Calculus Examples

$f (x) = (3 x - 5) (2 x + 4)$

Step 1

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 2

Set up the integral to solve.

$F (x) = \int (3 x - 5) (2 x + 4) d x$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$\int 3 x (2 x + 4) - 5 (2 x + 4) d x$

Step 3.2

Apply the distributive property.

$\int 3 x (2 x) + 3 x \cdot 4 - 5 (2 x + 4) d x$

Step 3.3

Apply the distributive property.

$\int 3 x (2 x) + 3 x \cdot 4 - 5 (2 x) - 5 \cdot 4 d x$

Step 3.4

Move $x$ .

$\int 3 \cdot 2 x \cdot x + 3 x \cdot 4 - 5 (2 x) - 5 \cdot 4 d x$

Step 3.5

Move $x$ .

$\int 3 \cdot 2 x \cdot x + 3 \cdot 4 x - 5 (2 x) - 5 \cdot 4 d x$

Step 3.6

Multiply $3$ by $2$ .

$\int 6 x \cdot x + 3 \cdot 4 x - 5 \cdot 2 x - 5 \cdot 4 d x$

Step 3.7

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

$\int 6 (x^{1} x) + 3 \cdot 4 x - 5 \cdot 2 x - 5 \cdot 4 d x$

Step 3.8

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

$\int 6 (x^{1} x^{1}) + 3 \cdot 4 x - 5 \cdot 2 x - 5 \cdot 4 d x$

Step 3.9

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

$\int 6 x^{1 + 1} + 3 \cdot 4 x - 5 \cdot 2 x - 5 \cdot 4 d x$

Step 3.10

Add $1$ and $1$ .

$\int 6 x^{2} + 3 \cdot 4 x - 5 \cdot 2 x - 5 \cdot 4 d x$

Step 3.11

Multiply $3$ by $4$ .

$\int 6 x^{2} + 12 x - 5 \cdot 2 x - 5 \cdot 4 d x$

Step 3.12

Multiply $- 5$ by $2$ .

$\int 6 x^{2} + 12 x - 10 x - 5 \cdot 4 d x$

Step 3.13

Multiply $- 5$ by $4$ .

$\int 6 x^{2} + 12 x - 10 x - 20 d x$

Step 3.14

Subtract $10 x$ from $12 x$ .

$\int 6 x^{2} + 2 x - 20 d x$

Step 4

Split the single integral into multiple integrals.

$\int 6 x^{2} d x + \int 2 x d x + \int - 20 d x$

Step 5

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$6 \int x^{2} d x + \int 2 x d x + \int - 20 d x$

Step 6

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

$6 (\frac{1}{3} x^{3} + C) + \int 2 x d x + \int - 20 d x$

Step 7

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$6 (\frac{1}{3} x^{3} + C) + 2 \int x d x + \int - 20 d x$

Step 8

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

$6 (\frac{1}{3} x^{3} + C) + 2 (\frac{1}{2} x^{2} + C) + \int - 20 d x$

Step 9

Apply the constant rule.

$6 (\frac{1}{3} x^{3} + C) + 2 (\frac{1}{2} x^{2} + C) - 20 x + C$

Step 10

Simplify.

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Step 10.1

Simplify.

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Step 10.1.1

Combine $\frac{1}{3}$ and $x^{3}$ .

$6 (\frac{x^{3}}{3} + C) + 2 (\frac{1}{2} x^{2} + C) - 20 x + C$

Step 10.1.2

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

$6 (\frac{x^{3}}{3} + C) + 2 (\frac{x^{2}}{2} + C) - 20 x + C$

Step 10.2

Simplify.

$2 x^{3} + x^{2} - 20 x + C$

Step 11

The answer is the antiderivative of the function $f (x) = (3 x - 5) (2 x + 4)$ .

$F (x) =$ $2 x^{3} + x^{2} - 20 x + C$