Calculus Examples

$\int_{0}^{\infty} 8 e^{- 8 x} d x$

Step 1

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

$lim_{t \to \infty} \int_{0}^{t} 8 e^{- 8 x} d x$

Step 2

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$lim_{t \to \infty} 8 \int_{0}^{t} e^{- 8 x} d x$

Step 3

Let $u = - 8 x$ . Then $d u = - 8 d x$ , so $- \frac{1}{8} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.1

Let $u = - 8 x$ . Find $\frac{d u}{d x}$ .

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Step 3.1.1

Differentiate $- 8 x$ .

$\frac{d}{d x} [- 8 x]$

Step 3.1.2

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x$ with respect to $x$ is $- 8 \frac{d}{d x} [x]$ .

$- 8 \frac{d}{d x} [x]$

Step 3.1.3

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

$- 8 \cdot 1$

Step 3.1.4

Multiply $- 8$ by $1$ .

$- 8$

Step 3.2

Substitute the lower limit in for $x$ in $u = - 8 x$ .

$u_{lower} = - 8 \cdot 0$

Step 3.3

Multiply $- 8$ by $0$ .

$u_{lower} = 0$

Step 3.4

Substitute the upper limit in for $x$ in $u = - 8 x$ .

$u_{upper} = - 8 t$

Step 3.5

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

$u_{lower} = 0$

$u_{upper} = - 8 t$

Step 3.6

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

$lim_{t \to \infty} 8 \int_{0}^{- 8 t} e^{u} \frac{1}{- 8} d u$

Step 4

Simplify.

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Step 4.1

Move the negative in front of the fraction.

$lim_{t \to \infty} 8 \int_{0}^{- 8 t} e^{u} (- \frac{1}{8}) d u$

Step 4.2

Combine $e^{u}$ and $\frac{1}{8}$ .

$lim_{t \to \infty} 8 \int_{0}^{- 8 t} - \frac{e^{u}}{8} d u$

Step 5

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

$lim_{t \to \infty} 8 (- \int_{0}^{- 8 t} \frac{e^{u}}{8} d u)$

Step 6

Multiply $- 1$ by $8$ .

$lim_{t \to \infty} - 8 \int_{0}^{- 8 t} \frac{e^{u}}{8} d u$

Step 7

Since $\frac{1}{8}$ is constant with respect to $u$ , move $\frac{1}{8}$ out of the integral.

$lim_{t \to \infty} - 8 (\frac{1}{8} \int_{0}^{- 8 t} e^{u} d u)$

Step 8

Simplify.

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Step 8.1

Combine $\frac{1}{8}$ and $- 8$ .

$lim_{t \to \infty} \frac{- 8}{8} \int_{0}^{- 8 t} e^{u} d u$

Step 8.2

Cancel the common factor of $- 8$ and $8$ .

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Step 8.2.1

Factor $8$ out of $- 8$ .

$lim_{t \to \infty} \frac{8 \cdot - 1}{8} \int_{0}^{- 8 t} e^{u} d u$

Step 8.2.2

Cancel the common factors.

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Step 8.2.2.1

Factor $8$ out of $8$ .

$lim_{t \to \infty} \frac{8 \cdot - 1}{8 (1)} \int_{0}^{- 8 t} e^{u} d u$

Step 8.2.2.2

Cancel the common factor.

$lim_{t \to \infty} \frac{8 \cdot - 1}{8 \cdot 1} \int_{0}^{- 8 t} e^{u} d u$

Step 8.2.2.3

Rewrite the expression.

$lim_{t \to \infty} \frac{- 1}{1} \int_{0}^{- 8 t} e^{u} d u$

Step 8.2.2.4

Divide $- 1$ by $1$ .

$lim_{t \to \infty} - \int_{0}^{- 8 t} e^{u} d u$

Step 9

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

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

Step 10

Substitute and simplify.

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Step 10.1

Evaluate $e^{u}$ at $- 8 t$ and at $0$ .

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

Step 10.2

Simplify.

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Step 10.2.1

Anything raised to $0$ is $1$ .

$lim_{t \to \infty} - (e^{- 8 t} - 1 \cdot 1)$

Step 10.2.2

Multiply $- 1$ by $1$ .

$lim_{t \to \infty} - (e^{- 8 t} - 1)$

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 $t$ .

$- lim_{t \to \infty} e^{- 8 t} - 1$

Step 11.1.2

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

$- (lim_{t \to \infty} e^{- 8 t} - lim_{t \to \infty} 1)$

Step 11.2

Since the exponent $- 8 t$ approaches $- \infty$ , the quantity $e^{- 8 t}$ approaches $0$ .

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

Step 11.3

Evaluate the limit.

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Step 11.3.1

Evaluate the limit of $1$ which is constant as $t$ approaches $\infty$ .

$- (0 - 1 \cdot 1)$

Step 11.3.2

Simplify the answer.

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Step 11.3.2.1

Multiply $- 1$ by $1$ .

$- (0 - 1)$

Step 11.3.2.2

Subtract $1$ from $0$ .

$- - 1$

Step 11.3.2.3

Multiply $- 1$ by $- 1$ .

$1$