What is the integral from-4 to 4 of sqrt(25-x^2) ?

The integral from-4 to 4 of sqrt(25-x^2) is 25arcsin(4/5)+12

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integral from-4 to 4 of sqrt(25-x^2)

\int_{- 4}^{4} 25 - x^{2} d x

25 arcsin (\frac{4}{5}) + 12

Decimal

35.18238 \dots

Solution steps

\int_{- 4}^{4} 25 - x^{2} d x

Apply Trigonometric Substitution

: \int_{- a r c s i n (\frac{4}{5})}^{a r c s i n (\frac{4}{5})} 25 cos^{2} (u) d u

= \int_{- a r c s i n (\frac{4}{5})}^{a r c s i n (\frac{4}{5})} 25 cos^{2} (u) d u

Take the constant out:

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

= 25 \cdot \int_{- a r c s i n (\frac{4}{5})}^{a r c s i n (\frac{4}{5})} cos^{2} (u) d u

Rewrite using trig identities

= 25 \cdot \int_{- a r c s i n (\frac{4}{5})}^{a r c s i n (\frac{4}{5})} \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

= 25 \cdot \frac{1}{2} \cdot \int_{- a r c s i n (\frac{4}{5})}^{a r c s i n (\frac{4}{5})} 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

= 25 \cdot \frac{1}{2} (\int_{- a r c s i n (\frac{4}{5})}^{a r c s i n (\frac{4}{5})} 1 d u + \int_{- a r c s i n (\frac{4}{5})}^{a r c s i n (\frac{4}{5})} cos (2 u) d u)

\int_{- a r c s i n (\frac{4}{5})}^{a r c s i n (\frac{4}{5})} 1 d u = 2 arcsin (\frac{4}{5})

\int_{- a r c s i n (\frac{4}{5})}^{a r c s i n (\frac{4}{5})} cos (2 u) d u = \frac{24}{25}

= 25 \cdot \frac{1}{2} (2 arcsin (\frac{4}{5}) + \frac{24}{25})

Simplify

25 \cdot \frac{1}{2} (2 arcsin (\frac{4}{5}) + \frac{24}{25}) : 25 arcsin (\frac{4}{5}) + 12

= 25 arcsin (\frac{4}{5}) + 12

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What is the integral from-4 to 4 of sqrt(25-x^2) ?
The integral from-4 to 4 of sqrt(25-x^2) is 25arcsin(4/5)+12