Corrected exercises: areas between curves and average value (MATH 141)

a) Set the two heights equal: $x^2 - 2x = x + 4$, so $x^2 - 3x - 4 = 0$, which factors as $(x - 4)(x + 1) = 0$. The curves meet at $x = -1$ and $x = 4$. The points are $(-1, 3)$ and $(4, 8)$, and both check in BOTH equations: $(-1)^2 - 2(-1) = 3 = -1 + 4$ and $16 - 8 = 8 = 4 + 4$. These two numbers are the bounds of the integral. On an exam, write the equation, the factored form and the list of roots: the set-up marks depend on them, and a bound read off a sketch is not a justification.

b) Take a test value inside $(-1, 4)$, say $x = 0$: the line gives $4$, the parabola gives $0$, so the line is on top. Why one test is enough: the difference $(x + 4) - (x^2 - 2x) = -x^2 + 3x + 4 = -(x - 4)(x + 1)$ is continuous and vanishes only at $-1$ and $4$, so it cannot change sign in between. Now the $x$-axis. The slice at $x$ has length $(\text{top}) - (\text{bottom})$, a DIFFERENCE of two heights. At $x = 1$ the top is $5$ and the bottom is $-1$, and the slice has length $5 - (-1) = 6$, exactly what a ruler would measure from $y = -1$ to $y = 5$. A negative bottom height is simply subtracted, which adds. Nothing in the formula refers to the axis, so there is no reason to split at $x = 0$ or $x = 2$, where the parabola crosses it.

c) $A = \int_{-1}^{4} \left[(x + 4) - (x^2 - 2x)\right] dx = \int_{-1}^{4} (-x^2 + 3x + 4)\,dx = \left[-\frac{x^3}{3} + \frac{3x^2}{2} + 4x\right]_{-1}^{4}$. At $x = 4$: $-\frac{64}{3} + 24 + 16 = \frac{56}{3}$. At $x = -1$: $\frac{1}{3} + \frac{3}{2} - 4 = -\frac{13}{6}$. So $A = \frac{56}{3} + \frac{13}{6} = \frac{112 + 13}{6} = \frac{125}{6}$, a little under $21$. Two checks that cost nothing. The longest slice is at $x = \frac{3}{2}$, of length $\frac{25}{4}$, and the region is $5$ wide, so $A$ must be below $5 \cdot \frac{25}{4} = \frac{125}{4}$: it is. And Archimedes: a region cut from a parabola by a chord has $\frac{2}{3}$ of the enclosing rectangle, here $\frac{2}{3} \cdot \frac{125}{4} = \frac{125}{6}$, the same number.

d) The integrand $(x^2 - 2x) - (x + 4)$ is bottom minus top, negative at every point of $(-1, 4)$, so the integral returns the area with a minus sign. An area is never negative: a negative result is an ALARM that the order was reversed, never an answer. Taking the absolute value at the end happens to give $\frac{125}{6}$ here, but only because the integrand keeps ONE sign on the whole interval, the curves never crossing inside it. As soon as they cross (Exercise 2), the positive and negative pieces cancel inside the integral, and no absolute value applied afterwards can undo that. The safe habit is the one of part b): decide top and bottom FIRST, with a test value, on each piece.

e) The signed integrals are $\int_{-1}^{4} (x + 4)\,dx = \left[\frac{x^2}{2} + 4x\right]_{-1}^{4} = 24 - \left(-\frac{7}{2}\right) = \frac{55}{2}$ and $\int_{-1}^{4} (x^2 - 2x)\,dx = \left[\frac{x^3}{3} - x^2\right]_{-1}^{4} = \frac{16}{3} - \left(-\frac{4}{3}\right) = \frac{20}{3}$. Their difference is $\frac{165 - 40}{6} = \frac{125}{6}$, the right answer. The student replaced $\frac{20}{3}$ by the GEOMETRIC area between the parabola and the axis, $\frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}$, where the middle $\frac{4}{3}$ comes from $0 \le x \le 2$, BELOW the axis. On that stretch the slice of $R$ goes from the negative height of the parabola up to the line: that piece must be ADDED to the area under the line, and the signed integral does exactly that, since subtracting its negative contribution $-\frac{4}{3}$ adds $\frac{4}{3}$. Subtracting it instead costs $2 \cdot \frac{4}{3} = \frac{8}{3}$, and indeed $\frac{125}{6} - \frac{109}{6} = \frac{16}{6} = \frac{8}{3}$. Moral: $\int (f - g)$ needs no case analysis for the $x$-axis; only the crossings of the two curves matter.

