Chapter 5: Q. 84 (page 465)

Solve given definite integral.
${鈭�}_{鈭� 1}^{1} 鈥� x^{3} \sqrt{9 x^{2} 鈭� 1} d x$

Short Answer

Expert verified

$0$

Step by step solution

Step1. Given Information

The integral is as follows.

${鈭�}_{鈭� 1}^{1} 鈥� x^{3} \sqrt{9 x^{2} 鈭� 1} d x$

The objective is to solve the integral.

Step2. Assumptions

To solve the integral, let $u = x^{2}$

Now, derivation of above equation is solved below.

$u = x^{2} d u = 2 x d x$

Step3. Solution

The integral is solved below.

$\begin{aligned} 鈭� x^{3} \sqrt{9 x^{2} 鈭� 1} d x \\ = 鈭� \frac{u}{2} \sqrt{9 u 鈭� 1} d u \\ = \frac{1}{2} 鈭� u \sqrt{9 u 鈭� 1} d u \end{aligned}$

$= \frac{1}{2} 鈭� \frac{2}{9} v^{2} (\frac{v^{2}}{9} + \frac{1}{9}) d v [v = \sqrt{9 u 鈭� 1}, d v = \frac{9}{2 \sqrt{9 u 鈭� 1}} d x]$

$= \frac{1}{9} 鈭� \frac{v^{2}}{9} (v^{2} + 1) d v$

$= \frac{1}{81} 鈭� (v^{4} + v^{2}) d v$

$= \frac{1}{81} 鈭� v^{4} d v + \frac{1}{81} 鈭� v^{2} d v$

$= \frac{1}{405} (v)^{5} + \frac{1}{243} (v)^{3}$

$= \frac{1}{405} (\sqrt{9 u 鈭� 1})^{5} + \frac{1}{243} (\sqrt{9 u 鈭� 1})^{3}$

$= \frac{1}{405} {(\sqrt{9 x^{2} 鈭� 1})}^{5} + \frac{1}{243} {(\sqrt{9 x^{2} 鈭� 1})}^{3}$

$= (\frac{x^{2}}{45} + \frac{2}{1215}) {(9 x^{2} 鈭� 1)}^{\frac{3}{2}}$

Step4. Solution

The definite integral is solved below.

${鈭�}_{鈭� 1}^{1} 鈥� x^{3} \sqrt{9 x^{2} 鈭� 1} d x$

$= {[(\frac{x^{2}}{45} + \frac{2}{1215}) {(9 x^{2} 鈭� 1)}^{\frac{3}{2}}]}_{鈭� 1}^{1}$

$= [(\frac{1^{2}}{45} + \frac{2}{1215}) {(9 (1)^{2} 鈭� 1)}^{\frac{3}{2}}] 鈭� [(\frac{(鈭� 1)^{2}}{45} + \frac{2}{1215}) {(9 (鈭� 1)^{2} 鈭� 1)}^{\frac{3}{2}}]$

$= 0$

Therefore, the value is 0

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Solve given definite integral.
${鈭�}_{鈭� 1}^{1} 鈥� x^{3} \sqrt{9 x^{2} 鈭� 1} d x$

Short Answer

Step by step solution

Step1. Given Information

Step2. Assumptions

Step3. Solution

Step4. Solution

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Step2. Assumptions

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Solve given definite integral.
${鈭�}_{鈭� 1}^{1} 鈥� x^{3} \sqrt{9 x^{2} 鈭� 1} d x$