Calculus Examples

$2^{x + 1}$

Step 1

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

$f (x) = 2^{x + 1}$

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.

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

Step 4

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

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

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Step 4.1.1

Differentiate $x + 1$ .

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

Step 4.1.2

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

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

Step 4.1.3

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

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

Step 4.1.4

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

$1 + 0$

Step 4.1.5

Add $1$ and $0$ .

$1$

Step 4.2

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

$\int 2^{u} d u$

Step 5

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

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

Step 6

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

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

Step 7

Replace all occurrences of $u$ with $x + 1$ .

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

Step 8

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

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