If int_{sin x}^1 {{t^2}f(t);dt = 1 - sin x} , | vedclass

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(A) Given the equation: $\int_{\sin x}^1 {t^2}f(t) dt = 1 - \sin x$. Applying the Leibniz integral rule to differentiate both sides with respect to $x

Vedclass · Accepted Answer

$3$

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(A) Given the equation: $\int_{\sin x}^1 {t^2}f(t) dt = 1 - \sin x$. Applying the Leibniz integral rule to differentiate both sides with respect to $x$: $\frac{d}{dx} \left( \int_{\sin x}^1 {t^2}f(t) dt ight) = \frac{d}{dx} (1 - \sin x)$. Using the Fundamental Theorem of Calculus,the derivative of $\int_{g(x)}^a h(t) dt$ is $-h(g(x)) \cdot g'(x)$. So,$-(\sin x)^2 f(\sin x) \cdot \cos x = -\cos x$. Since $x \in (0, \frac{\pi}{2})$,$\cos x 
eq 0$,we can divide both sides by $-\cos x$: $(\sin x)^2 f(\sin x) = 1$. Thus,$f(\sin x) = \frac{1}{\sin^2 x}$. Replacing $\sin x$ with $t$,we get $f(t) = \frac{1}{t^2}$. Therefore,$f\left( \frac{1}{\sqrt{3}} ight) = \frac{1}{(1/\sqrt{3})^2} = \frac{1}{1/3} = 3$.

If $\int_{\sin x}^1 {{t^2}f(t)\;dt = 1 - \sin x} $,$x \in \left( {0,\frac{\pi }{2}} \right)$,then $f\left( {\frac{1}{{\sqrt 3 }}} \right)$ is equal to:

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