Practice: The Definite Integral

(a) Riemann sums at $n=4$. Width $\mathrm{\Delta }x=\frac{1}{4}$. Left endpoints $0,0.25,0.5,0.75$; right endpoints $0.25,0.5,0.75,1$.

Because $x^2$ is increasing on $[0,1]$, the left endpoint is the minimum of $f$ on each subinterval and the right endpoint its maximum. So $L_4$ underestimates and $R_4$ overestimates, and

The true value $\frac{1}{3}\approx 0.3333$ lies between them ✓. The gap. $R_n-L_n=0.25$ here. In general the two sums share every rectangle except the first and last, so

which for this integral is $\frac{1-0}{n}=\frac{1}{n}$. At $n=4$ that is $0.25$ ✓, and it tends to 0, so the two sums converge to a common value, which is what makes the limit exist. (b) ${\int }_1^4(x-6x^{-2})dx$. Antiderivative, term by term: $\int xdx=\frac{x^2}{2}$ and $\int -6x^{-2}dx=-6\cdot \frac{x^{-1}}{-1}=\frac{6}{x}$. So

Check: $F^{\prime }(x)=x-\frac{6}{x^2}$ ✓.

The integrand is continuous on $[1,4]$, the singularity at $x=0$ lies outside, so Part 2 applies. (c) Signed value against total area. With $F(x)=\frac{x^3}{3}-2x^2$: $F(3)=9-18=-9$, $F(0)=0$, so

Total area. The integrand $x(x-4)$ is negative throughout $(0,4)$, hence on all of $(0,3)$, for instance $f(1)=-3$, $f(2)=-4$, $f(3)=-3$. There is no sign change to split at, so the region lies entirely below the axis and

Why they differ. Each Riemann term is $f(x_i^{\ast })\mathrm{\Delta }x$ with , so its sign is the sign of $f$. Nothing in the construction takes an absolute value. The integral therefore reports area above the axis minus area below; the two agree only when $f\ge 0$ throughout. Here every contribution is negative, so the integral is $-9$ while the geometric area is $9$. (d) ${\int }_1^{3}x\sqrt{x^2-1}dx$ by substitution. Let $u=x^2-1$, so $du=2xdx$ and $xdx=\frac{1}{2}du$. The factor $x$ is present, which is what makes the substitution work. Convert the limits: $x=1\Rightarrow u=0$; $x=3\Rightarrow u=8$.

Since $8^{3/2}=(2^3{)}^{3/2}=2^{9/2}=16\sqrt{2}$,

Check: a midpoint Riemann sum with $\mathrm{400,000}$ subintervals gives $7.54247$ ✓. Note the lower limit became $u=0$, not $u=1$: converting is not optional, and here it happens to make the lower evaluation vanish. (e) The invalid evaluation. The computation is mechanically consistent but the theorem does not apply. Two independent reasons the answer cannot be right. 1. The sign is impossible. everywhere it is defined, so every Riemann sum is a sum of positive terms and any value must be positive. A negative answer contradicts the construction before any theorem is consulted.
2. The integral diverges. Near $x=0$ the integrand is unbounded. Splitting at the singularity, ${\int }_0^2{x}^{-2}dx=\underset{t\to 0^{+}}{lim}{[-\frac{1}{x}]}_t^2=\underset{t\to 0^{+}}{lim}(-\frac{1}{2}+\frac{1}{t})=+\mathrm{\infty }$. The integral has no finite value, so no finite answer can be correct. The hypothesis violated. Part 2 requires $F$ to be an antiderivative of $f$ on the whole closed interval $[a,b]$, that is, $F^{\prime }=f$ at every point of it. Here neither $f(x)=x^{-2}$ nor $F(x)=-1/x$ is even defined at $x=0\in [-2,2]$, so $F$ is not an antiderivative on that interval. The proof of Part 2 applies the mean value theorem across the interval, which requires differentiability throughout; that step fails. How to catch it in advance. Before applying Part 2, check the integrand for singularities inside the limits. Where one exists the integral is improper and must be split and evaluated as a limit, which here shows the divergence rather than concealing it behind a finite number.