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Q.
If $\int\limits_1^y x \ln x dx =\frac{1}{4}$, then the value of $y$ is equal to
Integrals
Solution:
Applying integration by parts with $u =\ln x$ and $v =\frac{ x ^2}{2}$, we get
$\int\limits_1^y x \ln x d x=\left.\frac{x^2}{2} \ln x\right|_1 ^y-\frac{1}{2} \int\limits_1^y x d x=\frac{y^2 \ln y}{2}-\frac{y^2}{4}+\frac{1}{4} .$
So, $y ^2 \ln y =\frac{ y ^2}{2}$. Since $y >1$, we obtain $\ln y =\frac{1}{2}$ and thus $y =\sqrt{ e }$.
Note: $y = 0$ is also acceptable