What is the integral from 0 to 2 of sqrt(16-x^2) ?

The integral from 0 to 2 of sqrt(16-x^2) is 8(pi/6+(sqrt(3))/4)

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integral from 0 to 2 of sqrt(16-x^2)

\int_{0}^{2} 16 - x^{2} d x

8 (\frac{π}{6} + \frac{3}{4})

Decimal

7.65289 \dots

Solution steps

\int_{0}^{2} 16 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 16 cos^{2} (u) d u

= \int_{0}^{\frac{π}{6}} 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_{0}^{\frac{π}{6}} cos^{2} (u) d u

Rewrite using trig identities

= 16 \cdot \int_{0}^{\frac{π}{6}} \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_{0}^{\frac{π}{6}} 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_{0}^{\frac{π}{6}} 1 d u + \int_{0}^{\frac{π}{6}} cos (2 u) d u)

\int_{0}^{\frac{π}{6}} 1 d u = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (2 u) d u = \frac{3}{4}

= 16 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4})

Simplify

16 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4}) : 8 (\frac{π}{6} + \frac{3}{4})

= 8 (\frac{π}{6} + \frac{3}{4})

\int_{0}^{\frac{π}{6}} \frac{1}{cos ( x )} d x

\int_{0}^{\frac{π}{2}} 2 d x

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\int_{0}^{1} \frac{1}{3 y + 1} d y

\int_{0}^{3} (- x^{2} + 4 x) d x

What is the integral from 0 to 2 of sqrt(16-x^2) ?
The integral from 0 to 2 of sqrt(16-x^2) is 8(pi/6+(sqrt(3))/4)