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

The integral from 2 to 4 of sqrt(25-x^2) is -sqrt(21)+(25arcsin(4/5)-25arcsin(2/5))/2+6

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

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

- 21 + \frac{25 arcsin ( \frac{4}{5} ) - 25 arcsin ( \frac{2}{5} )}{2} + 6

Decimal

7.86465 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

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

\int_{a r c s i n (\frac{2}{5})}^{a r c s i n (\frac{4}{5})} cos (2 u) d u = \frac{2 ( 6 - 21 )}{25}

= 25 \cdot \frac{1}{2} (arcsin (\frac{4}{5}) - arcsin (\frac{2}{5}) + \frac{2 ( 6 - 21 )}{25})

Simplify

25 \cdot \frac{1}{2} (arcsin (\frac{4}{5}) - arcsin (\frac{2}{5}) + \frac{2 ( 6 - 21 )}{25}) : - 21 + \frac{25 arcsin ( \frac{4}{5} ) - 25 arcsin ( \frac{2}{5} )}{2} + 6

= - 21 + \frac{25 arcsin ( \frac{4}{5} ) - 25 arcsin ( \frac{2}{5} )}{2} + 6

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What is the integral from 2 to 4 of sqrt(25-x^2) ?
The integral from 2 to 4 of sqrt(25-x^2) is -sqrt(21)+(25arcsin(4/5)-25arcsin(2/5))/2+6