a) Bring everything to one side and FACTOR: $x^3 - 2x^2 - 3x = x(x^2 - 2x - 3) = x(x - 3)(x + 1) = 0$. The roots are $x = -1$, $x = 0$ and $x = 3$. With $g(x) = 3x$, the points are $(-1, -3)$, $(0, 0)$ and $(3, 9)$; check on $f$: $f(-1) = -1 - 2 = -3$ and $f(3) = 27 - 18 = 9$. Three intersections cut the region into two pieces, which the figure shows as two lobes of very different sizes.

b) Let $h = f - g = x(x - 3)(x + 1)$. It is continuous and vanishes only at the three roots, so its sign is constant between them and one test value per interval decides. On $(-1, 0)$, at $x = -\frac{1}{2}$: $h = \left(-\frac{1}{2}\right)\left(-\frac{7}{2}\right)\left(\frac{1}{2}\right) = \frac{7}{8} > 0$, so $f$ is on top. On $(0, 3)$, at $x = 1$: $h = 1 \cdot (-2) \cdot 2 = -4 < 0$, so $g$ is on top. The sign flips at each simple root, as the sign table of a product of three linear factors predicts. Write this sign study on the copy: it is what justifies the split in c).

c) One integral per piece, top minus bottom on each: $A = \int_{-1}^{0} (f - g)\,dx + \int_{0}^{3} (g - f)\,dx$. An antiderivative of $f - g = x^3 - 2x^2 - 3x$ is $H(x) = \frac{x^4}{4} - \frac{2x^3}{3} - \frac{3x^2}{2}$, and $H(0) = 0$, $H(-1) = \frac{1}{4} + \frac{2}{3} - \frac{3}{2} = -\frac{7}{12}$, $H(3) = \frac{81}{4} - 18 - \frac{27}{2} = -\frac{45}{4}$. First piece: $H(0) - H(-1) = \frac{7}{12}$. Second piece: $-\left(H(3) - H(0)\right) = \frac{45}{4} = \frac{135}{12}$. Both are positive, as areas must be. $A = \frac{7 + 135}{12} = \frac{142}{12} = \frac{71}{6}$, a little under $12$.

d) $\int_{-1}^{3} (f - g)\,dx = H(3) - H(-1) = -\frac{45}{4} + \frac{7}{12} = -\frac{128}{12} = -\frac{32}{3}$. This is the NET signed area: the small lobe counts $+\frac{7}{12}$, the large one $-\frac{45}{4}$, and they partly cancel. Its absolute value $\frac{32}{3} = \frac{45}{4} - \frac{7}{12}$ is the DIFFERENCE of the two lobes, while the area is their SUM $\frac{45}{4} + \frac{7}{12}$. The gap between the two answers is $\frac{71}{6} - \frac{32}{3} = \frac{7}{6}$, exactly twice the small lobe, subtracted instead of added. In general $\left|\int h\right| \le \int |h|$, with equality only when $h$ keeps one sign: an absolute value placed OUTSIDE the integral repairs nothing once the curves cross.

