JEE Advance - Mathematics (2021 - Paper 2 Online - No. 11)
Let $${g_i}:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R,i = 1,2$$, and $$f:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R$$ be functions such that $${g_1}(x) = 1,{g_2}(x) = |4x - \pi |$$ and $$f(x) = {\sin ^2}x$$, for all $$x \in \left[ {{\pi \over 8},{{3\pi } \over 8}} \right]$$. Define $${S_i} = \int\limits_{{\pi \over 8}}^{{{3\pi } \over 8}} {f(x).{g_i}(x)dx} $$, i = 1, 2
The value of $${{16{S_1}} \over \pi }$$ is _____________.
The value of $${{16{S_1}} \over \pi }$$ is _____________.
Answer
2.00
Explanation
$${S_1} = \int_{\pi /8}^{3\pi /8} {{{\sin }^2}x.1\,dx = \int_{\pi /8}^{3\pi /8} {\left( {{{1 - \cos 2x} \over 2}} \right)dx} } $$
$$ = \left( {{x \over 2} - {{\sin 2x} \over 4}} \right)_{\pi /8}^{3\pi /8} = {\pi \over 8}$$
$$\therefore$$ $${{16{S_1}} \over \pi } = {{16} \over \pi } \times {\pi \over 8} = 2$$
$$ = \left( {{x \over 2} - {{\sin 2x} \over 4}} \right)_{\pi /8}^{3\pi /8} = {\pi \over 8}$$
$$\therefore$$ $${{16{S_1}} \over \pi } = {{16} \over \pi } \times {\pi \over 8} = 2$$
Comments (0)
