What is the integral from 0 to 1 of sqrt(9-12x+4x^2) ?

The integral from 0 to 1 of sqrt(9-12x+4x^2) is 2

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integral from 0 to 1 of sqrt(9-12x+4x^2)

\int_{0}^{1} 9 - 12 x + 4 x^{2} d x

2

Solution steps

\int_{0}^{1} 9 - 12 x + 4 x^{2} d x

Factor

9 - 12 x + 4 x^{2} : (2 x - 3)^{2}

= \int_{0}^{1} (2 x - 3)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{1} (2 x - 3)^{2} d x

Eliminate Absolutes

= \int_{0}^{1} (2 x - 3)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{1} ∣ (2 x - 3) ∣ d x

Eliminate Absolutes

= \int_{0}^{1} - 2 x + 3 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} 2 x d x + \int_{0}^{1} 3 d x

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

\int_{0}^{1} 3 d x = 3

= - 1 + 3

Simplify

= 2

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What is the integral from 0 to 1 of sqrt(9-12x+4x^2) ?
The integral from 0 to 1 of sqrt(9-12x+4x^2) is 2