Calculus Examples

$\frac{d y}{d x} = {(x - 2)}^{2}$

Step 1

Rewrite the equation.

$d y = {(x - 2)}^{2} d x$

Step 2

Integrate both sides.

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Set up an integral on each side.

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

Apply the constant rule.

$y + C_{1} = \int {(x - 2)}^{2} d x$

Integrate the right side.

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Let $u = x - 2$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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

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Differentiate $x - 2$ .

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

By the Sum Rule, the derivative of $x - 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 2]$ .

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

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

$1 + \frac{d}{d x} [- 2]$

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$1 + 0$

Add $1$ and $0$ .

$1$

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

$y + C_{1} = \int u^{2} d u$

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

$y + C_{1} = \frac{1}{3} u^{3} + C_{2}$

Replace all occurrences of $u$ with $x - 2$ .

$y + C_{1} = \frac{1}{3} {(x - 2)}^{3} + C_{2}$

Group the constant of integration on the right side as $K$ .

$y = \frac{1}{3} {(x - 2)}^{3} + K$

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