ₙ→∞} n^4 ≤ft[ 1/n^5 + {1}{(n²+1)^{5/2}} + {1}{(n²+4)^{5/2}} + {1}{(n²+9)^{5/2}} + + {1}{4{2}n^5} ] = [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : 562{5}{6{2}}625

Correct Answer is : 562{5}{6{2}}625

Test Index

Definite Integration

Section: Mathematics

Question : 54 of 54

Marks: +1, -0

\lim\limits_{n\to\infty} n^4 \left[ \frac{1}{n^5} + \frac{1}{(n^2+1)^{\frac{5}{2}}} + \frac{1}{(n^2+4)^{\frac{5}{2}}} + \frac{1}{(n^2+9)^{\frac{5}{2}}} + \dots + \frac{1}{4\sqrt{2}n^5} \right] =

Solution:

\;\lim\limits_{n\to\infty} x^4 \left[ \begin{array}{c} \frac{1}{x^5} + \frac{1}{(x^2+1)^{\frac{5}{2}}} + \frac{1}{(x^2+4)^{\frac{5}{2}}} \\ + \frac{1}{(x^2+9)^{\frac{5}{2}}} + \dots + \frac{1}{4\sqrt{2}x^5} \end{array} \right]

\;\lim\limits_{n\to\infty} \sum\limits_{r=1}^{n} \frac{x^4}{(x^2+r^2)^{\frac{5}{2}}} \left[ \because \lim\limits_{n\to\infty} \frac{x^4}{x^5}=0 \right]

\;\lim\limits_{n\to\infty} \sum\limits_{r=1}^{n} \frac{1}{n} \left[ \frac{1}{\left(1+\left(\frac{r}{n}\right)^2\right)^{\frac{5}{2}}} \right]

=\int\limits_{0}^{1} \frac{1}{(1+x^2)^{\frac{5}{2}}} dx

\;\;\;\text{ Let } x = \tan \theta

dx = \sec^2 \theta d\theta

\;x \to 0, \theta \to 0 \;\text{ and }\;

\;x \to 1, \theta \to \frac{\pi}{4}

=\int\limits_{0}^{\frac{\pi}{4}} \frac{1}{(1+\tan^2 \theta)^{\frac{5}{2}}} \times \sec^2 \theta d\theta

=\int\limits_{0}^{\frac{\pi}{4}} \frac{1}{\sec^3 \theta} d\theta

=\int\limits_{0}^{\frac{\pi}{4}} \cos^3 \theta d\theta

=\frac{1}{4} \int\limits_{0}^{\frac{\pi}{4}} (\cos 3\theta + 3\cos \theta) d\theta

=\frac{1}{4} \left[ \frac{\sin 3\theta}{3} + 3 \sin \theta \right]_{0}^{\frac{\pi}{4}}

=\frac{1}{4} \left[ \frac{1}{3} \sin \left(\pi - \frac{\pi}{4}\right) + 3 \sin \frac{\pi}{4} \right]

=\frac{1}{4} \left[ \frac{1}{3} \times \frac{1}{\sqrt{2}} + \frac{3}{\sqrt{2}} \right]

=\frac{1+9}{4 \times 3 \sqrt{2}} \Rightarrow \frac{10}{4 \times 3 \sqrt{2}} = \frac{5}{6 \sqrt{2}}

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