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  • Provide solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 6

Provide solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 6

Answers (1)

Answer:

 (a)

Hint:

 Find

A_{n}>A_{n+1}\: or\: A_{n}<A_{n+1}

Given:

 y=(tan\, x)^{n}, x=0, y=0 \: and\: x=\frac{\pi }{4}

Explanation:

\begin{aligned} &A_{n}=\int_{0}^{\frac{\pi }{4}}(tan\, x)^{n}dx \\ &A_{n-2}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, xdx \qquad\qquad\dots(i) \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n}\, xdx \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: tan^{2}dx \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: (sec^{2}x-1)dx \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: sec^{2}x dx-\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, xdx \\ &A_{n}+A_{n-2}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: (sec^{2}x)dx \end{aligned}

Let,

\begin{aligned} &t=tan\: x \\ &dt=sec^{2}xdx\\ &x=0\: \: x=\frac{\pi }{4}\\ &t=0\: \: t=1\\ &A_{n}+A_{n-2}=\int_{0}^{1}t^{n-2}dt\\ &=\left [ \frac{t^{n-1}}{n-1} \right ]_{0}^{1}\\ &=\frac{1}{n-1} \end{aligned}

Posted by

Gurleen Kaur

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