Find the curve y = f(x) where f(x)≥0, f(0)= 0, bounding a curfilinear 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.
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1 Answers
62so what you are saying is that \int_{0}^{x}{f(t)dt}=kf^{n+1}(x)
differentiate with respect to x, you will get
\\f(x)=k(n+1)f^n(x)f'(x) \\\Rightarrow f^{n-1}(x)f'(x)=1/k(n+1) \\\Rightarrow \int nf^{n-1}(x)f'(x)dx=\int n/k(n+1)dx \\\Rightarrow f^{n}(x)=1/k(n+1)x+c \\\Rightarrow f(x)=(1/k(n+1)x+c)^{1/n}
f(0)=0 hence c=0
and f(1) = 1
1= 1/k(n+1)
k=1/(n+1)
hence the function is f(x) = x1/n
