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Sia ? 6 years, 4 months ago
Let {tex}I = \int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}xdx} {/tex}
{tex}= 2\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}xdx} {/tex} ...(i)
{tex}{\because \int\limits_{ - a}^a {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx,} } }{/tex} when f(x) is even function]
{tex}\Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}\left( {\frac{\pi }{2} - x} \right)dx} {/tex}
{tex}\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx = } } } \right]{/tex}
{tex}\Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} {/tex} …(ii)
Adding eq. (i) and (ii),
{tex}2I = 2\int\limits_0^{\frac{\pi }{2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)dx} {/tex}
{tex}= 2\int\limits_0^{\frac{\pi }{2}} {1dx} {/tex}
{tex}= 2\left( x \right)_0^{\frac{\pi }{2}}{/tex}
{tex} = 2.\frac{\pi }{2} = \pi {/tex}
{tex}\Rightarrow I = \frac{\pi }{2}{/tex}
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Tripti Rawat 6 years, 4 months ago
The given curve is y=[x(x-2)]{tex}^2{/tex}
Then, y=[x{tex}^2{/tex}-2x]{tex}^2{/tex}
{tex} \frac { d y } { d x } = 0{/tex}
{tex} \Rightarrow \quad \frac { d } { d x } \left( x ^ { 2 } - 2 x \right) ^ { 2 } = 0{/tex}
{tex} \Rightarrow{/tex}2(x2 - 2x)(2x - 2) = 0
{tex} \Rightarrow{/tex} x = 0, 1, 2
When x = 0, then y = [0 -(-2)]2 = 0
When x = 1, then y = [1 - 2(1)]2 = 1
When x = 2, then y = [22 - 2 {tex} \times{/tex} 2]2 = 0
Hence, the tangent is parallel to X-axis at the points (0, 0), ( 1, 1) and (2, 0).
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Check NCERT solutions here : <a href="https://mycbseguide.com/ncert-solutions.html">https://mycbseguide.com/ncert-solutions.html</a>
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Kumar Vishal Behera 6 years, 4 months ago
1Thank You