Corrected exercises: areas, Riemann sums and the definite integral (MATH 141)

a) Split the sum and pull out the constant: $\sum_{i=1}^{20} (3i - 2) = 3\sum_{i=1}^{20} i - \sum_{i=1}^{20} 2 = 3 \cdot \frac{20 \cdot 21}{2} - 2 \cdot 20 = 630 - 40 = 590$. The trap is the last term: $\sum_{i=1}^{20} 2$ is twenty copies of $2$, that is $40$, not $2$. The constant does not depend on $i$, but it is still added once per term. Writing $630 - 2 = 628$ is the most common slip on this line, and it reappears in every Riemann sum where $f$ has a constant term.

b) The formula needs a sum starting at $1$, so complete it and remove what was added: $\sum_{i=4}^{10} i^2 = \sum_{i=1}^{10} i^2 - \sum_{i=1}^{3} i^2 = \frac{10 \cdot 11 \cdot 21}{6} - (1 + 4 + 9) = 385 - 14 = 371$. The subtracted sum stops at $3$, the index just BEFORE the first term kept. Subtracting $\sum_{i=1}^{4} i^2 = 30$ removes the term $16$ that belongs to the sum and gives $355$. Check by counting terms: from $4$ to $10$ there are $10 - 4 + 1 = 7$ terms, $16 + 25 + 36 + 49 + 64 + 81 + 100 = 371$.

c) First route, expand: $(i + 1)^2 = i^2 + 2i + 1$, so $\sum_{i=1}^{n} (i + 1)^2 = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} + n$. Over the common denominator $6$: $\frac{(2n^3 + 3n^2 + n) + (6n^2 + 6n) + 6n}{6} = \frac{2n^3 + 9n^2 + 13n}{6}$. Second route, rename: with $j = i + 1$, $j$ runs from $2$ to $n + 1$, so the sum is $\sum_{j=2}^{n+1} j^2 = \sum_{j=1}^{n+1} j^2 - 1 = \frac{(n+1)(n+2)(2n+3)}{6} - 1$. The formula is applied with $n + 1$ in place of $n$, in all three factors. For $n = 2$: $\frac{16 + 36 + 26}{6} = 13$ and $\frac{3 \cdot 4 \cdot 7}{6} - 1 = 14 - 1 = 13$, which is $2^2 + 3^2$. The trap to avoid: $\sum (i + 1)^2 \neq \sum i^2 + 1$; the square must be expanded before the sum is split.

d) Divide the closed form by $n^3$: $\frac{2n^3 + 9n^2 + 13n}{6n^3} = \frac{1}{3} + \frac{3}{2n} + \frac{13}{6n^2} \to \frac{1}{3}$. The shift by $1$ changed the lower terms but not the leading one, $\frac{2n^3}{6}$, and only the leading term survives division by $n^3$. Next, $\frac{1}{n^4}\left[\frac{n(n+1)}{2}\right]^2 = \frac{1}{4}\left(\frac{n+1}{n}\right)^2 \to \frac{1}{4}$. A frequent slip is to keep $\frac{n \cdot n \cdot n}{6}$ as the leading part of $\sum i^2$ and announce $\frac{1}{6}$: the factor $2n + 1$ contributes $2n$, not $n$.

e) The shaded staircase has columns of heights $1, 2, \ldots, n$, so its area is $\sum_{i=1}^{n} i$. The turned-over copy fills exactly the rest of a rectangle $n$ wide and $n + 1$ high, so $2\sum_{i=1}^{n} i = n(n + 1)$, which is the formula. Now $\frac{n(n+1)}{2} = \frac{n^2}{2} + \frac{n}{2}$: the staircase is the triangle plus $\frac{n}{2}$, one half-square of overshoot on each of the $n$ steps. Relative to the size of the picture, $\frac{n/2}{n^2} = \frac{1}{2n} \to 0$. This is the whole idea of the chapter in one figure: rectangles overshoot a slanted boundary, and the error, shared among more and thinner rectangles, becomes negligible. Rescaled to the unit square, it says $\frac{1}{n^2}\sum_{i=1}^{n} i \to \frac{1}{2}$, the area of the triangle under $y = x$ on $[0, 1]$.