e) Dividing by $x$ is allowed only when $x \ne 0$, so the division silently deletes the root $x = 0$. Yet $f(0) = 0 = g(0)$: the curves do meet at the origin, and that is precisely where the top curve changes. Without it the student integrates $f - g$ over $[-1, 3]$ in one piece and lands on the $-\frac{32}{3}$ of part d), losing the whole question. The rule is the one of part a): bring everything to one side and factor, never divide by an expression that can vanish. The same trap returns with trigonometric curves, where dividing by $\sin x$ deletes the multiples of $\pi$ (Exercise 10).

a) Write the line as $x = y + 1$ and set the two $x$ equal: $3 - y^2 = y + 1$, so $y^2 + y - 2 = 0$, that is $(y + 2)(y - 1) = 0$. The curves meet at $y = -2$ and $y = 1$, at the points $(-1, -2)$ and $(2, 1)$; check on the parabola: $3 - 4 = -1$ and $3 - 1 = 2$. Solving in $y$ is the natural move here, because the parabola is already written as $x$ in terms of $y$. Note the trap waiting in b): the bounds of an integral in $y$ are $-2$ and $1$, NOT the $x$-coordinates $-1$ and $2$.

b) Test $y = 0$: the parabola gives $x = 3$, the line gives $x = 1$, so the parabola is on the RIGHT on the whole interval $(-2, 1)$, where the two expressions never meet. Then $A = \int_{-2}^{1} \left[(3 - y^2) - (y + 1)\right] dy = \int_{-2}^{1} (2 - y - y^2)\,dy = \left[2y - \frac{y^2}{2} - \frac{y^3}{3}\right]_{-2}^{1}$. At $y = 1$: $2 - \frac{1}{2} - \frac{1}{3} = \frac{7}{6}$. At $y = -2$: $-4 - 2 + \frac{8}{3} = -\frac{10}{3}$. So $A = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2}$. One integral, a polynomial, no square root: horizontal slices see the same two boundaries, the line on the left and the parabola on the right, from the bottom of the region to its top.

c) Solve the parabola for $y$: $y^2 = 3 - x$ gives TWO functions, the upper branch $y = \sqrt{3 - x}$ and the lower branch $y = -\sqrt{3 - x}$, joined at the vertex $(3, 0)$. For $-1 \le x \le 2$, the vertical slice starts on the LOWER branch and ends on the line $y = x - 1$, which is the top of the region there (see the figure of the solution). For $2 \le x \le 3$, the line has left the region, having crossed the upper branch at $(2, 1)$, and the slice runs from the lower branch to the upper one. The top boundary changes formula at $x = 2$, hence two integrals: $A = \int_{-1}^{2} \left[(x - 1) + \sqrt{3 - x}\right] dx + \int_{2}^{3} 2\sqrt{3 - x}\,dx$. The classic slip is to take the line as the bottom: the line is below the UPPER branch, but the region lies below the line.

d) With $u = 3 - x$, $\int \sqrt{3 - x}\,dx = -\frac{2}{3}(3 - x)^{3/2} + C$, and differentiating gives back $\sqrt{3 - x}$. First integral: $\int_{-1}^{2} (x - 1)\,dx = \left[\frac{x^2}{2} - x\right]_{-1}^{2} = 0 - \frac{3}{2} = -\frac{3}{2}$, and $\int_{-1}^{2} \sqrt{3 - x}\,dx = -\frac{2}{3}(1) + \frac{2}{3}(8) = \frac{14}{3}$, so the first piece is $\frac{14}{3} - \frac{3}{2} = \frac{19}{6}$. Second: $2\left[-\frac{2}{3}(3 - x)^{3/2}\right]_{2}^{3} = 2 \cdot \frac{2}{3} = \frac{4}{3}$. Total $\frac{19}{6} + \frac{8}{6} = \frac{27}{6} = \frac{9}{2}$, as in b). Do not be alarmed by $-\frac{3}{2}$: it is one term of a slice length, not an area, and the slice length $(x - 1) + \sqrt{3 - x}$ itself is positive. The comparison of the two methods is the lesson: same area, but three antiderivatives and a split in $x$ against one polynomial in $y$.

