"Evaluate the integrals: ∫x349−x2dx" was answered by our experts

Rui Baldwin

Answered question

2021-10-12

Evaluate the integrals:

\int x^{3} \sqrt{49 - x^{2}} d x

Answer & Explanation

SabadisO

Skilled2021-10-13Added 108 answers

Given:
$\int x^{3} \sqrt{49 - x^{2}} d x$
To evaluate:
The value of the integral.
Consider,
$I = \int x^{3} \sqrt{49 - x^{2}} d x$
$\Rightarrow d u = \frac{1}{2 \sqrt{49 - x^{2}}} \cdot \frac{d}{d x} (- x^{2})$
$d u = \frac{1}{2 \sqrt{49 - x^{2}}} \cdot (- 2 x) d x$
$d u = - \frac{x}{\sqrt{49 - x^{2}}} d x$
$\Rightarrow x d x = - \sqrt{49 - x^{2}} d u$
$x d x = - u d u$
From equation (2),
$u^{2} = 49 - x^{2}$
$x^{2} = 49 - u^{2}$
Equation (1) becomes,
$I = \int x^{3} (- u) d u$
$= \int x^{2} \cdot x (- u) d u$
$= \int (49 - u^{2}) \cdot (- u^{2}) d u$
$= \int (- 49 u^{2}) d u + \int u^{4} d u$
$= - 49 \int u^{2} d u + \int u^{4} d u$
$= - 49 [\frac{u^{3}}{3}] + [\frac{u^{5}}{5}]$
Back substitute:
$u = \sqrt{49 - x^{2}}$
$\Rightarrow I = - 49 [\frac{{(\sqrt{49 - x^{2}})}^{3}}{3}] + [\frac{{(\sqrt{49 - x^{2}})}^{5}}{5}]$
$= - \frac{49}{3} {[{(49 - x^{2})}^{\frac{1}{2}}]}^{3} + \frac{1}{5} {[{(49 - x^{2})}^{\frac{1}{2}}]}^{5}$

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