Not that Snake again!

Calculus Level 5

f ( x ) = x + 0 1 ( x y 2 + x 2 y ) f ( y ) d y \large f(x) = x + \int_0^1 (xy^2 + x^2 y) f(y) \, dy

Let x x and y y be real and independent variables. Find f ( 1 ) f(1) to 3 decimal places.


See its sister problem: Not that Snake!


The answer is 2.185.

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1 solution

Rachit Shukla
Jul 11, 2016

f ( x ) = x + x 0 1 y 2 f ( y ) d y + x 2 0 1 y f ( y ) d y = x + x c 1 + x 2 c 2 \displaystyle f(x)=x+x\int_{0}^{1}y^2f(y)dy+x^2\int_{0}^{1}yf(y)dy= x+xc_{1}+x^2c_{2}

where, c 1 = 0 1 y 2 f ( y ) d y = 0 1 y 2 ( y + y c 1 + y 2 c 2 ) d y = 1 4 + c 1 4 + c 2 5 \displaystyle c_{1}=\int_{0}^{1}y^2f(y)dy=\int_{0}^{1}y^2(y+yc_{1}+y^2c_{2})dy=\frac{1}{4}+\frac{c_{1}}{4}+\frac{c_{2}}{5}

and, c 2 = 0 1 y f ( y ) d y = 0 1 y ( y + y c 1 + y 2 c 2 ) d y = 1 3 + c 1 3 + c 2 4 \displaystyle c_{2}=\int_{0}^{1}yf(y)dy=\int_{0}^{1}y(y+yc_{1}+y^2c_{2})dy=\frac{1}{3}+\frac{c_{1}}{3}+\frac{c_{2}}{4}

Solve to get: c 1 = 61 119 c_{1}=\frac{61}{119} and c 2 = 80 119 c_{2}=\frac{80}{119}

f ( 1 ) = 1 + 61 119 + 80 119 2.185 \Rightarrow f(1)=1+\frac{61}{119}+\frac{80}{119} \approx 2.185

Excellent problem and an even better solution

space sizzlers - 4 years, 3 months ago

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