What is the integral from 0 to 1 of sqrt(x^2-2x+1) ?

The integral from 0 to 1 of sqrt(x^2-2x+1) is 1/2

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integral from 0 to 1 of sqrt(x^2-2x+1)

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Factor

x^{2} - 2 x + 1 : (x - 1)^{2}

= \int_{0}^{1} (x - 1)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{1} (x - 1)^{2} d x

Eliminate Absolutes

= \int_{0}^{1} (x - 1)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{1} ∣ (x - 1) ∣ d x

Eliminate Absolutes

= \int_{0}^{1} - x + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{1} x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 1 d x = 1

= - \frac{1}{2} + 1

Simplify

- \frac{1}{2} + 1 : \frac{1}{2}

= \frac{1}{2}

f (x) = 3 x + 1, (5, 4)

\frac{d}{d x} (\frac{x ^{2} + 1}{x - 1})

t^{2} y^{^{''}} + 2 t y^{^{'}} - 6 y = 0

\int \frac{1}{3 + t} d t

y^{^{'}} = 3 y x^{2} - 3 x^{2}

What is the integral from 0 to 1 of sqrt(x^2-2x+1) ?
The integral from 0 to 1 of sqrt(x^2-2x+1) is 1/2