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31/01/2022 1:58 pm
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Let f be a non–negative function in [0, 1] and twice differentiable in (0, 1). If \(\int_0^x \sqrt{(1-(f'(t))^2}\)dt = \(\int_0^x f(t)\), 0 ≤ x ≤ 1 and f(0), then \(\lim\limits_{x \to 0}\frac{1}{2} \int_0^x f(t) dt\):
(1) equals 0
(2) equals 1
(3) does not exist
(4) equals \(\frac{1}{2}\)
1 Answer
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31/01/2022 2:19 pm
Correct answer: (4) equals \(\frac{1}{2}\)
Explanation:
\(\int_0^x \sqrt{(1-(f'(t))^2}\)dt = \(\int_0^x f(t)\), 0 ≤ x ≤ 1
differentiating both the sides
\(\sqrt{1 - (f'(x))^2}\) = f(x)
⇒ 1 - (f'(x))2 = f2(x)
\(\frac{f'(x)}{\sqrt{1-f^2(x)}}\) = 1
sin-1 f(x) = x + C
∵ f(0) = 0 ⇒ C = 0 ⇒ f(x) = sin x
Now \(\lim\limits_{x \to 0} \frac{\int_0^x sin \ t \ dt}{x^2}\) \(\Big(\frac{0}{0}\Big)\)
= \(\frac{1}{2}\)