What is the integral from 0 to 1 of sqrt(16-9x^2) ?

The integral from 0 to 1 of sqrt(16-9x^2) is (4(2arcsin(3/4)+sin(2arcsin(3/4))))/3

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integral from 0 to 1 of sqrt(16-9x^2)

\int_{0}^{1} 16 - 9 x^{2} d x

\frac{4 ( 2 arcsin ( \frac{3}{4} ) + sin ( 2 arcsin ( \frac{3}{4} ) ) )}{3}

Decimal Notation

3.58437 \dots

Solution steps

\int_{0}^{1} 16 - 9 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{3}{4})} \frac{16}{3} cos^{2} (u) d u

= \int_{0}^{a r c s i n (\frac{3}{4})} \frac{16}{3} cos^{2} (u) d u

Take the constant out:

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

= \frac{16}{3} \cdot \int_{0}^{a r c s i n (\frac{3}{4})} cos^{2} (u) d u

Rewrite using trig identities

= \frac{16}{3} \cdot \int_{0}^{a r c s i n (\frac{3}{4})} \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

= \frac{16}{3} \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{3}{4})} 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

= \frac{16}{3} \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{3}{4})} 1 d u + \int_{0}^{a r c s i n (\frac{3}{4})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{3}{4})} 1 d u = arcsin (\frac{3}{4})

\int_{0}^{a r c s i n (\frac{3}{4})} cos (2 u) d u = \frac{1}{2} sin (2 arcsin (\frac{3}{4}))

= \frac{16}{3} \cdot \frac{1}{2} (arcsin (\frac{3}{4}) + \frac{1}{2} sin (2 arcsin (\frac{3}{4})))

Simplify

\frac{16}{3} \cdot \frac{1}{2} (arcsin (\frac{3}{4}) + \frac{1}{2} sin (2 arcsin (\frac{3}{4}))) : \frac{4 ( 2 arcsin ( \frac{3}{4} ) + sin ( 2 arcsin ( \frac{3}{4} ) ) )}{3}

= \frac{4 ( 2 arcsin ( \frac{3}{4} ) + sin ( 2 arcsin ( \frac{3}{4} ) ) )}{3}

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What is the integral from 0 to 1 of sqrt(16-9x^2) ?
The integral from 0 to 1 of sqrt(16-9x^2) is (4(2arcsin(3/4)+sin(2arcsin(3/4))))/3