a) $\Delta x = \frac{b - a}{n} = \frac{4 - 0}{4} = 1$, and the division points are $x_i = 0 + i\Delta x = i$, for $i = 0, 1, 2, 3, 4$. The left sum uses the left end of each strip, $x_0, \ldots, x_3$, that is $0, 1, 2, 3$. The right sum uses $x_1, \ldots, x_4$, that is $1, 2, 3, 4$. The midpoint sum uses the centres $\bar{x}_i = \frac{x_{i-1} + x_i}{2}$, that is $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{7}{2}$. Each sum has exactly $n = 4$ terms: listing five sample points, $0$ to $4$, for $L_4$ is the error that adds a fifth rectangle standing outside the interval.

b) The values needed are $f(0) = 16$, $f(1) = 15$, $f(2) = 12$, $f(3) = 7$, $f(4) = 0$. So $L_4 = 1 \cdot (16 + 15 + 12 + 7) = 50$ and $R_4 = 1 \cdot (15 + 12 + 7 + 0) = 34$. For the midpoints, $f(\frac{1}{2}) = \frac{63}{4}$, $f(\frac{3}{2}) = \frac{55}{4}$, $f(\frac{5}{2}) = \frac{39}{4}$, $f(\frac{7}{2}) = \frac{15}{4}$, so $M_4 = \frac{63 + 55 + 39 + 15}{4} = \frac{172}{4} = 43$. Note that $M_4$ is NOT the average of $L_4$ and $R_4$, which is $42$: the midpoint sum evaluates $f$ at new points, it does not average old values.

c) $f'(x) = -2x \le 0$ on $[0, 4]$, so $f$ is decreasing there. On each strip $[x_{i-1}, x_i]$ the largest value of $f$ is therefore at the LEFT end: the left rectangle contains the region above that strip, and the right rectangle is contained in it. Summing over the four strips, $R_4 \le A \le L_4$, and since $f$ is strictly decreasing the inequalities are strict: $34 < A < 50$. The figure shows the left rectangles sticking out above the curve, but a picture is an illustration, not a proof: the marks go to the word decreasing and to its justification.

d) Write both sums out: $L_n = \Delta x\left[f(x_0) + f(x_1) + \cdots + f(x_{n-1})\right]$ and $R_n = \Delta x\left[f(x_1) + \cdots + f(x_{n-1}) + f(x_n)\right]$. Every interior value appears in both and cancels, leaving $L_n - R_n = \Delta x\left[f(x_0) - f(x_n)\right] = \frac{4}{n}\left[f(0) - f(4)\right] = \frac{4}{n} \cdot 16 = \frac{64}{n}$. For $n = 4$ this gives $16 = 50 - 34$, as it must. Since $A$ lies between $R_n$ and $L_n$, it is known to within $1$ as soon as $\frac{64}{n} \le 1$, that is $n \ge 64$ strips. The formula works only because $f$ is monotonic: it is the bracket of c), measured.

e) $\frac{128}{3} = 42\frac{2}{3}$, which lies between $R_4 = 34$ and $L_4 = 50$, as c) demands. The errors are $50 - \frac{128}{3} = \frac{22}{3}$ for $L_4$, $\frac{128}{3} - 34 = \frac{26}{3}$ for $R_4$, and only $43 - \frac{128}{3} = \frac{1}{3}$ for $M_4$. Draw a midpoint rectangle on one strip: on the left half of the strip the curve is above the top of the rectangle, on the right half it is below. The missing piece and the extra piece nearly cancel, which a left or right rectangle cannot do, since on a decreasing function it is entirely above or entirely below. That is an observation about this figure, not a rule to quote: nothing in this chapter says on which side of $A$ a midpoint sum falls.

