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Definite Integration

Section: Mathematics

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Question : 4 of 54

Marks: +1, -0

\int\limits_{0}^{2} x^8 \left( \frac{4}{x^2} - 1 \right)^{5/2} dx =

[AP EAPCET 24 May 2025 S1]

$\frac{2^{15}}{63}$
$\frac{2^{16}}{315}$
$\frac{2^{16}}{189}$
$\frac{2^{10}}{63}$

Solution:

I = \int\limits_{0}^{2} x^8 \left( \frac{4}{x^2} - 1 \right)^{5/2} dx

= \int\limits_{0}^{2} x^8 \left( \frac{4-x^2}{x^2} \right)^{5/2} dx

= \int\limits_{0}^{2} x^8 \cdot \frac{(4-x^2)^{5/2}}{x^5} dx

= \int\limits_{0}^{2} x^3 (4-x^2)^{5/2} dx

Let

u = 4 - x^2

Then,

du = -2x \, dx

, so

x \, dx = -\frac{1}{2} du \quad x^2 = 4 - u

and

x^3 = x \cdot x^2 = x(4-u)

when

x = 0, \quad u = 4

and when

x = 2, \quad u = 0

\therefore \int\limits_{0}^{2} x^8 \left( \frac{4}{x^2} - 1 \right)^{5/2} dx

= \int\limits_{0}^{2} x^3 (4-x^2)^{5/2} dx

= \int\limits_{4}^{0(4-u)} \cdot u^{5/2} \cdot \left( -\frac{1}{2} \right) du

= \frac{1}{2} \left[ 4 \cdot \frac{u^{7/2}}{7/2} - \frac{u^{9/2}}{9/2} \right]_{0}^{4}

= \frac{1}{2} \left[ \frac{8}{7} u^{7/2} - \frac{2}{9} u^{9/2} \right]_{0}^{4}

= \frac{1}{2} \left[ \frac{8}{7} (u)^{7/2} - \frac{2}{9} (4)^{9/2} - 0 \right]

= \frac{1}{2} \left[ \frac{8}{7} (128) - \frac{2}{9} (512) \right]

= \frac{1}{2} \left[ \frac{1024}{7} - \frac{1024}{9} \right]

= \frac{1024}{2} \cdot \frac{2}{63} = \frac{1024}{63}

\therefore \int\limits_{0}^{2} x^8 \left( \frac{4}{x^2} - 1 \right)^{5/2} dx = \frac{1024}{63} = \frac{2^{10}}{63}

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