Calculus Examples

$e^{- s t}$

Step 1

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

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Step 1.1

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

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Step 1.1.1

Differentiate $- s t$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

$- t \cdot 1$

Step 1.1.4

Multiply $- 1$ by $1$ .

$- t$

Step 1.2

Rewrite the problem using $u$ and $d u$ .

$\int - e^{u} \frac{- 1}{- t} d u$

Step 2

Simplify.

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Step 2.1

Dividing two negative values results in a positive value.

$\int - e^{u} \frac{1}{t} d u$

Step 2.2

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

$\int - \frac{e^{u}}{t} d u$

Step 3

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

$- \int \frac{e^{u}}{t} d u$

Step 4

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

$- (\frac{1}{t} \int e^{u} d u)$

Step 5

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

$- \frac{1}{t} (e^{u} + C)$

Step 6

Simplify.

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Step 6.1

Simplify.

$- \frac{1}{t} e^{u} + C$

Step 6.2

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

$- \frac{e^{u}}{t} + C$

Step 7

Replace all occurrences of $u$ with $- s t$ .

$- \frac{e^{- s t}}{t} + C$