Calculus Examples

$\int_{0}^{\infty} e^{- s t} d t$

Step 1

Write the integral as a limit as $t_{1}$ approaches $\infty$ .

$lim_{t_{1} \to \infty} \int_{0}^{t_{1}} e^{- s t} d t$

Step 2

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

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Step 2.1

Let $u = - s t$ . Find $\frac{d u}{d t}$ .

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Step 2.1.1

Differentiate $- s t$ .

$\frac{d}{d t} [- s t]$

Step 2.1.2

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

$- s \frac{d}{d t} [t]$

Step 2.1.3

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

$- s \cdot 1$

Step 2.1.4

Multiply $- 1$ by $1$ .

$- s$

Step 2.2

Substitute the lower limit in for $t$ in $u = - s t$ .

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

Step 2.3

Multiply $- s \cdot 0$ .

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Step 2.3.1

Multiply $0$ by $- 1$ .

$u_{lower} = 0 s$

Step 2.3.2

Multiply $0$ by $s$ .

$u_{lower} = 0$

Step 2.4

Substitute the upper limit in for $t$ in $u = - s t$ .

$u_{upper} = - s t_{1}$

Step 2.5

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

$u_{lower} = 0$

$u_{upper} = - s t_{1}$

Step 2.6

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

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

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

Step 3

Simplify.

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Step 3.1

Dividing two negative values results in a positive value.

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

Step 3.2

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

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

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

Step 4

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

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

Step 5

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

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

Step 6

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

$lim_{t_{1} \to \infty} - \frac{1}{s} e^{u}]_{0}^{- s t_{1}}$

Step 7

Combine $e^{u}]_{0}^{- s t_{1}}$ and $\frac{1}{s}$ .

$lim_{t_{1} \to \infty} - \frac{e^{u}]_{0}^{- s t_{1}}}{s}$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $e^{u}$ at $- s t_{1}$ and at $0$ .

$lim_{t_{1} \to \infty} - \frac{(e^{- s t_{1}}) - e^{0}}{s}$

Step 8.2

Simplify.

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Step 8.2.1

Anything raised to $0$ is $1$ .

$lim_{t_{1} \to \infty} - \frac{e^{- s t_{1}} - 1 \cdot 1}{s}$

Step 8.2.2

Multiply $- 1$ by $1$ .

$lim_{t_{1} \to \infty} - \frac{e^{- s t_{1}} - 1}{s}$

Step 9

Evaluate the limit.

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Step 9.1

Evaluate the limit.

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Step 9.1.1

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

$- lim_{t_{1} \to \infty} \frac{e^{- s t_{1}} - 1}{s}$

Step 9.1.2

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

$- \frac{1}{s} lim_{t_{1} \to \infty} e^{- s t_{1}} - 1$

Step 9.1.3

Split the limit using the Sum of Limits Rule on the limit as $t_{1}$ approaches $\infty$ .

$- \frac{1}{s} (lim_{t_{1} \to \infty} e^{- s t_{1}} - lim_{t_{1} \to \infty} 1)$

Step 9.2

Since the exponent $- s t_{1}$ approaches $- \infty$ , the quantity $e^{- s t_{1}}$ approaches $0$ .

$- \frac{1}{s} (0 - lim_{t_{1} \to \infty} 1)$

Step 9.3

Evaluate the limit.

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Step 9.3.1

Evaluate the limit of $1$ which is constant as $t_{1}$ approaches $\infty$ .

$- \frac{1}{s} (0 - 1 \cdot 1)$

Step 9.3.2

Simplify the answer.

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Step 9.3.2.1

Multiply $- 1$ by $1$ .

$- \frac{1}{s} (0 - 1)$

Step 9.3.2.2

Subtract $1$ from $0$ .

$- \frac{1}{s} \cdot - 1$

Step 9.3.2.3

Multiply $- \frac{1}{s} \cdot - 1$ .

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Step 9.3.2.3.1

Multiply $- 1$ by $- 1$ .

$1 \frac{1}{s}$

Step 9.3.2.3.2

Multiply $\frac{1}{s}$ by $1$ .

$\frac{1}{s}$