Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Let $u = - 2 x - 4$ . Then $d u = - 2 d x$ , so $- \frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.1

Let $u = - 2 x - 4$ . Find $\frac{d u}{d x}$ .

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Step 3.1.1

Differentiate $- 2 x - 4$ .

$\frac{d}{d x} [- 2 x - 4]$

Step 3.1.2

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

$\frac{d}{d x} [- 2 x] + \frac{d}{d x} [- 4]$

Step 3.1.3

Evaluate $\frac{d}{d x} [- 2 x]$ .

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Step 3.1.3.1

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

$- 2 \frac{d}{d x} [x] + \frac{d}{d x} [- 4]$

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 $n = 1$ .

$- 2 \cdot 1 + \frac{d}{d x} [- 4]$

Step 3.1.3.3

Multiply $- 2$ by $1$ .

$- 2 + \frac{d}{d x} [- 4]$

Step 3.1.4

Differentiate using the Constant Rule.

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Step 3.1.4.1

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

$- 2 + 0$

Step 3.1.4.2

Add $- 2$ and $0$ .

$- 2$

Step 3.2

Substitute the lower limit in for $x$ in $u = - 2 x - 4$ .

$u_{lower} = - 2 \cdot 2 - 4$

Step 3.3

Simplify.

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Step 3.3.1

Multiply $- 2$ by $2$ .

$u_{lower} = - 4 - 4$

Step 3.3.2

Subtract $4$ from $- 4$ .

$u_{lower} = - 8$

Step 3.4

Substitute the upper limit in for $x$ in $u = - 2 x - 4$ .

$u_{upper} = - 2 t - 4$

Step 3.5

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

$u_{lower} = - 8$

$u_{upper} = - 2 t - 4$

Step 3.6

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

$lim_{t \to \infty} 2 \int_{- 8}^{- 2 t - 4} e^{u} \frac{1}{- 2} d u$

Step 4

Simplify.

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Step 4.1

Move the negative in front of the fraction.

$lim_{t \to \infty} 2 \int_{- 8}^{- 2 t - 4} e^{u} (- \frac{1}{2}) d u$

Step 4.2

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

$lim_{t \to \infty} 2 \int_{- 8}^{- 2 t - 4} - \frac{e^{u}}{2} d u$

Step 5

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$lim_{t \to \infty} 2 (- \int_{- 8}^{- 2 t - 4} \frac{e^{u}}{2} d u)$

Step 6

Multiply $- 1$ by $2$ .

$lim_{t \to \infty} - 2 \int_{- 8}^{- 2 t - 4} \frac{e^{u}}{2} d u$

Step 7

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

$lim_{t \to \infty} - 2 (\frac{1}{2} \int_{- 8}^{- 2 t - 4} e^{u} d u)$

Step 8

Simplify.

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Step 8.1

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

$lim_{t \to \infty} \frac{- 2}{2} \int_{- 8}^{- 2 t - 4} e^{u} d u$

Step 8.2

Cancel the common factor of $- 2$ and $2$ .

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Step 8.2.1

Factor $2$ out of $- 2$ .

$lim_{t \to \infty} \frac{2 \cdot - 1}{2} \int_{- 8}^{- 2 t - 4} e^{u} d u$

Step 8.2.2

Cancel the common factors.

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Step 8.2.2.1

Factor $2$ out of $2$ .

$lim_{t \to \infty} \frac{2 \cdot - 1}{2 (1)} \int_{- 8}^{- 2 t - 4} e^{u} d u$

Step 8.2.2.2

Cancel the common factor.

$lim_{t \to \infty} \frac{2 \cdot - 1}{2 \cdot 1} \int_{- 8}^{- 2 t - 4} e^{u} d u$

Step 8.2.2.3

Rewrite the expression.

$lim_{t \to \infty} \frac{- 1}{1} \int_{- 8}^{- 2 t - 4} e^{u} d u$

Step 8.2.2.4

Divide $- 1$ by $1$ .

$lim_{t \to \infty} - \int_{- 8}^{- 2 t - 4} e^{u} d u$

Step 9

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

$lim_{t \to \infty} - (e^{u}]_{- 8}^{- 2 t - 4})$

Step 10

Evaluate $e^{u}$ at $- 2 t - 4$ and at $- 8$ .

$lim_{t \to \infty} - (e^{- 2 t - 4} - e^{- 8})$

Step 11

Evaluate the limit.

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Step 11.1

Evaluate the limit.

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Step 11.1.1

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

$- lim_{t \to \infty} e^{- 2 t - 4} - e^{- 8}$

Step 11.1.2

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

$- (lim_{t \to \infty} e^{- 2 t - 4} - lim_{t \to \infty} e^{- 8})$

Step 11.2

Since the exponent $- 2 t - 4$ approaches $- \infty$ , the quantity $e^{- 2 t - 4}$ approaches $0$ .

$- (0 - lim_{t \to \infty} e^{- 8})$

Step 11.3

Evaluate the limit.

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Step 11.3.1

Evaluate the limit of $e^{- 8}$ which is constant as $t$ approaches $\infty$ .

$- (0 - e^{- 8})$

Step 11.3.2

Simplify the answer.

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Step 11.3.2.1

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

$- (0 - \frac{1}{e^{8}})$

Step 11.3.2.2

Subtract $\frac{1}{e^{8}}$ from $0$ .

$- - \frac{1}{e^{8}}$

Step 11.3.2.3

Multiply $- - \frac{1}{e^{8}}$ .

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Step 11.3.2.3.1

Multiply $- 1$ by $- 1$ .

$1 \frac{1}{e^{8}}$

Step 11.3.2.3.2

Multiply $\frac{1}{e^{8}}$ by $1$ .

$\frac{1}{e^{8}}$

Step 12

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{e^{8}}$

Decimal Form:

$0.00033546 \dots$