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Find the curv...

Question

Find the curve $y = f (x)$ where $f (x) \geq 0, f (0) = 0$ bounding a curvilinear trapezoid with the base [0, x] whose area is proportional to $(n + 1)^{t h}$ power of $f (x)$ . It is known that $f (1) = 1$ .

y = x^{1 / n}

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y = x^{- 1 / n}

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y = x^{- n}

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y = x^{n}

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Solution

The correct option is D $y = x^{1 / n}$
$A = \int_{0}^{x} f (t) d t = λ (f (x))^{n + 1} = λ y^{n + 1}$
$\Rightarrow y = λ (n + 1) y^{n} y^{'}$
$\Rightarrow d x = λ (n + 1) y^{n - 1} d y$
$\Rightarrow x = \frac{λ (n + 1)}{n} y^{n} + C$
$\Rightarrow C = 0, λ = \frac{n}{n + 1} (f (0) = 0, f (1) = 1)$
$\Rightarrow x = y^{n} \Rightarrow y = x^{1 / n}$

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Find the curve y=f(x) where f(x)≥0,f(0)=0 bounding a curvilinear trapezoid with the base [0, x] whose area is proportional to (n+1)th power of f(x). It is known that f(1)=1.

The correct option is D y=x1/nA=∫x0f(t)dt=λ(f(x))n+1=λyn+1⇒y=λ(n+1)yny′⇒dx=λ(n+1)yn−1dy⇒x=λ(n+1)nyn+C⇒C=0,λ=nn+1(f(0)=0,f(1)=1)⇒x=yn⇒y=x1/n

Find the curve $y = f (x)$ where $f (x) \geq 0, f (0) = 0$ bounding a curvilinear trapezoid with the base [0, x] whose area is proportional to $(n + 1)^{t h}$ power of $f (x)$ . It is known that $f (1) = 1$ .