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anemone
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Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.
I presume you know that to determine maximum or minimum values for a differentiable function, you set the derivative equal to 0. You should also know that if [tex]f(y)= \int_a^y g(x,y) dx[/tex] then [tex]df/dy= g(y,y)[/tex]. So you should find y such that [tex]\sqrt{y^4+ (y- y^2)^2}= 0[/tex]. (If that y is not between 0 and 1 then look at the value at 0 and 1.)anemone said:Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.
HallsofIvy said:I presume you know that to determine maximum or minimum values for a differentiable function, you set the derivative equal to 0. You should also know that if [tex]f(y)= \int_a^y g(x,y) dx[/tex] then [tex]df/dy= g(y,y)[/tex]. So you should find y such that [tex]\sqrt{y^4+ (y- y^2)^2}= 0[/tex]. (If that y is not between 0 and 1 then look at the value at 0 and 1.)
Random Variable said:If $ \displaystyle f(y)= \int_a^y g(x,y) \ dx$, then $ \displaystyle \frac{df}{dy}= \int_{a}^{y} g_{y}(x,y) \ dx + g(y,y)$.
Random Variable said:If $ \displaystyle f(y)= \int_a^y g(x,y) \ dx$, then $ \displaystyle \frac{df}{dy}= \int_{a}^{y} g_{y}(x,y) \ dx + g(y,y)$.
ZaidAlyafey said:That integral doesn't seem to be solvable in terms of elementary functions.
anemone said:Hmm...that's not quite right, Zaid!
ZaidAlyafey said:Well, I don't know whether I am messing something but I still cannot think how to find an elementary anti-derivative for that integral.
[sp]Let \(\displaystyle f(y) = \int_0^y \sqrt{x^4+(y-y^2)^2} dx.\) My instinct is that the maximum value of $f$ over the interval $[0,1]$ must be \(\displaystyle f(1) = \int_0^1 x^2\, dx = 1/3.\) Following anemone's hint about using an inequality, it occurred to me that if $a,b \geqslant 0$ then $\sqrt{a^2+b^2} \leqslant a+b.$ Applying that with $a=x^2$ and $b= y-y^2$, you see that \(\displaystyle f(y) \leqslant \int_0^y(x^2 + y - y^2)\,dx = \tfrac13y^3 + y^2 - y^3 = y^2 - \tfrac23y^3.\) But $y^2 - \tfrac23y^3$ is an increasing function on the interval $[0,1]$, with a maximum value $1/3$ when $y=1$. Therefore the maximum value of $f(y)$ is $f(1) = 1/3.$[/sp]anemone said:Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.
Opalg said:[sp]Let \(\displaystyle f(y) = \int_0^y \sqrt{x^4+(y-y^2)^2} dx.\) My instinct is that the maximum value of $f$ over the interval $[0,1]$ must be \(\displaystyle f(1) = \int_0^1 x^2\, dx = 1/3.\) Following anemone's hint about using an inequality, it occurred to me that if $a,b \geqslant 0$ then $\sqrt{a^2+b^2} \leqslant a+b.$ Applying that with $a=x^2$ and $b= y-y^2$, you see that \(\displaystyle f(y) \leqslant \int_0^y(x^2 + y - y^2)\,dx = \tfrac13y^3 + y^2 - y^3 = y^2 - \tfrac23y^3.\) But $y^2 - \tfrac23y^3$ is an increasing function on the interval $[0,1]$, with a maximum value $1/3$ when $y=1$. Therefore the maximum value of $f(y)$ is $f(1) = 1/3.$[/sp]
The exact value of the given integral can be solved by using the substitution method or by using integration by parts. First, substitute the given limits of integration into the given integral. Then, use the appropriate integration techniques to simplify the integral and solve for the exact value.
The given integral is significant in mathematics because it represents the area under the curve of a function. It is also used to calculate the volume of a solid of revolution and to solve many real-world problems in physics, engineering, and economics.
Yes, the given integral can also be solved numerically using methods such as Simpson's rule or the trapezoidal rule. These methods approximate the exact value of the integral by dividing the region under the curve into smaller subintervals and calculating the area of each subinterval.
To solve this type of integral, it is helpful to first identify the type of function involved. In this case, the given integral is a square root function, which can be simplified by using algebraic techniques. Additionally, using trigonometric substitutions can also be helpful in solving this type of integral.
The exact value of the given integral can be verified by using online integration calculators or by checking the solution with the help of a computer software program such as MATLAB or Wolfram Alpha. Additionally, the solution can also be verified by hand by using differentiation to check if the result is equal to the original function.