Area via an improper integral

Text

Compute $\displaystyle\int_{0}^{1}\left(x^{2}-\ln x\right)dx$ (the integral is improper at $x=0$, where $\ln x\to-\infty$).

Solution

Split the integral. For the polynomial part $\displaystyle\int_{0}^{1}x^{2}\,dx=\frac{1}{3}$. For the logarithm, an antiderivative is $x\ln x-x$, and $\int_{0}^{1}\ln x\,dx=\big[x\ln x-x\big]_{0}^{1}=(0-1)-\lim_{x\to0^{+}}\big(x\ln x-x\big)=-1.$ Hence $\int_{0}^{1}\left(x^{2}-\ln x\right)dx=\frac{1}{3}+1=\frac{4}{3}\approx1.3333.$ $\boxed{\dfrac{4}{3}\approx1.3333}$