What is the integral from 2 to 4 of sqrt(16-x^2) ?

The integral from 2 to 4 of sqrt(16-x^2) is 8(pi/3-(sqrt(3))/4)

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integral from 2 to 4 of sqrt(16-x^2)

\int_{2}^{4} 16 - x^{2} d x

8 (\frac{π}{3} - \frac{3}{4})

Solution steps

\int_{2}^{4} 16 - x^{2} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 16 cos^{2} (u) d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 16 cos^{2} (u) d u

Take the constant out:

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

= 16 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 16 \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

= 16 \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

= 16 \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}

= 16 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

Simplify

16 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : 8 (\frac{π}{3} - \frac{3}{4})

= 8 (\frac{π}{3} - \frac{3}{4})

6 e^{2 t}

\frac{d}{d x} (sin^{2} (cos (x)))

\frac{\partial}{\partial y} (6 y)

\int \frac{z}{z + 1} d z

12 (2 x - 3)^{5}

What is the integral from 2 to 4 of sqrt(16-x^2) ?
The integral from 2 to 4 of sqrt(16-x^2) is 8(pi/3-(sqrt(3))/4)