a) The time step is $\Delta t = 5$ s and there are $n = 6$ intervals: $[0, 5], [5, 10], \ldots, [25, 30]$. The left estimate uses the speed at the START of each interval: $L_6 = 5(0 + 6 + 11 + 15 + 18 + 20) = 5 \cdot 70 = 350$ m. The right estimate uses the speed at the END: $R_6 = 5(6 + 11 + 15 + 18 + 20 + 21) = 5 \cdot 91 = 455$ m. Seven readings give six intervals, hence six terms in each sum: the left sum drops the last reading and the right sum drops the first. Using all seven readings in one sum is the classic error, and here it gives $5 \cdot 91 = 455$ for the left sum, the right answer to a different question.

b) On each interval the speed is at least its value at the start and at most its value at the end, since it never decreases. So the distance covered on that interval lies between (starting speed) $\times\, 5$ and (final speed) $\times\, 5$. Adding the six intervals, $L_6 \le d \le R_6$: $350 \le d \le 455$ m. This is the monotonicity argument of Exercise 2 with the roles swapped, because the function now INCREASES: for an increasing function the left sum is the underestimate. The statement gives this information in words, and the answer must quote it; the table alone could not rule out a dip between two readings.

c) A midpoint sum needs the speed at the centre of each interval. With intervals of $5$ s the centres are $2.5, 7.5, \ldots$ s, and the table has no reading there: $M_6$ is out of reach. With $n = 3$ intervals of $10$ s, $[0, 10]$, $[10, 20]$ and $[20, 30]$, the centres are $5$, $15$ and $25$ s, which ARE in the table. So $M_3 = 10\left[v(5) + v(15) + v(25)\right] = 10(6 + 15 + 20) = 410$ m. It lies inside the bracket of b), as any sensible estimate must. The width is $10$ s, not $5$: using $\Delta t = 5$ with three terms counts only half of the thirty seconds.

d) For $n$ equal intervals of width $\Delta t$, the interior readings appear in both sums and cancel, as in Exercise 2: $R_n - L_n = \Delta t\left[v(30) - v(0)\right] = 21\,\Delta t$. The distance is known to within $10$ m, by the bracket alone, when $21\,\Delta t \le 10$, that is $\Delta t \le \frac{10}{21}$ s, slightly under half a second. Over thirty seconds this means $n = \frac{30}{\Delta t} \ge 30 \cdot \frac{21}{10} = 63$ intervals, that is $64$ readings counting the one at $t = 0$. The argument uses only the two end speeds and the monotonicity: nothing about the shape of the curve in between.

e) Each term $v(t_i)\,\Delta t$ is a speed in metres per second times a duration in seconds, so its unit is the metre: a rectangle drawn on a velocity graph carries the unit of its height times the unit of its width. The area under a velocity curve is a distance, and here, since the speed is never negative, it is the distance travelled. For a braking car the speed is largest at the start of each interval, so the LEFT estimate becomes the upper one and the right estimate the lower one. Which sum overestimates is never a property of left or right: it is a property of the direction in which the function moves.

a) $\Delta x = \frac{4 - 0}{n} = \frac{4}{n}$ and $x_i = 0 + \frac{4i}{n} = \frac{4i}{n}$. Then $f(x_i) = 16 - \frac{16i^2}{n^2}$, and $R_n = \sum_{i=1}^{n}\left(16 - \frac{16i^2}{n^2}\right)\frac{4}{n} = \frac{64}{n}\sum_{i=1}^{n} 1 - \frac{64}{n^3}\sum_{i=1}^{n} i^2 = 64 - \frac{64}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = 64 - \frac{32(n+1)(2n+1)}{3n^2}$. The first sum is $n$ copies of $\frac{64}{n}$, that is $64$. Check at $n = 4$: $64 - \frac{32 \cdot 5 \cdot 9}{48} = 64 - 30 = 34$, exactly the $R_4$ of Exercise 2, so the closed form is right before the limit is even taken. Now $\frac{(n+1)(2n+1)}{n^2} \to 2$, so $R_n \to 64 - \frac{64}{3} = \frac{128}{3}$. Everything is pulled out of the sum BEFORE the formulas are applied: $\frac{4}{n}$ and $\frac{16}{n^2}$ do not depend on $i$.

