Calculus Examples

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

Step 1

Let $u = 1 - x - y$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 1 - x - y$ . Find $\frac{d u}{d y}$ .

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Step 1.1.1

Differentiate $1 - x - y$ .

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

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $1 - x - y$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [- x] + \frac{d}{d y} [- y]$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

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

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

Step 1.1.3.2

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

$0 + 0 - 1 \cdot 1$

Step 1.1.3.3

Multiply $- 1$ by $1$ .

$0 + 0 - 1$

Step 1.1.4

Combine terms.

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Step 1.1.4.1

Add $0$ and $0$ .

$0 - 1$

Step 1.1.4.2

Subtract $1$ from $0$ .

$- 1$

Step 1.2

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

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

Step 2

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

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

Step 3

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

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

Step 4

Rewrite $- (\frac{1}{3} u^{3} + C)$ as $- \frac{1}{3} u^{3} + C$ .

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

Step 5

Replace all occurrences of $u$ with $1 - x - y$ .

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