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

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

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

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

2 π

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 64 sin^{4} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 64 sin^{4} (u) cos^{2} (u) d u

Take the constant out:

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

= 64 \cdot \int_{0}^{\frac{π}{2}} sin^{4} (u) cos^{2} (u) d u

Rewrite using trig identities

= 64 \cdot \int_{0}^{\frac{π}{2}} sin^{4} (u) (1 - sin^{2} (u)) d u

Expand

sin^{4} (u) (1 - sin^{2} (u)) : sin^{4} (u) - sin^{6} (u)

= 64 \cdot \int_{0}^{\frac{π}{2}} sin^{4} (u) - sin^{6} (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

= 64 (\int_{0}^{\frac{π}{2}} sin^{4} (u) d u - \int_{0}^{\frac{π}{2}} sin^{6} (u) d u)

\int_{0}^{\frac{π}{2}} sin^{4} (u) d u = \frac{3 π}{16}

\int_{0}^{\frac{π}{2}} sin^{6} (u) d u = \frac{5 π}{32}

= 64 (\frac{3 π}{16} - \frac{5 π}{32})

Simplify

64 (\frac{3 π}{16} - \frac{5 π}{32}) : 2 π

= 2 π

\int_{- 1}^{3} (3 x - 8) (x + 3) d x

\int_{0}^{4} 5 x^{2} d x

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

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

\int_{0}^{1 - x} 1 d y

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