What is the integral from 0 to 4 of xsqrt(21-x^2) ?

The integral from 0 to 4 of xsqrt(21-x^2) is (21sqrt(21)-5sqrt(5))/3

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integral from 0 to 4 of xsqrt(21-x^2)

\int_{0}^{4} x 21 - x^{2} d x

\frac{21 21 - 5 5}{3}

Decimal Notation

28.35124 \dots

Solution steps

\int_{0}^{4} x 21 - x^{2} d x

Apply u-substitution

: \int_{21}^{5} - \frac{u}{2} d u

= \int_{21}^{5} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{5}^{21} - \frac{u}{2} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{5}^{21} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{21}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{21})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{21}

Compute the boundaries:

1421 - \frac{10 5}{3}

= \frac{1}{2} (1421 - \frac{10 5}{3})

Simplify

= \frac{21 21 - 5 5}{3}

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

\frac{d}{d x} (1 - e^{- 5 x})

x \frac{d y}{d x} = y^{2} - 4

x y^{^{'}} + y = 5 x y^{2}, y (1) = 3

\frac{( x - 1 ) ^{3}}{x ^{2}}

What is the integral from 0 to 4 of xsqrt(21-x^2) ?
The integral from 0 to 4 of xsqrt(21-x^2) is (21sqrt(21)-5sqrt(5))/3