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

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

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

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

\frac{19 π}{12}

Decimal

4.97418 \dots

Solution steps

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

Apply u-substitution

: \int_{\frac{3}{2}}^{1} - 2 u^{2} d u

= π \cdot \int_{\frac{3}{2}}^{1} - 2 u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

= π (- (- 2 \cdot \int_{1}^{\frac{3}{2}} u^{2} d u))

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{1}^{\frac{3}{2}}

= π (- (- 2 [\frac{u ^{3}}{3}]_{1}^{\frac{3}{2}}))

Simplify

= 2 π [\frac{u ^{3}}{3}]_{1}^{\frac{3}{2}}

Compute the boundaries:

\frac{19}{24}

= 2 π \frac{19}{24}

Simplify

= \frac{19 π}{12}

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