e) The first student has left minus right: at $y = 0$ the integrand is $1 - 3 = -2$, and the integral is $-\frac{9}{2}$. In $y$, the slice is RIGHT minus LEFT, the horizontal analogue of top minus bottom. The second student took the upper branch as the top and the line as the bottom on $[-1, 2]$. That slice is a real length, but of the region lying ABOVE the line and below the upper branch, a different region: $\int_{-1}^{2} \sqrt{3 - x}\,dx - \int_{-1}^{2} (x - 1)\,dx = \frac{14}{3} + \frac{3}{2} = \frac{37}{6} \ne \frac{9}{2}$. Moral: with vertical slices, a sideways parabola is TWO functions, and each slice must be read on the figure, from the curve it starts on to the curve it ends on.

a) $f_{\text{ave}} = \frac{1}{4 - 1}\int_1^4 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_1^4 = \frac{1}{3} \cdot \frac{64 - 1}{3} = \frac{1}{3} \cdot 21 = 7$. The endpoint average is $\frac{1 + 16}{2} = \frac{17}{2}$ and the midpoint value is $\frac{25}{4}$: neither is $7$, and the order $\frac{25}{4} < 7 < \frac{17}{2}$ is no accident. For a curve that bends upward, the chord joining the endpoints lies ABOVE the graph, so the endpoint average overestimates, and the tangent at the midpoint lies BELOW it, so the midpoint value underestimates. The formula $\frac{f(a) + f(b)}{2}$ is the average value only for a linear $f$; for anything else it is a guess, and it costs the whole part.

b) On $[0, \pi]$: $\frac{1}{\pi}\int_0^\pi \sin x\,dx = \frac{1}{\pi}\left[-\cos x\right]_0^\pi = \frac{1}{\pi}(1 + 1) = \frac{2}{\pi}$, about $0.64$. On $[0, 2\pi]$: $\frac{1}{2\pi}\left[-\cos x\right]_0^{2\pi} = \frac{1}{2\pi}(-1 + 1) = 0$. The average is $0$ although $\sin x$ is not identically $0$: the hump of signed area $+2$ and the hump of signed area $-2$ in the figure cancel. The area between the curve and the axis, on the other hand, counts both humps as positive: $\int_0^{2\pi} |\sin x|\,dx = 2 + 2 = 4$. This is the one place in the chapter where the sign is KEPT on purpose: an average temperature of $-5$ degrees means something, a negative area does not. Never put an absolute value into an average value unless the question asks for an average DISTANCE.

c) The width is $1$, so $f_{\text{ave}} = \int_0^1 \frac{dx}{1 + x^2} = \arctan 1 - \arctan 0 = \frac{\pi}{4}$, about $0.785$. Dividing by $b - a = 1$ changes nothing here, which is exactly why students forget the factor in the next part, where the width is not $1$.

d) Integration by parts with $u = x$ (it becomes simpler when differentiated) and $dv = \cos x\,dx$: $du = dx$, $v = \sin x$, so $\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C$. Check by differentiating: $\sin x + x\cos x - \sin x = x\cos x$. Then $\int_0^{\pi/2} x\cos x\,dx = \left[x\sin x + \cos x\right]_0^{\pi/2} = \left(\frac{\pi}{2} + 0\right) - (0 + 1) = \frac{\pi}{2} - 1$. Divide by the width $\frac{\pi}{2}$: $f_{\text{ave}} = \frac{2}{\pi}\left(\frac{\pi}{2} - 1\right) = 1 - \frac{2}{\pi}$, about $1 - 0.64 = 0.36$. The two classic losses: reporting $\frac{\pi}{2} - 1$, the integral, as the average; and MULTIPLYING by the width instead of dividing, which gives $\frac{\pi}{2}\left(\frac{\pi}{2} - 1\right)$. The factor is $\frac{1}{b - a} = \frac{2}{\pi}$.