b) $\Delta x = \frac{3 - 1}{n} = \frac{2}{n}$ and $x_i = 1 + \frac{2i}{n}$: the left end $a = 1$ must be there. Expanding, $x_i^2 - 2x_i = 1 + \frac{4i}{n} + \frac{4i^2}{n^2} - 2 - \frac{4i}{n} = \frac{4i^2}{n^2} - 1$; the terms in $i$ cancel, which is what writing $x^2 - 2x = (x - 1)^2 - 1$ predicts. So $R_n = \sum_{i=1}^{n}\left(\frac{4i^2}{n^2} - 1\right)\frac{2}{n} = \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} - 2 = \frac{4(n+1)(2n+1)}{3n^2} - 2$. The $-2$ is $n$ copies of $-\frac{2}{n}$, not $-\frac{2}{n}$ once. Check at $n = 1$: one rectangle of width $2$ and height $f(3) = 3$, so $R_1 = 6$, and the formula gives $\frac{4 \cdot 2 \cdot 3}{3} - 2 = 6$. Finally $R_n \to \frac{8}{3} - 2 = \frac{2}{3}$.

c) $x^2 - 2x = x(x - 2)$ is negative on $(1, 2)$ and positive on $(2, 3)$, as the figure of the solution shows. A Riemann sum adds $f(x_i)\,\Delta x$ with its sign, so every rectangle standing below the axis is SUBTRACTED. In the limit, $\int_1^3 (x^2 - 2x)\,dx$ is the area of the part above the axis minus the area of the part below: a net, or signed, area. The geometric area of the whole region counts both parts positively and is therefore larger than $\frac{2}{3}$. A problem that asks for an area of a region whose boundary crosses the axis must be split at the crossing, here $x = 2$; a problem that asks for an integral must not be.

d) The left sum uses $x_{i-1} = \frac{4(i-1)}{n}$, so with $k = i - 1$ it runs over $k = 0, \ldots, n - 1$: $L_n = \frac{4}{n}\sum_{k=0}^{n-1}\left(16 - \frac{16k^2}{n^2}\right) = 64 - \frac{64}{n^3}\sum_{k=1}^{n-1} k^2 = 64 - \frac{64}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} = 64 - \frac{32(n-1)(2n-1)}{3n^2}$. The term $k = 0$ contributes $0$ to $\sum k^2$ but $16$ to the first sum, which still has $n$ terms and still gives $64$; the formula for $\sum k^2$ is then used with $n - 1$ in place of $n$. At $n = 4$: $64 - \frac{32 \cdot 3 \cdot 7}{48} = 64 - 14 = 50$, the $L_4$ of Exercise 2. And $\frac{(n-1)(2n-1)}{n^2} \to 2$, so $L_n \to \frac{128}{3}$, the same limit. For a continuous function the choice of sample points changes every approximation and never the integral.

e) With $x_i = \frac{2i}{n}$ and $\Delta x = \frac{2}{n}$, the sample points run from $\frac{2}{n}$ to $2$: they cover $[0, 2]$, not $[1, 3]$. His sum is $\sum_{i=1}^{n}\left(\frac{4i^2}{n^2} - \frac{4i}{n}\right)\frac{2}{n} = \frac{8}{n^3}\sum i^2 - \frac{8}{n^2}\sum i \to \frac{8}{3} - 4 = -\frac{4}{3}$, which is $\int_0^2 (x^2 - 2x)\,dx$, the integral over the part where the parabola is entirely below the axis. Even the sign is wrong. The quickest defence is the $n = 1$ check of b): his formula at $n = 1$ gives $8 \cdot \frac{1}{1} - 8 \cdot 1 = 0$, which is $2f(2)$, not $2f(3) = 6$.

