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

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

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

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

2 π

Decimal Notation

6.28318 \dots

Solution steps

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

Apply Trigonometric Substitution

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

= \int_{- \frac{π}{2}}^{\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_{- \frac{π}{2}}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

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

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

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

= 16 \cdot \frac{1}{8} (π - 0)

Simplify

16 \cdot \frac{1}{8} (π - 0) : 2 π

= 2 π

5 x + 3 x

s \to 1 lim (\frac{s ^{3} - 1}{s - 1})

x \to 6 lim (5 x)

\int_{0}^{2} (x - 1)^{25} d x

\int_{- \infty}^{0} (x e^{2 x}) d x

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