What is the integral from 0 to 2 of x^3sqrt(7-x^2) ?

The integral from 0 to 2 of x^3sqrt(7-x^2) is -7sqrt(3)+(98sqrt(7)+27sqrt(3))/(15)

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integral from 0 to 2 of x^3sqrt(7-x^2)

\int_{0}^{2} x^{3} 7 - x^{2} d x

- 73 + \frac{98 7 + 27 3}{15}

Decimal Notation

8.27891 \dots

Solution steps

\int_{0}^{2} x^{3} 7 - x^{2} d x

Apply u-substitution

: \int_{7}^{3} - (- u^{2} + 7) u^{2} d u

= \int_{7}^{3} - (- u^{2} + 7) u^{2} d u

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

= - \int_{3}^{7} - (- u^{2} + 7) u^{2} d u

Take the constant out:

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

= - (- \int_{3}^{7} (- u^{2} + 7) u^{2} d u)

Expand

(- u^{2} + 7) u^{2} : - u^{4} + 7 u^{2}

= - (- \int_{3}^{7} - u^{4} + 7 u^{2} 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

= - (- (- \int_{3}^{7} u^{4} d u + \int_{3}^{7} 7 u^{2} d u))

\int_{3}^{7} u^{4} d u = \frac{49 7 - 9 3}{5}

\int_{3}^{7} 7 u^{2} d u = 7 (\frac{7 7}{3} - 3)

= - (- (- \frac{49 7 - 9 3}{5} + 7 (\frac{7 7}{3} - 3)))

Simplify

- (- (- \frac{49 7 - 9 3}{5} + 7 (\frac{7 7}{3} - 3))) : - 73 + \frac{98 7 + 27 3}{15}

= - 73 + \frac{98 7 + 27 3}{15}

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

(4 x + 4 y)^{3} = 64 x^{3} + 64 y^{3} (- 1.1)

\int 8 x^{3} + 3 sin (x) d x

What is the integral from 0 to 2 of x^3sqrt(7-x^2) ?
The integral from 0 to 2 of x^3sqrt(7-x^2) is -7sqrt(3)+(98sqrt(7)+27sqrt(3))/(15)