Practice: Techniques of Integration

(a) ${\int }_1^{2}x\ln xdx$ by parts. Take $u=\ln x$ and $dv=xdx$. The reason is that $\ln x$ simplifies when differentiated, becoming $1/x$, while $x$ is trivial to integrate. The reverse choice would leave a logarithm to integrate and a higher power of $x$, which is worse. So $du={\displaystyle \frac{dx}{x}}$ and $v={\displaystyle \frac{x^2}{2}}$:

The $1/x$ cancelled against $x^2$, which is the whole point of the choice. Finishing:

Check: differentiating $\frac{x^2}{2}\ln x-\frac{x^2}{4}$ gives $x\ln x+\frac{x}{2}-\frac{x}{2}=x\ln x$ ✓. A midpoint sum returns $0.6362943611$ ✓. (b) ${\int }_3^5{\displaystyle \frac{dx}{(x-1)(x+2)}}$ by partial fractions. The integrand is proper, numerator degree 0, denominator degree 2, so decompose directly:

Substituting the roots isolates each coefficient: at $x=1$, $1=3A$ so $A=\frac{1}{3}$; at $x=-2$, $1=-3B$ so $B=-\frac{1}{3}$. Verification at $x=3$: the original is ${\displaystyle \frac{1}{2\cdot 5}}=0.1$; the decomposition gives ${\displaystyle \frac{1/3}{2}}-{\displaystyle \frac{1/3}{5}}=\frac{1}{6}-\frac{1}{15}=0.1$ ✓. Integrating term by term:

Check: a midpoint sum returns $0.1188916480$ ✓. (c) ${\int }_0^1{\displaystyle \frac{dx}{\sqrt{x}}}$ — improper at the left endpoint. The integrand is unbounded as $x\to 0^{+}$, so the integral is improper and is defined as a limit:

It converges, to 2. Check: the partial values are $1.3675$ at $s=0.1$, $1.9368$ at $s=0.001$, $1.9980$ at $s={10}^{-6}$ ✓. By the p-test near zero: ${\int }_0^1{x}^{-p}dx$ converges when , and here $p=\frac{1}{2}$ ✓. the region is unbounded in height and has finite area. Unboundedness alone never decides convergence. (d) ${\int }_1^{\mathrm{\infty }}{\displaystyle \frac{dx}{x^3}}$ — improper at infinity.

It converges, to $\frac{1}{2}$. The threshold. ${\int }_1^{\mathrm{\infty }}{x}^{-p}dx$ converges exactly when , with value ${\displaystyle \frac{1}{p-1}}$; here $p=3$ gives $\frac{1}{2}$ ✓. The case $p=1$ diverges, since the antiderivative $\ln t$ grows without bound. The two thresholds run opposite ways, at infinity, at zero, because at infinity the integrand must decay fast enough, while near zero it must blow up slowly enough. $p=1$ fails at both ends. (e) The invalid evaluation of ${\int }_{-1}^1{x}^{-2}dx$. Two independent reasons the answer cannot be right. 1. The sign is impossible. everywhere it is defined, so every Riemann term is positive and any value must be positive. The reported $-2$ contradicts the construction before any theorem is consulted.
2. The integral diverges. Near 0 the integrand is unbounded, and the accumulated area grows without limit, so no finite number can be the answer. The hypothesis violated. Part 2 of the fundamental theorem requires $F$ to be an antiderivative of $f$ on the whole closed interval. Neither $f(x)=x^{-2}$ nor $F(x)=-1/x$ is defined at $x=0\in [-1,1]$, so $F$ is not an antiderivative there and the theorem does not apply. The arithmetic proceeded regardless, which is what produced a number. The correct evaluation. The integrand is unbounded at an interior point, so split there and treat each piece as a limit:

For the right piece,

since $\frac{1}{s}-1$ reads $9$ at $s=0.1$, $99$ at $s=0.01$ and $9999$ at $s={10}^{-4}$. That piece diverges, and by symmetry so does the left one. Since one piece diverges, the whole integral diverges. The divergences do not cancel: convergence requires each piece to converge on its own. Pairing the two symmetrically to obtain 0 computes the Cauchy principal value, which is a different quantity and is not the integral. By the p-test: near zero, ${\int }_0^1{x}^{-p}dx$ needs , and here $p=2$ ✓ diverges. The same conclusion without computing anything.