Find Upper Bound of Integral Here's a problem a friend sent me:
$$\int_1^y x\log x\,dx = \frac14$$
Any fun solutions I can give them?
 A: We have
$$\log(x)=\int_1^x\frac1t\ dt$$
Thus,
$$\begin{align}\int_1^yx\log(x)\ dx&=\int_1^yx\int_1^x\frac1t\ dt\ dx\\\{1<x<y,1<t<x\}&=\int_1^y\int_1^x\frac xt\ dt\ dx\\\{1<t<y,t<x<y\}&=\int_1^y\int_t^y\frac xt\ dx\ dt\\&=\int_1^y\frac{x^2}{2t}\bigg|_{x=t}^{x=y}\ dt\\&=\frac12\int_1^y\frac{y^2}t-t\ dt\\&=\frac12\left[y^2\ln(t)-\frac12t^2\right]_{t=1}^y\\&=\frac12y^2\ln(y)-\frac14y^2+\frac14\end{align}$$
where $\{\}$ represents the coordinates of the integration variables.
It's then trivial to see that as $y\to0^+$, we have
$$\int_1^{0+}x\log(x)\ dx=\lim_{y\to0^+}\frac12y^2\ln(y)-\frac14y^2+\frac14=\frac14$$
The second solution occurs around 3.162, which you may prove its existence through intermediate value theorem.

We can attempt to algebraically solve this of course:
$$\frac14=\frac12y^2\ln(y)-\frac14y^2+\frac14$$
$$0=\frac12y^2\ln(y)-\frac14y^2=\frac14y^2(2\ln(y)-1)$$
Thus, either $y=0$ or
$$0=2\ln(y)-1\\2\ln(y)=1\\\ln(y)=\frac12\\y=e^{1/2}$$
which are your two solutions.
A: You may want to try integrating the left-hand side by parts; that is, applying the rule
$$ \int_a^b u(x)v(x) \mbox{d}x = U(x)v(x)\big|_a^b - \int_a^b U(x)\frac{\mbox{d}v}{\mbox{d}x}\mbox{d}x,$$
where $U(x)$ is an antiderivative of $u(x)$. 
In this case, letting $u = x$ and $v = \log x,$ you will get a much easier integral on the left-hand side, and thus the expression for $y$ you need to solve this.
A: An easy way to figure out the indefinite integral $\int x\log x \,dx$ is to start by ignoring the fact that  $\log x$ is not constant, and then adding a term to correct for that lie afterward.
Thus we try 
$$
\int x\log x \,dx = \frac{x^2}{2} \log x ? \\
\frac{d}{dx}\left( \frac{x^2}{2} \log x \right) = x\log x + \color{red}{\frac12 x} \\
\int x\log x \,dx = \frac{x^2}{2} \log x -\color{red}{\frac14 x^2}
$$
(This, by the way, is just what you are doing when you integrate by parts.  The technique works precisely when integration by parts works.)
Then $$\frac14 = \int_1^y x\log x \,dx = \frac{y^2}{2} \log y -\frac14 y^2 -0
+\frac14 1^2
\\\frac{y^2}{2} \log y = \frac14 y^2 \mbox{    (or  }y=0)\\
\log y = \frac12 \\ y = \sqrt{e}
$$

But note that another solution is $y=0$.

A: HINT
Try to compute the integral by parts, differentiating the log and integrating $x$, then evaluate to an arbitrary $y$ and solve the resulting equation.
