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

The integral from 1 to 2 of pi(2-1/2 x)^2 is pi 19/12

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

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

π \frac{19}{12}

Decimal Notation

4.97418 \dots

Solution steps

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

Take the constant out:

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

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

Simplify

(2 - \frac{1}{2} x)^{2} : 4 - 2 x + \frac{1}{4} x^{2}

= π \cdot \int_{1}^{2} 4 - 2 x + \frac{1}{4} 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

= π (\int_{1}^{2} 4 d x - \int_{1}^{2} 2 x d x + \int_{1}^{2} \frac{1}{4} x^{2} d x)

\int_{1}^{2} 4 d x = 4

\int_{1}^{2} 2 x d x = 3

\int_{1}^{2} \frac{1}{4} x^{2} d x = \frac{7}{12}

= π (4 - 3 + \frac{7}{12})

Simplify

π (4 - 3 + \frac{7}{12}) : π \frac{19}{12}

= π \frac{19}{12}

x = f x x^{3} - 4 x, (- 2, 0)

x \to 10 - lim (\frac{9}{x - 10})

\frac{d}{d x} ((7 - x^{2}))

1 - x

f (x) = (6 x - 4)^{4}

What is the integral from 1 to 2 of pi(2-1/2 x)^2 ?
The integral from 1 to 2 of pi(2-1/2 x)^2 is pi 19/12