Calculus Examples

$\int 16 x^{2} \sqrt[3]{7 - 4 x^{3}} d x$

Step 1

Since $16$ is constant with respect to $x$ , move $16$ out of the integral.

$16 \int x^{2} \sqrt[3]{7 - 4 x^{3}} d x$

Step 2

Let $u = 7 - 4 x^{3}$ . Then $d u = - 12 x^{2} d x$ , so $- \frac{1}{12} d u = x^{2} d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = 7 - 4 x^{3}$ . Find $\frac{d u}{d x}$ .

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Step 2.1.1

Differentiate $7 - 4 x^{3}$ .

$\frac{d}{d x} [7 - 4 x^{3}]$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

By the Sum Rule, the derivative of $7 - 4 x^{3}$ with respect to $x$ is $\frac{d}{d x} [7] + \frac{d}{d x} [- 4 x^{3}]$ .

$\frac{d}{d x} [7] + \frac{d}{d x} [- 4 x^{3}]$

Step 2.1.2.2

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- 4 x^{3}]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- 4 x^{3}]$ .

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Step 2.1.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x^{3}$ with respect to $x$ is $- 4 \frac{d}{d x} [x^{3}]$ .

$0 - 4 \frac{d}{d x} [x^{3}]$

Step 2.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$0 - 4 (3 x^{2})$

Step 2.1.3.3

Multiply $3$ by $- 4$ .

$0 - 12 x^{2}$

Step 2.1.4

Subtract $12 x^{2}$ from $0$ .

$- 12 x^{2}$

Step 2.2

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

$16 \int \sqrt[3]{u} \frac{1}{- 12} d u$

Step 3

Simplify.

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Step 3.1

Move the negative in front of the fraction.

$16 \int \sqrt[3]{u} (- \frac{1}{12}) d u$

Step 3.2

Combine $\sqrt[3]{u}$ and $\frac{1}{12}$ .

$16 \int - \frac{\sqrt[3]{u}}{12} d u$

Step 4

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

$16 (- \int \frac{\sqrt[3]{u}}{12} d u)$

Step 5

Multiply $- 1$ by $16$ .

$- 16 \int \frac{\sqrt[3]{u}}{12} d u$

Step 6

Since $\frac{1}{12}$ is constant with respect to $u$ , move $\frac{1}{12}$ out of the integral.

$- 16 (\frac{1}{12} \int \sqrt[3]{u} d u)$

Step 7

Simplify the expression.

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Step 7.1

Simplify.

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Step 7.1.1

Combine $\frac{1}{12}$ and $- 16$ .

$\frac{- 16}{12} \int \sqrt[3]{u} d u$

Step 7.1.2

Cancel the common factor of $- 16$ and $12$ .

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Step 7.1.2.1

Factor $4$ out of $- 16$ .

$\frac{4 (- 4)}{12} \int \sqrt[3]{u} d u$

Step 7.1.2.2

Cancel the common factors.

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Step 7.1.2.2.1

Factor $4$ out of $12$ .

$\frac{4 \cdot - 4}{4 \cdot 3} \int \sqrt[3]{u} d u$

Step 7.1.2.2.2

Cancel the common factor.

$\frac{4 \cdot - 4}{4 \cdot 3} \int \sqrt[3]{u} d u$

Step 7.1.2.2.3

Rewrite the expression.

$\frac{- 4}{3} \int \sqrt[3]{u} d u$

Step 7.1.3

Move the negative in front of the fraction.

$- \frac{4}{3} \int \sqrt[3]{u} d u$

Step 7.2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{u}$ as $u^{\frac{1}{3}}$ .

$- \frac{4}{3} \int u^{\frac{1}{3}} d u$

Step 8

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

$- \frac{4}{3} (\frac{3}{4} u^{\frac{4}{3}} + C)$

Step 9

Simplify.

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Step 9.1

Rewrite $- \frac{4}{3} (\frac{3}{4} u^{\frac{4}{3}} + C)$ as $- \frac{4}{3} \cdot \frac{3}{4} u^{\frac{4}{3}} + C$ .

$- \frac{4}{3} \cdot \frac{3}{4} u^{\frac{4}{3}} + C$

Step 9.2

Simplify.

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Step 9.2.1

Multiply $\frac{3}{4}$ by $\frac{4}{3}$ .

$- \frac{3 \cdot 4}{4 \cdot 3} u^{\frac{4}{3}} + C$

Step 9.2.2

Multiply $3$ by $4$ .

$- \frac{12}{4 \cdot 3} u^{\frac{4}{3}} + C$

Step 9.2.3

Multiply $4$ by $3$ .

$- \frac{12}{12} u^{\frac{4}{3}} + C$

Step 9.2.4

Cancel the common factor of $12$ .

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Step 9.2.4.1

Cancel the common factor.

$- \frac{12}{12} u^{\frac{4}{3}} + C$

Step 9.2.4.2

Rewrite the expression.

$- 1 \cdot 1 u^{\frac{4}{3}} + C$

Step 9.2.5

Multiply $- 1$ by $1$ .

$- u^{\frac{4}{3}} + C$

Step 10

Replace all occurrences of $u$ with $7 - 4 x^{3}$ .

$- {(7 - 4 x^{3})}^{\frac{4}{3}} + C$