Calculus Examples

$\int 2^{\frac{- x}{2}} d x$

Step 1

Move the negative in front of the fraction.

$\int 2^{- \frac{x}{2}} d x$

Step 2

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

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Step 2.1

Let $u_{1} = - \frac{x}{2}$ . Find $\frac{d u_{1}}{d x}$ .

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Step 2.1.1

Differentiate $- \frac{x}{2}$ .

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

Step 2.1.2

Since $- \frac{1}{2}$ is constant with respect to $x$ , the derivative of $- \frac{x}{2}$ with respect to $x$ is $- \frac{1}{2} \frac{d}{d x} [x]$ .

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

Step 2.1.3

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

$- \frac{1}{2} \cdot 1$

Step 2.1.4

Multiply $- 1$ by $1$ .

$- \frac{1}{2}$

Step 2.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int - 2^{u_{1}} \cdot \frac{- 1}{- \frac{1}{2}} d u_{1}$

Step 3

Simplify.

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Step 3.1

Dividing two negative values results in a positive value.

$\int - 2^{u_{1}} \cdot \frac{1}{\frac{1}{2}} d u_{1}$

Step 3.2

Multiply by the reciprocal of the fraction to divide by $\frac{1}{2}$ .

$\int - 2^{u_{1}} \cdot (1 \cdot 2) d u_{1}$

Step 3.3

Multiply $2$ by $1$ .

$\int - 2^{u_{1}} \cdot 2 d u_{1}$

Step 3.4

Multiply $2$ by $- 1$ .

$\int - 2 \cdot 2^{u_{1}} d u_{1}$

Step 3.5

Factor out negative.

$\int - (2 \cdot 2^{u_{1}}) d u_{1}$

Step 3.6

Raise $2$ to the power of $1$ .

$\int - (2^{1} \cdot 2^{u_{1}}) d u_{1}$

Step 3.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int - 2^{1 + u_{1}} d u_{1}$

Step 4

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

$- \int 2^{1 + u_{1}} d u_{1}$

Step 5

Let $u_{2} = 1 + u_{1}$ . Then $d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 5.1

Let $u_{2} = 1 + u_{1}$ . Find $\frac{d u_{2}}{d u_{1}}$ .

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Step 5.1.1

Differentiate $1 + u_{1}$ .

$\frac{d}{d u_{1}} [1 + u_{1}]$

Step 5.1.2

By the Sum Rule, the derivative of $1 + u_{1}$ with respect to $u_{1}$ is $\frac{d}{d u_{1}} [1] + \frac{d}{d u_{1}} [u_{1}]$ .

$\frac{d}{d u_{1}} [1] + \frac{d}{d u_{1}} [u_{1}]$

Step 5.1.3

Since $1$ is constant with respect to $u_{1}$ , the derivative of $1$ with respect to $u_{1}$ is $0$ .

$0 + \frac{d}{d u_{1}} [u_{1}]$

Step 5.1.4

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

$0 + 1$

Step 5.1.5

Add $0$ and $1$ .

$1$

Step 5.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$- \int 2^{u_{2}} d u_{2}$

Step 6

The integral of $2^{u_{2}}$ with respect to $u_{2}$ is $\frac{2^{u_{2}}}{\ln (2)}$ .

$- (\frac{2^{u_{2}}}{\ln (2)} + C)$

Step 7

Rewrite $- (\frac{2^{u_{2}}}{\ln (2)} + C)$ as $- \frac{1}{\ln (2)} \cdot 2^{u_{2}} + C$ .

$- \frac{1}{\ln (2)} \cdot 2^{u_{2}} + C$

Step 8

Substitute back in for each integration substitution variable.

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Step 8.1

Replace all occurrences of $u_{2}$ with $1 + u_{1}$ .

$- \frac{1}{\ln (2)} \cdot 2^{1 + u_{1}} + C$

Step 8.2

Replace all occurrences of $u_{1}$ with $- \frac{x}{2}$ .

$- \frac{1}{\ln (2)} \cdot 2^{1 - \frac{x}{2}} + C$

Step 9

Combine $2^{1 - \frac{x}{2}}$ and $\frac{1}{\ln (2)}$ .

$- \frac{2^{1 - \frac{x}{2}}}{\ln (2)} + C$

Step 10

Reorder terms.

$- \frac{1}{\ln (2)} \cdot 2^{1 - \frac{1}{2} x} + C$