e) If $m \le f(x) \le M$ on $[a, b]$, the comparison property of the integral gives $\int_a^b m\,dx \le \int_a^b f(x)\,dx \le \int_a^b M\,dx$, that is $m(b - a) \le \int_a^b f \le M(b - a)$. Dividing by $b - a > 0$ keeps the inequalities: $m \le f_{\text{ave}} \le M$. An average lies between the smallest and the largest value, as any average should. For a): $x^2$ runs from $1$ to $16$ on $[1, 4]$, and $1 \le 7 \le 16$. For c): $\frac{1}{1 + x^2}$ decreases from $1$ to $\frac{1}{2}$ on $[0, 1]$, so $\frac{1}{2} \le \frac{\pi}{4} \le 1$, that is $2 \le \pi \le 4$: true, and a genuine check on c). This test takes five seconds and catches the forgotten factor $\frac{1}{b - a}$ whenever the interval is long.

a) $f_{\text{ave}} = \frac{1}{2}\int_0^2 (3x^2 - 2x)\,dx = \frac{1}{2}\left[x^3 - x^2\right]_0^2 = \frac{1}{2}(8 - 4) = 2$. Then solve $f(c) = 2$: $3c^2 - 2c - 2 = 0$, so $c = \frac{2 \pm \sqrt{4 + 24}}{6} = \frac{2 \pm 2\sqrt 7}{6} = \frac{1 \pm \sqrt 7}{3}$. Now locate each root without a calculator: $4 < 7 < 9$ gives $2 < \sqrt 7 < 3$. So $\frac{1 + \sqrt 7}{3}$ lies between $1$ and $\frac{4}{3}$, inside $[0, 2]$: it is the answer. And $\frac{1 - \sqrt 7}{3}$ lies between $-\frac{2}{3}$ and $-\frac{1}{3}$, outside $[0, 2]$: it solves the equation but is NOT a value of $c$, since the theorem locates $c$ in $[a, b]$. Reporting both roots, or the wrong one, costs the mark for the conclusion.

b) $g_{\text{ave}} = \frac{1}{3}\int_0^3 (x^2 - 4x + 5)\,dx = \frac{1}{3}\left[\frac{x^3}{3} - 2x^2 + 5x\right]_0^3 = \frac{1}{3}(9 - 18 + 15) = 2$. Solve $c^2 - 4c + 5 = 2$: $c^2 - 4c + 3 = (c - 1)(c - 3) = 0$, so $c = 1$ or $c = 3$, BOTH in $[0, 3]$. The theorem guarantees at least one $c$; here there are two, one of them an endpoint, and a complete answer lists both. The phrase the number $c$ is a habit of speech, not a claim of uniqueness: a continuous function can cross its average height as many times as it likes.

c) The figure of the solution shows the graph of $f$, the line $y = 2$ and the rectangle $[0, 2] \times [0, 2]$. Its area is $2 \cdot 2 = 4 = \int_0^2 f(x)\,dx$. But $f$ is negative on $\left(0, \frac{2}{3}\right)$, so $4$ is a SIGNED area: the part of the region above the axis counts positively, the small part below counts negatively, and their combination equals the rectangle. Equivalently, the piece of the graph that rises above the line $y = 2$ (to the right of $c$) exactly fills the gap between the line and the graph to the left of $c$, since $\int_0^2 (f - 2)\,dx = 4 - 4 = 0$. That is the picture the theorem draws: level the graph off at height $f(c)$ without changing the signed area.

d) $\lfloor x \rfloor = 0$ on $[0, 1)$ and $1$ on $[1, 2)$, with $\lfloor 2 \rfloor = 2$ at the single point $x = 2$, which changes no integral. So $\int_0^2 \lfloor x \rfloor\,dx = 0 \cdot 1 + 1 \cdot 1 = 1$ and the average is $\frac{1}{2}$. But $\lfloor x \rfloor$ only takes the values $0$, $1$ and $2$: the average $\frac{1}{2}$ is never reached. The function is integrable, so its average exists, but it is not continuous (it jumps at $x = 1$ and $x = 2$), and without continuity the conclusion of the theorem fails. Before invoking the theorem on an exam, write the hypothesis: $f$ is continuous on $[a, b]$.

