Calculus Examples

$4 e^{- 2 x} + {(x - 1)}^{3}$

Step 1

Write $4 e^{- 2 x} + {(x - 1)}^{3}$ as a function.

$f (x) = 4 e^{- 2 x} + {(x - 1)}^{3}$

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.

$F (x) = \int 4 e^{- 2 x} + {(x - 1)}^{3} d x$

Step 4

Split the single integral into multiple integrals.

$\int 4 e^{- 2 x} d x + \int {(x - 1)}^{3} d x$

Step 5

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$4 \int e^{- 2 x} d x + \int {(x - 1)}^{3} d x$

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $- 2 x$ .

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

Step 6.1.2

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

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

Step 6.1.3

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

$- 2 \cdot 1$

Step 6.1.4

Multiply $- 2$ by $1$ .

$- 2$

Step 6.2

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

$4 \int e^{u_{1}} \frac{1}{- 2} d u_{1} + \int {(x - 1)}^{3} d x$

Step 7

Simplify.

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Step 7.1

Move the negative in front of the fraction.

$4 \int e^{u_{1}} (- \frac{1}{2}) d u_{1} + \int {(x - 1)}^{3} d x$

Step 7.2

Combine $e^{u_{1}}$ and $\frac{1}{2}$ .

$4 \int - \frac{e^{u_{1}}}{2} d u_{1} + \int {(x - 1)}^{3} d x$

Step 8

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

$4 (- \int \frac{e^{u_{1}}}{2} d u_{1}) + \int {(x - 1)}^{3} d x$

Step 9

Multiply $- 1$ by $4$ .

$- 4 \int \frac{e^{u_{1}}}{2} d u_{1} + \int {(x - 1)}^{3} d x$

Step 10

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

$- 4 (\frac{1}{2} \int e^{u_{1}} d u_{1}) + \int {(x - 1)}^{3} d x$

Step 11

Simplify.

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Step 11.1

Combine $\frac{1}{2}$ and $- 4$ .

$\frac{- 4}{2} \int e^{u_{1}} d u_{1} + \int {(x - 1)}^{3} d x$

Step 11.2

Cancel the common factor of $- 4$ and $2$ .

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Step 11.2.1

Factor $2$ out of $- 4$ .

$\frac{2 \cdot - 2}{2} \int e^{u_{1}} d u_{1} + \int {(x - 1)}^{3} d x$

Step 11.2.2

Cancel the common factors.

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Step 11.2.2.1

Factor $2$ out of $2$ .

$\frac{2 \cdot - 2}{2 (1)} \int e^{u_{1}} d u_{1} + \int {(x - 1)}^{3} d x$

Step 11.2.2.2

Cancel the common factor.

$\frac{2 \cdot - 2}{2 \cdot 1} \int e^{u_{1}} d u_{1} + \int {(x - 1)}^{3} d x$

Step 11.2.2.3

Rewrite the expression.

$\frac{- 2}{1} \int e^{u_{1}} d u_{1} + \int {(x - 1)}^{3} d x$

Step 11.2.2.4

Divide $- 2$ by $1$ .

$- 2 \int e^{u_{1}} d u_{1} + \int {(x - 1)}^{3} d x$

Step 12

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

$- 2 (e^{u_{1}} + C) + \int {(x - 1)}^{3} d x$

Step 13

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

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Step 13.1

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

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Step 13.1.1

Differentiate $x - 1$ .

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

Step 13.1.2

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 13.1.3

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 13.1.4

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

$1 + 0$

Step 13.1.5

Add $1$ and $0$ .

$1$

Step 13.2

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

$- 2 (e^{u_{1}} + C) + \int {u_{2}}^{3} d u_{2}$

Step 14

By the Power Rule, the integral of ${u_{2}}^{3}$ with respect to $u_{2}$ is $\frac{1}{4} {u_{2}}^{4}$ .

$- 2 (e^{u_{1}} + C) + \frac{1}{4} {u_{2}}^{4} + C$

Step 15

Simplify.

$- 2 e^{u_{1}} + \frac{1}{4} {u_{2}}^{4} + C$

Step 16

Substitute back in for each integration substitution variable.

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Step 16.1

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

$- 2 e^{- 2 x} + \frac{1}{4} {u_{2}}^{4} + C$

Step 16.2

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

$- 2 e^{- 2 x} + \frac{1}{4} {(x - 1)}^{4} + C$

Step 17

The answer is the antiderivative of the function $f (x) = 4 e^{- 2 x} + {(x - 1)}^{3}$ .

$F (x) =$ $- 2 e^{- 2 x} + \frac{1}{4} {(x - 1)}^{4} + C$