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:

$0.00033546 \dots$