a) The factor $\frac{3}{n}$ is $\Delta x$, so $b - a = 3$. Inside the root, $\frac{9i^2}{n^2} = \left(\frac{3i}{n}\right)^2$, so $x_i = \frac{3i}{n} = 0 + i\,\Delta x$: $a = 0$, $b = 3$ and $f(x) = \sqrt{9 - x^2}$. The limit is $\int_0^3 \sqrt{9 - x^2}\,dx$. Squaring $y = \sqrt{9 - x^2}$ gives $x^2 + y^2 = 9$ with $y \ge 0$: the curve is the upper half of the circle of radius $3$, and between $x = 0$ and $x = 3$ the region under it is a QUARTER disc. Its area is $\frac{1}{4}\pi \cdot 3^2 = \frac{9\pi}{4}$, about $7.07$ with $\pi \approx 3.14$. The figure of the solution shows six of the rectangles of the sum: since $f$ decreases on $[0, 3]$, the right rectangles sit inside the quarter disc and every $R_n$ is below $\frac{9\pi}{4}$.

b) First reading: $\Delta x = \frac{2}{n}$ and $x_i = \frac{2i}{n}$, so $[a, b] = [0, 2]$ and $f(x) = 1 + x$. Second reading: $x_i = 1 + \frac{2i}{n}$, so $[a, b] = [1, 3]$ and $f(x) = x$. The same sum is both a right sum of $1 + x$ on $[0, 2]$ and of $x$ on $[1, 3]$: the second graph is the first one shifted right by $1$. Both regions are trapezoids, with parallel sides $1$ and $3$ and width $2$, so each integral equals $\frac{1 + 3}{2} \cdot 2 = 4$. With the formulas: $\frac{2}{n}\sum_{i=1}^{n} 1 + \frac{4}{n^2}\sum_{i=1}^{n} i = 2 + \frac{4}{n^2} \cdot \frac{n(n+1)}{2} = 2 + \frac{2(n+1)}{n} \to 4$. The reading is not unique, the value is: any correct reading earns the marks, as long as $a$, $b$ and $f$ are consistent with each other.

c) $\Delta x = \frac{4}{n}$, $x_i = \frac{4i}{n}$, so the limit is $\int_0^4 (x - 3)\,dx$. The line $y = x - 3$ is below the axis on $[0, 3]$, a triangle of legs $3$ and $3$ with area $\frac{9}{2}$, and above on $[3, 4]$, a triangle of legs $1$ and $1$ with area $\frac{1}{2}$. The integral counts the first negatively: $-\frac{9}{2} + \frac{1}{2} = -4$. With the formulas: $\frac{16}{n^2}\sum_{i=1}^{n} i - \frac{12}{n}\sum_{i=1}^{n} 1 = \frac{8(n+1)}{n} - 12 \to 8 - 12 = -4$. A negative limit of sums is no anomaly: it is the signal that the region lies mostly below the axis, and the geometric answer must carry the same sign.

d) $\frac{2}{n} = \Delta x$, and in the exponent $-1 + \frac{2i}{n} = a + i\,\Delta x$ with $a = -1$, so $b = -1 + 2 = 1$: the limit is $\int_{-1}^{1} e^x\,dx$. The reading $x_i = \frac{2i}{n}$ on $[0, 2]$ with $f(x) = e^{x - 1}$ is equally correct. Its value is left for the next chapter. For the second request, $\Delta x = \frac{5 - 2}{n} = \frac{3}{n}$ and $x_i = 2 + \frac{3i}{n}$, so $\int_2^5 \sqrt{1 + x^3}\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} \frac{3}{n}\sqrt{1 + \left(2 + \frac{3i}{n}\right)^3}$. The $2 +$ inside the cube is the part most often dropped, and without it the sum describes $[0, 3]$.

e) Here $\Delta x = \frac{1}{n}$ and $x_i = \frac{i}{n}$, but there are $2n$ terms: the last sample point is $\frac{2n}{n} = 2$. The interval is $[0, 2]$, of length $2n \cdot \frac{1}{n} = 2$, and the limit is $\int_0^2 x\,dx$, a triangle of base $2$ and height $2$, area $2$. With the formula, now applied with $2n$ in place of $n$: $\frac{1}{n^2} \cdot \frac{2n(2n + 1)}{2} = \frac{2n + 1}{n} \to 2$. The answer $\frac{1}{2}$, which is $\int_0^1 x\,dx$, comes from assuming that every sum runs over $[0, 1]$; the upper index of the sum is part of the reading.