e) Let $F(x) = \int_a^x f(t)\,dt$. Since $f$ is continuous, the Fundamental Theorem of Calculus (part 1) says $F$ is differentiable with $F'(x) = f(x)$; in particular $F$ is continuous on $[a, b]$ and differentiable on $(a, b)$, the two hypotheses of the mean value theorem. It gives $c$ in $(a, b)$ with $F'(c) = \frac{F(b) - F(a)}{b - a}$. Now $F(a) = 0$ and $F(b) = \int_a^b f$, so $f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx = f_{\text{ave}}$. This proof even places $c$ in the OPEN interval, which is consistent with b): besides the endpoint $3$, the interior value $c = 1$ is there.

a) The curve $y = \sqrt x$ meets the axis at $(0, 0)$ and the line $y = x - 2$ meets it at $(2, 0)$. For the curve and the line: $\sqrt x = x - 2$ requires $x - 2 \ge 0$, since a square root is never negative. Squaring gives $x = x^2 - 4x + 4$, so $x^2 - 5x + 4 = (x - 1)(x - 4) = 0$. Test both in the ORIGINAL equation: $x = 4$ gives $2 = 2$, kept; $x = 1$ gives $\sqrt 1 = 1$ against $1 - 2 = -1$, rejected. Squaring has merged $\sqrt x = x - 2$ with $-\sqrt x = x - 2$, and $x = 1$ solves the second one: it is where the line meets the LOWER half $y = -\sqrt x$, a curve that is not part of this problem. Corners: $(0, 0)$, $(2, 0)$, $(4, 2)$.

b) Vertical slices: the top is $\sqrt x$ all the way from $x = 0$ to $x = 4$, but the bottom is the axis $y = 0$ for $0 \le x \le 2$ and the line $y = x - 2$ for $2 \le x \le 4$. Two formulas for the bottom, so two integrals: $A_1 = \int_0^2 \sqrt x\,dx + \int_2^4 (\sqrt x - x + 2)\,dx$. Horizontal slices: $y$ runs from $0$ to $2$; the left end is always on the curve, $x = y^2$, and the right end always on the line, $x = y + 2$. One formula each, so one integral: $A_1 = \int_0^2 (y + 2 - y^2)\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_0^2 = 2 + 4 - \frac{8}{3} = \frac{10}{3}$. Confirmation: the two vertical pieces combine into $\int_0^4 \sqrt x\,dx - \int_2^4 (x - 2)\,dx = \frac{2}{3} \cdot 8 - 2 = \frac{10}{3}$, the area under the curve minus the triangle of legs $2$ and $2$ under the line.

c) $e^x = 3$ at $x = \ln 3$, so the corners of $R_2$ are $(0, 1)$, $(0, 3)$ and $(\ln 3, 3)$. Vertical slices: from $x = 0$ to $\ln 3$, top $y = 3$, bottom $y = e^x$, one integral: $A_2 = \int_0^{\ln 3} (3 - e^x)\,dx = \left[3x - e^x\right]_0^{\ln 3} = (3\ln 3 - 3) - (0 - 1) = 3\ln 3 - 2$. Horizontal slices: $y$ runs from $1$, NOT from $0$, since the region starts where the curve meets the $y$-axis, up to $3$; left end $x = 0$, right end $x = \ln y$: $A_2 = \int_1^3 \ln y\,dy$. By parts, $\int \ln y\,dy = y\ln y - y + C$, so $A_2 = (3\ln 3 - 3) - (0 - 1) = 3\ln 3 - 2$, the same. Here both set-ups are single integrals, and the tie is broken by the antiderivative: $e^x$ is immediate, $\ln y$ needs parts. So this time $x$ wins. Starting the $y$-integral at $0$ would describe another region and produce an improper integral, a sure sign the bounds are wrong.