Calculus Examples

$f (x) = x {(8 - x)}^{2}$

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Set up the integral to solve.

$F (x) = \int x {(8 - x)}^{2} d x$

Let $u = 8 - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = 8 - x$ . Find $\frac{d u}{d x}$ .

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Differentiate $8 - x$ .

$\frac{d}{d x} [8 - x]$

Differentiate.

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By the Sum Rule, the derivative of $8 - x$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [- x]$ .

$\frac{d}{d x} [8] + \frac{d}{d x} [- x]$

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

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

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

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Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$0 - \frac{d}{d x} [x]$

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

$0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$0 - 1$

Subtract $1$ from $0$ .

$- 1$

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

$\int - (- u + 8) u^{2} d u$

Expand $- (- u + 8) u^{2}$ .

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Apply the distributive property.

$\int (- - u - 1 \cdot 8) u^{2} d u$

Apply the distributive property.

$\int - - u u^{2} - 1 \cdot 8 u^{2} d u$

Move parentheses.

$\int - - 1 u \cdot u^{2} - 1 \cdot 8 u^{2} d u$

Multiply $- 1$ by $- 1$ .

$\int 1 u \cdot u^{2} - 1 \cdot 8 u^{2} d u$

Multiply $u$ by $1$ .

$\int u \cdot u^{2} - 1 \cdot 8 u^{2} d u$

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

$\int u^{1} u^{2} - 1 \cdot 8 u^{2} d u$

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

$\int u^{1 + 2} - 1 \cdot 8 u^{2} d u$

Add $1$ and $2$ .

$\int u^{3} - 1 \cdot 8 u^{2} d u$

Multiply $- 1$ by $8$ .

$\int u^{3} - 8 u^{2} d u$

Split the single integral into multiple integrals.

$\int u^{3} d u + \int - 8 u^{2} d u$

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

$\frac{1}{4} u^{4} + C + \int - 8 u^{2} d u$

Since $- 8$ is constant with respect to $u$ , move $- 8$ out of the integral.

$\frac{1}{4} u^{4} + C - 8 \int u^{2} d u$

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

$\frac{1}{4} u^{4} + C - 8 (\frac{1}{3} u^{3} + C)$

Simplify.

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Simplify.

$\frac{1}{4} u^{4} - 8 (\frac{1}{3}) u^{3} + C$

Simplify.

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Combine $- 8$ and $\frac{1}{3}$ .

$\frac{1}{4} u^{4} + \frac{- 8}{3} u^{3} + C$

Move the negative in front of the fraction.

$\frac{1}{4} u^{4} - \frac{8}{3} u^{3} + C$

Replace all occurrences of $u$ with $8 - x$ .

$\frac{1}{4} {(8 - x)}^{4} - \frac{8}{3} {(8 - x)}^{3} + C$

The answer is the antiderivative of the function $f (x) = x {(8 - x)}^{2}$ .

$F (x) =$ $\frac{1}{4} {(8 - x)}^{4} - \frac{8}{3} {(8 - x)}^{3} + C$