What is the integral from 0 to 4 of x^3sqrt(7-x^2) ?

The integral from 0 to 4 of x^3sqrt(7-x^2) is (98sqrt(7))/(15)+i 558/5

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integral from 0 to 4 of x^3sqrt(7-x^2)

\int_{0}^{4} x^{3} 7 - x^{2} d x

\frac{98 7}{15} + i \frac{558}{5}

Solution steps

\int_{0}^{4} x^{3} 7 - x^{2} d x

Apply u-substitution

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

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

Expand

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

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

\int_{7}^{3 i} u^{4} d u = - \frac{49 7}{5} + i \frac{243}{5}

\int_{7}^{3 i} 7 u^{2} d u = - \frac{49 7}{3} - 63 i

= - (- (- \frac{49 7}{5} + i \frac{243}{5}) - \frac{49 7}{3} - 63 i)

Simplify

- (- (- \frac{49 7}{5} + i \frac{243}{5}) - \frac{49 7}{3} - 63 i) : \frac{98 7}{15} + i \frac{558}{5}

= \frac{98 7}{15} + i \frac{558}{5}

f (x) = x^{3} - \frac{2}{x ^{2}} + 2, (1, 3)

\frac{\partial}{\partial x} (x^{1})

\frac{\partial}{\partial x} (\frac{2 x + y}{z})

\int_{0}^{\frac{3 π}{2}} cos^{3} (29 x) d x

\frac{d}{d x} (\frac{x ^{3} + 3 x}{x ^{2} - 4 x + 3})

What is the integral from 0 to 4 of x^3sqrt(7-x^2) ?
The integral from 0 to 4 of x^3sqrt(7-x^2) is (98sqrt(7))/(15)+i 558/5