What is the integral from 0 to 2 of 3/2 (1-x)^2 ?

The integral from 0 to 2 of 3/2 (1-x)^2 is 1

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integral from 0 to 2 of 3/2 (1-x)^2

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

1

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

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

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

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

\int_{0}^{2} 2 x d x = 4

\int_{0}^{2} x^{2} d x = \frac{8}{3}

= \frac{3}{2} (2 - 4 + \frac{8}{3})

Simplify

\frac{3}{2} (2 - 4 + \frac{8}{3}) : 1

= 1

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What is the integral from 0 to 2 of 3/2 (1-x)^2 ?
The integral from 0 to 2 of 3/2 (1-x)^2 is 1