What is the integral from 6 to 12 of sqrt(144-x^2) ?

The integral from 6 to 12 of sqrt(144-x^2) is 72(pi/3-(sqrt(3))/4)

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integral from 6 to 12 of sqrt(144-x^2)

\int_{6}^{12} 144 - x^{2} d x

72 (\frac{π}{3} - \frac{3}{4})

Decimal Notation

44.22130 \dots

Solution steps

\int_{6}^{12} 144 - x^{2} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 144 cos^{2} (u) d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 144 cos^{2} (u) d u

Take the constant out:

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

= 144 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 144 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 144 \cdot \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} 1 + cos (2 u) d u

Apply the Sum Rule:

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

= 144 \cdot \frac{1}{2} (\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u + \int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u)

\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u = \frac{π}{3}

\int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u = - \frac{3}{4}

= 144 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

Simplify

144 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : 72 (\frac{π}{3} - \frac{3}{4})

= 72 (\frac{π}{3} - \frac{3}{4})

x^{2} y^{^{''}} - 6 y = 0, y (1) = 6, y^{^{'}} (1) = - 2

\frac{\partial}{\partial x} (e^{x y} y^{2} + \frac{x}{y})

\int e^{- \frac{x ^{2}}{2}} d x

\int_{0}^{1} \frac{x ^{4} + 3}{x} d x

y = \frac{1}{( 3 x + 8 ) ^{2}}

What is the integral from 6 to 12 of sqrt(144-x^2) ?
The integral from 6 to 12 of sqrt(144-x^2) is 72(pi/3-(sqrt(3))/4)