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

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

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

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

π

Decimal Notation

3.14159 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 16 sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 16 sin^{2} (u) cos^{2} (u) d u

Take the constant out:

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

= 16 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 16 \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= 16 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 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

= 16 \cdot \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (4 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

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

= 16 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Simplify

16 \cdot \frac{1}{8} (\frac{π}{2} - 0) : π

= π

f (x) = \frac{- 8 x}{x ^{2} + 1}

3 t + 2

x \to - 2 lim (x^{2} + 2 x - 3)

4 cos (x)

2 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

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