What is the integral from-10 to 10 of sqrt(256-x^2) ?

The integral from-10 to 10 of sqrt(256-x^2) is 128(2arcsin(5/8)+(5sqrt(39))/(32))

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integral from-10 to 10 of sqrt(256-x^2)

\int_{- 10}^{10} 256 - x^{2} d x

128 (2 arcsin (\frac{5}{8}) + \frac{5 39}{32})

Decimal

297.73363 \dots

Solution steps

\int_{- 10}^{10} 256 - x^{2} d x

Apply Trigonometric Substitution

: \int_{- a r c s i n (\frac{5}{8})}^{a r c s i n (\frac{5}{8})} 256 cos^{2} (u) d u

= \int_{- a r c s i n (\frac{5}{8})}^{a r c s i n (\frac{5}{8})} 256 cos^{2} (u) d u

Take the constant out:

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

= 256 \cdot \int_{- a r c s i n (\frac{5}{8})}^{a r c s i n (\frac{5}{8})} cos^{2} (u) d u

Rewrite using trig identities

= 256 \cdot \int_{- a r c s i n (\frac{5}{8})}^{a r c s i n (\frac{5}{8})} \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

= 256 \cdot \frac{1}{2} \cdot \int_{- a r c s i n (\frac{5}{8})}^{a r c s i n (\frac{5}{8})} 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

= 256 \cdot \frac{1}{2} (\int_{- a r c s i n (\frac{5}{8})}^{a r c s i n (\frac{5}{8})} 1 d u + \int_{- a r c s i n (\frac{5}{8})}^{a r c s i n (\frac{5}{8})} cos (2 u) d u)

\int_{- a r c s i n (\frac{5}{8})}^{a r c s i n (\frac{5}{8})} 1 d u = 2 arcsin (\frac{5}{8})

\int_{- a r c s i n (\frac{5}{8})}^{a r c s i n (\frac{5}{8})} cos (2 u) d u = \frac{5 39}{32}

= 256 \cdot \frac{1}{2} (2 arcsin (\frac{5}{8}) + \frac{5 39}{32})

Simplify

256 \cdot \frac{1}{2} (2 arcsin (\frac{5}{8}) + \frac{5 39}{32}) : 128 (2 arcsin (\frac{5}{8}) + \frac{5 39}{32})

= 128 (2 arcsin (\frac{5}{8}) + \frac{5 39}{32})

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What is the integral from-10 to 10 of sqrt(256-x^2) ?
The integral from-10 to 10 of sqrt(256-x^2) is 128(2arcsin(5/8)+(5sqrt(39))/(32))