int_{1}^{t} 4 x e^{- x^{2}} d x$

Step 2

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$lim_{t \to \infty} 4 \int_{1}^{t} x e^{- x^{2}} d x$

Step 3

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

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Step 3.1

Let $u_{2} = e^{- x^{2}}$ . Find $\frac{d u_{2}}{d x}$ .

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Step 3.1.1

Differentiate $e^{- x^{2}}$ .

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

Step 3.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{2}$ .

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Step 3.1.2.1

To apply the Chain Rule, set $u_{1}$ as $- x^{2}$ .

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

Step 3.1.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 3.1.2.3

Replace all occurrences of $u_{1}$ with $- x^{2}$ .

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

Step 3.1.3

Differentiate.

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Step 3.1.3.1

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

$e^{- x^{2}} (- \frac{d}{d x} [x^{2}])$

Step 3.1.3.2

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

$e^{- x^{2}} (- (2 x))$

Step 3.1.3.3

Multiply $2$ by $- 1$ .

$e^{- x^{2}} (- 2 x)$

Step 3.1.4

Simplify.

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Step 3.1.4.1

Reorder the factors of $e^{- x^{2}} (- 2 x)$ .

$- 2 e^{- x^{2}} x$

Step 3.1.4.2

Reorder factors in $- 2 e^{- x^{2}} x$ .

$- 2 x e^{- x^{2}}$

Step 3.2

Substitute the lower limit in for $x$ in $u_{2} = e^{- x^{2}}$ .

$u_{lower} = e^{- 1^{2}}$

Step 3.3

Simplify.

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Step 3.3.1

One to any power is one.

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

Step 3.3.2

Multiply $- 1$ by $1$ .

$u_{lower} = e^{- 1}$

Step 3.3.3

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

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

Step 3.4

Substitute the upper limit in for $x$ in $u_{2} = e^{- x^{2}}$ .

$u_{upper} = e^{- t^{2}}$

Step 3.5

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

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

$u_{upper} = e^{- t^{2}}$

Step 3.6

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$lim_{t \to \infty} 4 \int_{\frac{1}{e}}^{e^{- t^{2}}} \frac{1}{- 2} d u_{2}$

Step 4

Move the negative in front of the fraction.

$lim_{t \to \infty} 4 \int_{\frac{1}{e}}^{e^{- t^{2}}} - \frac{1}{2} d u_{2}$

Step 5

Apply the constant rule.

$lim_{t \to \infty} 4 (- \frac{1}{2} u_{2}]_{\frac{1}{e}}^{e^{- t^{2}}})$

Step 6

Simplify the answer.

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Step 6.1

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

$lim_{t \to \infty} 4 (- \frac{u_{2}}{2}]_{\frac{1}{e}}^{e^{- t^{2}}})$

Step 6.2

Substitute and simplify.

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Step 6.2.1

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

$lim_{t \to \infty} 4 ((- \frac{e^{- t^{2}}}{2}) + \frac{\frac{1}{e}}{2})$

Step 6.2.2

Simplify.

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Step 6.2.2.1

Rewrite $\frac{\frac{1}{e}}{2}$ as a product.

$lim_{t \to \infty} 4 (- \frac{e^{- t^{2}}}{2} + \frac{1}{e} \cdot \frac{1}{2})$

Step 6.2.2.2

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

$lim_{t \to \infty} 4 (- \frac{e^{- t^{2}}}{2} + \frac{1}{e \cdot 2})$

Step 6.2.2.3

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

$lim_{t \to \infty} 4 (- \frac{e^{- t^{2}}}{2} + \frac{1}{2 e})$

Step 7

Evaluate the limit.

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Step 7.1

Evaluate the limit.

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Step 7.1.1

Move the term $4$ outside of the limit because it is constant with respect to $t$ .

$4 lim_{t \to \infty} - \frac{e^{- t^{2}}}{2} + \frac{1}{2 e}$

Step 7.1.2

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

$4 (- lim_{t \to \infty} \frac{e^{- t^{2}}}{2} + lim_{t \to \infty} \frac{1}{2 e})$

Step 7.1.3

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $t$ .

$4 (- \frac{1}{2} lim_{t \to \infty} e^{- t^{2}} + lim_{t \to \infty} \frac{1}{2 e})$

Step 7.2

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

$4 (- \frac{1}{2} \cdot 0 + lim_{t \to \infty} \frac{1}{2 e})$

Step 7.3

Evaluate the limit.

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Step 7.3.1

Evaluate the limit of $\frac{1}{2 e}$ which is constant as $t$ approaches $\infty$ .

$4 (- \frac{1}{2} \cdot 0 + \frac{1}{2 e})$

Step 7.3.2

Simplify the answer.

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Step 7.3.2.1

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

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Step 7.3.2.1.1

Multiply $0$ by $- 1$ .

$4 (0 (\frac{1}{2}) + \frac{1}{2 e})$

Step 7.3.2.1.2

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

$4 (0 + \frac{1}{2 e})$

Step 7.3.2.2

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

$4 \frac{1}{2 e}$

Step 7.3.2.3

Cancel the common factor of $2$ .

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Step 7.3.2.3.1

Factor $2$ out of $4$ .

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

Step 7.3.2.3.2

Factor $2$ out of $2 e$ .

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

Step 7.3.2.3.3

Cancel the common factor.

$2 \cdot 2 \frac{1}{2 e}$

Step 7.3.2.3.4

Rewrite the expression.

$2 \frac{1}{e}$

Step 7.3.2.4

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

$\frac{2}{e}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{2}{e}$

Decimal Form:

$0.73575888 \dots$