What is the integral from 0 to sqrt(3 of)sqrt(4-x^2) ?

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

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integral from 0 to sqrt(3 of)sqrt(4-x^2)

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

2 (\frac{π}{3} + \frac{3}{4})

Decimal

2.96042 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= 4 \cdot \int_{0}^{\frac{π}{3}} \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

= 4 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} 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

= 4 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{3}} 1 d u + \int_{0}^{\frac{π}{3}} cos (2 u) d u)

\int_{0}^{\frac{π}{3}} 1 d u = \frac{π}{3}

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

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

Simplify

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

= 2 (\frac{π}{3} + \frac{3}{4})

\int_{0}^{1} 1 + 4 x^{2} x d x

\int_{1}^{3} 1 + \frac{1}{x} + \frac{1}{x ^{2}} d x

\int_{0}^{1} 2 x e^{- x} d x

\int_{1}^{- 1} (9 x + 8)^{2} d x

\int_{- 1}^{1} 4 x d x

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