Grade 10Integral Calculus

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Aditya Kartikeya

11 Years agoGrade 10

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello Student,
Please find answer to your question below

$L = \lim_{n\rightarrow \infty }\frac{\pi }{6n}[sec^{2}(\frac{\pi }{6n})+sec^{2}(\frac{2\pi }{6n})+....................+sec^{2}(\frac{n\pi }{6n})]$
$L = \lim_{n\rightarrow \infty }\frac{\pi }{6n}[\sum_{r = 1}^{n}sec^{2}(\frac{r\pi }{6n})]$
Apply the definition of definite integral we have
$L = \int_{0}^{1}\frac{\pi }{6}sec^{2}(\frac{\pi x}{6})dx$
$tan\frac{\pi x}{6} = t$
$x = 0\rightarrow t = 0$
$x = 1\rightarrow t = \frac{1}{\sqrt{3}}$
$\frac{\pi }{6}sec^{2}\frac{\pi x}{6}dx = dt$
$L = \int_{0}^{1/\sqrt{3}}dt$
$L =(t)_{0}^{1/\sqrt{3}}$
$L = \frac{1}{\sqrt{3}}$