Calculus Examples

$\int_{0}^{1} \sqrt{x^{2} - 2 x + 1} d x$

Step 1

Let $u = \sqrt{x^{2} - 2 x + 1}$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = \sqrt{x^{2} - 2 x + 1}$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $\sqrt{x^{2} - 2 x + 1}$ .

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

Step 1.1.2

Rewrite $\frac{d}{d x} [\sqrt{x^{2} - 2 x + 1}]$ as $\frac{d}{d x} [x - 1]$ .

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Step 1.1.2.1

Factor using the perfect square rule.

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Step 1.1.2.1.1

Rewrite $1$ as $1^{2}$ .

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

Step 1.1.2.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 1.1.2.1.3

Rewrite the polynomial.

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

Step 1.1.2.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

$\frac{d}{d x} [\sqrt{{(x - 1)}^{2}}]$

Step 1.1.2.2

Pull terms out from under the radical, assuming positive real numbers.

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

Step 1.1.3

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

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

Step 1.1.4

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

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

Step 1.1.5

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

$1 + 0$

Step 1.1.6

Add $1$ and $0$ .

$1$

Step 1.2

Substitute the lower limit in for $x$ in $u = \sqrt{x^{2} - 2 x + 1}$ .

$u_{lower} = \sqrt{0^{2} - 2 \cdot 0 + 1}$

Step 1.3

Simplify.

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Step 1.3.1

Raising $0$ to any positive power yields $0$ .

$u_{lower} = \sqrt{0 - 2 \cdot 0 + 1}$

Step 1.3.2

Multiply $- 2$ by $0$ .

$u_{lower} = \sqrt{0 + 0 + 1}$

Step 1.3.3

Add $0$ and $0$ .

$u_{lower} = \sqrt{0 + 1}$

Step 1.3.4

Add $0$ and $1$ .

$u_{lower} = \sqrt{1}$

Step 1.3.5

Any root of $1$ is $1$ .

$u_{lower} = 1$

Step 1.4

Substitute the upper limit in for $x$ in $u = \sqrt{x^{2} - 2 x + 1}$ .

$u_{upper} = \sqrt{1^{2} - 2 \cdot 1 + 1}$

Step 1.5

Simplify.

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Step 1.5.1

One to any power is one.

$u_{upper} = \sqrt{1 - 2 \cdot 1 + 1}$

Step 1.5.2

Multiply $- 2$ by $1$ .

$u_{upper} = \sqrt{1 - 2 + 1}$

Step 1.5.3

Subtract $2$ from $1$ .

$u_{upper} = \sqrt{- 1 + 1}$

Step 1.5.4

Add $- 1$ and $1$ .

$u_{upper} = \sqrt{0}$

Step 1.5.5

Rewrite $0$ as $0^{2}$ .

$u_{upper} = \sqrt{0^{2}}$

Step 1.5.6

Pull terms out from under the radical, assuming positive real numbers.

$u_{upper} = 0$

Step 1.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 0$

Step 1.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\int_{1}^{0} u d u$

Step 2

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

$\frac{1}{2} u^{2}]_{1}^{0}$

Step 3

Substitute and simplify.

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Step 3.1

Evaluate $\frac{1}{2} u^{2}$ at $0$ and at $1$ .

$(\frac{1}{2} \cdot 0^{2}) - \frac{1}{2} \cdot 1^{2}$

Step 3.2

Simplify.

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Step 3.2.1

Raising $0$ to any positive power yields $0$ .

$\frac{1}{2} \cdot 0 - \frac{1}{2} \cdot 1^{2}$

Step 3.2.2

Multiply $\frac{1}{2}$ by $0$ .

$0 - \frac{1}{2} \cdot 1^{2}$

Step 3.2.3

One to any power is one.

$0 - \frac{1}{2} \cdot 1$

Step 3.2.4

Multiply $- 1$ by $1$ .

$0 - \frac{1}{2}$

Step 3.2.5

Subtract $\frac{1}{2}$ from $0$ .

$- \frac{1}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{2}$

Decimal Form:

$- 0.5$

Step 5