Calculus Examples

$[8 (4^{x})]$

Step 1

Write $[8 (4^{x})]$ as a function.

$f (x) = [8 (4^{x})]$

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.

$F (x) = \int 8 (4^{x}) d x$

Step 4

Remove parentheses.

$\int 8 (4^{x}) d x$

Step 5

Simplify.

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Step 5.1

Rewrite $8$ as $2^{3}$ .

$\int 2^{3} \cdot 4^{x} d x$

Step 5.2

Rewrite $4$ as $2^{2}$ .

$\int 2^{3} \cdot {(2^{2})}^{x} d x$

Step 5.3

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int 2^{3} \cdot 2^{2 x} d x$

Step 5.4

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

$\int 2^{3 + 2 x} d x$

Step 6

Let $u = 3 + 2 x$ . 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 6.1

Let $u = 3 + 2 x$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1

Differentiate $3 + 2 x$ .

$\frac{d}{d x} [3 + 2 x]$

Step 6.1.2

Differentiate.

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Step 6.1.2.1

By the Sum Rule, the derivative of $3 + 2 x$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [2 x]$ .

$\frac{d}{d x} [3] + \frac{d}{d x} [2 x]$

Step 6.1.2.2

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

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

Step 6.1.3

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

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Step 6.1.3.1

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 6.1.3.2

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 6.1.3.3

Multiply $2$ by $1$ .

$0 + 2$

Step 6.1.4

Add $0$ and $2$ .

$2$

Step 6.2

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

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

Step 7

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

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

Step 8

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

$\frac{1}{2} \int 2^{u} d u$

Step 9

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

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

Step 10

Simplify.

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Step 10.1

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

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

Step 10.2

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

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

Step 11

Replace all occurrences of $u$ with $3 + 2 x$ .

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

Step 12

The answer is the antiderivative of the function $f (x) = 8 (4^{x})$ .

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