What is the integral from 0 to 4 of x^3sqrt(x^2+9) ?

The integral from 0 to 4 of x^3sqrt(x^2+9) is 1412/5

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integral from 0 to 4 of x^3sqrt(x^2+9)

\int_{0}^{4} x^{3} x^{2} + 9 d x

\frac{1412}{5}

Decimal

282.4

Solution steps

\int_{0}^{4} x^{3} x^{2} + 9 d x

Apply u-substitution

: \int_{3}^{5} (u^{2} - 9) u^{2} d u

= \int_{3}^{5} (u^{2} - 9) u^{2} d u

Expand

(u^{2} - 9) u^{2} : u^{4} - 9 u^{2}

= \int_{3}^{5} u^{4} - 9 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}^{5} u^{4} d u - \int_{3}^{5} 9 u^{2} d u

\int_{3}^{5} u^{4} d u = \frac{2882}{5}

\int_{3}^{5} 9 u^{2} d u = 294

= \frac{2882}{5} - 294

Simplify

\frac{2882}{5} - 294 : \frac{1412}{5}

= \frac{1412}{5}

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What is the integral from 0 to 4 of x^3sqrt(x^2+9) ?
The integral from 0 to 4 of x^3sqrt(x^2+9) is 1412/5