Calculus Examples

$3^{1 - 2 x}$

Step 1

Write $3^{1 - 2 x}$ as a function.

$f (x) = 3^{1 - 2 x}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int 3^{1 - 2 x} d x$

Step 4

Let $u = 1 - 2 x$ . 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 4.1

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

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Step 4.1.1

Differentiate $1 - 2 x$ .

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

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $1 - 2 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x]$ .

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

Step 4.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- 2 x]$

Step 4.1.3

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

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Step 4.1.3.1

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

$0 - 2 \frac{d}{d x} [x]$

Step 4.1.3.2

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

$0 - 2 \cdot 1$

Step 4.1.3.3

Multiply $- 2$ by $1$ .

$0 - 2$

Step 4.1.4

Subtract $2$ from $0$ .

$- 2$

Step 4.2

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

$\int 3^{u} \cdot \frac{1}{- 2} d u$

Step 5

Simplify.

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Step 5.1

Move the negative in front of the fraction.

$\int 3^{u} \cdot (- \frac{1}{2}) d u$

Step 5.2

Combine $3^{u}$ and $\frac{1}{2}$ .

$\int - \frac{3^{u}}{2} d u$

Step 6

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

$- \int \frac{3^{u}}{2} d u$

Step 7

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$- (\frac{1}{2} \int 3^{u} d u)$

Step 8

The integral of $3^{u}$ with respect to $u$ is $\frac{3^{u}}{\ln (3)}$ .

$- \frac{1}{2} (\frac{3^{u}}{\ln (3)} + C)$

Step 9

Simplify.

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Step 9.1

Rewrite $- \frac{1}{2} (\frac{3^{u}}{\ln (3)} + C)$ as $- \frac{1}{2} \cdot \frac{1}{\ln (3)} \cdot 3^{u} + C$ .

$- \frac{1}{2} \cdot \frac{1}{\ln (3)} \cdot 3^{u} + C$

Step 9.2

Simplify.

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Step 9.2.1

Multiply $\frac{1}{\ln (3)}$ by $\frac{1}{2}$ .

$- \frac{1}{\ln (3) \cdot 2} \cdot 3^{u} + C$

Step 9.2.2

Move $2$ to the left of $\ln (3)$ .

$- \frac{1}{2 \ln (3)} \cdot 3^{u} + C$

Step 10

Replace all occurrences of $u$ with $1 - 2 x$ .

$- \frac{1}{2 \ln (3)} \cdot 3^{1 - 2 x} + C$

Step 11

The answer is the antiderivative of the function $f (x) = 3^{1 - 2 x}$ .

$F (x) =$ $- \frac{1}{2 \ln (3)} \cdot 3^{1 - 2 x} + C$