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Please show if this expression is positive, negative or both under different

regions of positive real numbers?

$$

g\left(y\right)=\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

Given,

- What exactly is the difference between a derivative and a total derivative?
- Show that if $f(x)=\int_{0}^x f(t)dt$, then $f=0$
- Let $f(x)=\int_0^1|t-x|t~dt$ for all real $x$. Sketch the graph of $f(x)$, what is the minimum value of $f(x)$
- Proving a function is constant, under certain conditions?
- Do different methods of calculating fractional derivatives have to be equal?
- second derivative at point where there is no first derivative

$$

x,y,Q\geq0;\;F\left(x,y\right)>0;\;F\left(\infty,y\right)=1;\;\frac{\partial F\left(x,y\right)}{\partial y}\leq0;\;\frac{\partial F\left(x,y\right)}{\partial x}\geq0

$$

Please let me know if the question is not well formed or if anything

is not clear.

$$

g\left(y\right)\leq\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2F\left(Q,y\right)\,\left\{ \int_{0}^{Q}dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

g\left(y\right)\leq\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2QF\left(Q,y\right)\,\int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

g\left(y\right)\leq\int_{0}^{Q}\left[2\left(Q-x\right)-2QF\left(Q,y\right)\right]\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

g\left(y\right)\leq\int_{0}^{Q}\left[2\left(Q-x\right)-2QF\left(Q,y\right)\right]\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx<0

$$

If the below condition holds,

$$

\int_{0}^{Q}2\left(Q-x\right)\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx<\int_{0}^{Q}2QF\left(Q,y\right)\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

\int_{0}^{Q}2\left(Q-x\right)\,dx>\int_{0}^{Q}2QF\left(Q,y\right)\,dx

$$

$$

\left|\left(Qx-\frac{x^{2}}{2}\right)\right|_{0}^{Q}>QF\left(Q,y\right)\left|x\right|_{0}^{Q}

$$

$$

\frac{Q^{2}}{2}>Q^{2}F\left(Q,y\right)

$$

$$

F\left(Q,y\right)<\frac{1}{2}

$$

Alternately,

$$

g\left(y\right)\geq\int_{0}^{Q}2Q\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

g\left(y\right)\geq\int_{0}^{Q}\left[2Q-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \right]\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

Am unable to simplify this alternate condition further.

Inference Regarding Definite Integral of Product of Two Functions

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- What is a counterexample to the converse of this corollary related to the Dominated Convergence Theorem?
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- Calculating $\int_0^\infty \frac{\sin(x)}{x} \frac{\sin(x / 3)}{x / 3} \frac{\sin(x / 5)}{x / 5} \cdots \frac{\sin(x / 15)}{x / 15} \ dx$
- How do you integrate $e^{x^2}$?

$$

g\left(y\right)\leq\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2F\left(Q,y\right)\,\left\{ \int_{0}^{Q}dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

g\left(y\right)\leq\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2QF\left(Q,y\right)\,\int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

g\left(y\right)\leq\int_{0}^{Q}\left[2\left(Q-x\right)-2QF\left(Q,y\right)\right]\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

g\left(y\right)\leq\int_{0}^{Q}\left[2\left(Q-x\right)-2QF\left(Q,y\right)\right]\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx<0

$$

If the below condition holds,

$$

\int_{0}^{Q}2\left(Q-x\right)\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx<\int_{0}^{Q}2QF\left(Q,y\right)\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

\int_{0}^{Q}2\left(Q-x\right)\,dx>\int_{0}^{Q}2QF\left(Q,y\right)\,dx

$$

$$

\left|\left(Qx-\frac{x^{2}}{2}\right)\right|_{0}^{Q}>QF\left(Q,y\right)\left|x\right|_{0}^{Q}

$$

$$

\frac{Q^{2}}{2}>Q^{2}F\left(Q,y\right)

$$

$$

F\left(Q,y\right)<\frac{1}{2}

$$

Alternately,

$$

g\left(y\right)\geq\int_{0}^{Q}2Q\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

$$

g\left(y\right)\geq\int_{0}^{Q}\left[2Q-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \right]\frac{\partial F\left(x,y\right)}{\partial y}\,dx

$$

Am unable to simplify this alternate condition further.

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