What is the integral from 1 to 3 of sqrt((x-1)(3-x)) ?

The integral from 1 to 3 of sqrt((x-1)(3-x)) is pi/2

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integral from 1 to 3 of sqrt((x-1)(3-x))

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Apply u-substitution

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

Complete the square

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

= \int_{0}^{2} - (u - 1)^{2} + 1 d u

Apply u-substitution

= \int_{- 1}^{1} - v^{2} + 1 d v

Apply Trigonometric Substitution

= \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (w) d w

Rewrite using trig identities

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + cos ( 2 w )}{2} d w

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 + cos (2 w) d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d w + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 w) d w)

\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d w = π

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 w) d w = 0

= \frac{1}{2} (π + 0)

Simplify

\frac{1}{2} (π + 0) : \frac{π}{2}

= \frac{π}{2}

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