What is the integral from 0 to 3 of sqrt(4-4x+x^2) ?

The integral from 0 to 3 of sqrt(4-4x+x^2) is 5/2

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integral from 0 to 3 of sqrt(4-4x+x^2)

\int_{0}^{3} 4 - 4 x + x^{2} d x

\frac{5}{2}

Decimal Notation

2.5

Solution steps

\int_{0}^{3} 4 - 4 x + x^{2} d x

Factor

4 - 4 x + x^{2} : (x - 2)^{2}

= \int_{0}^{3} (x - 2)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{3} (x - 2)^{2} d x

Eliminate Absolutes

= \int_{0}^{2} (x - 2)^{2} d x + \int_{2}^{3} (x - 2)^{2} d x

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

\int_{2}^{3} (x - 2)^{2} d x = \frac{1}{2}

= 2 + \frac{1}{2}

Simplify

2 + \frac{1}{2} : \frac{5}{2}

= \frac{5}{2}

\int_{1}^{4} ∣ 3 x - 6 ∣ d x

\int_{- 3}^{0} 3 x^{2} (3 x^{3} + 3)^{6} d x

\int_{- 1}^{3} \frac{1}{x} d x

\int_{0}^{4} \frac{2}{2 x + 1} d x

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

What is the integral from 0 to 3 of sqrt(4-4x+x^2) ?
The integral from 0 to 3 of sqrt(4-4x+x^2) is 5/2