Chapter 7 Solutions

Authored worked explanation for the original source problem.

1. The width is (2−0)/2=1 and the left sample inputs are 0 and 1.

   Why: A left sum omits the rightmost endpoint.

2. $L_2=1(0^2+1^2)=1$.

   Why: Add width times height for each rectangle.

Reviewed answer: $1$ square unit.

Takeaway: Choose the samples before evaluating heights.

Authored worked explanation for the original source problem.

1. Width is 1; right sample inputs are 2, 3, and 4.

   Why: Three equal subintervals cover [1,4].

2. $R_3=1(\sqrt{2}+\sqrt{3}+\sqrt{4})$.

   Why: Multiply the sum of right-endpoint heights by the common width.

Reviewed answer: $\sqrt{2}+\sqrt{3}+2\approx 5.1463$ square units.

Takeaway: The source's 5.15 is the result rounded to two decimals.

Authored worked explanation for the original source problem.

1. Width is π/2, and midpoints are π/4, 3π/4, 5π/4, and 7π/4.

   Why: Sample halfway inside each quarter of the interval.

2. The four cosine values cancel in opposite pairs, leaving four heights' constant parts totaling 8.

   Why: The symmetry avoids separate decimal evaluations.

3. $M_4=(\pi /2)·8=4\pi$.

   Why: Use the interval width after summing heights.

Reviewed answer: $4\pi$ square units.

Takeaway: Periodic symmetry can simplify a numerical rule exactly.

Authored worked explanation for the original source problem.

1. Width is 2; sampled heights at −2, 0, 2 are 16, 0, 16.

   Why: Two trapezoids need three boundary heights.

2. $T_2=2(16/2+0+16/2)=32$.

   Why: The two endpoint heights receive half weight.

Reviewed answer: $32$ square units.

Takeaway: Interior boundaries are counted by both neighboring trapezoids.

Authored worked explanation for the original source problem.

1. On [0,1], √x is above x²; use their difference as the rectangle height.

   Why: An area between curves uses upper minus lower.

2. With width 1/4, left inputs 0, 1/4, 1/2, 3/4 give $L_4=(1/4)[0+(1/2-1/16)+(\sqrt{2}/2-1/4)+(\sqrt{3}/2-9/16\left)\right]$.

   Why: Each rectangle samples the gap, not just the upper curve.

3. This is about 0.299533, which rounds to the source's 0.3.

   Why: Keep precision until the completed sum.

Reviewed answer: $0.299533\dots$ square units.

Takeaway: Build the gap function before applying the numerical rule.

Authored worked explanation for the original source problem.

Source review note: The source gives 3.25; the four-trapezoid calculation for the printed curves gives 3.2395055…, rounding to 3.24.

1. Let $g\left(x\right)=e^{x-1}-1-\ln x$; g≥0 on [1,3].

   Why: The exponential curve lies above the logarithmic curve on this interval.

2. With width 1/2, use heights at 1, 1.5, 2, 2.5, 3 and compute $T_4=(1/2)\left[g\right(1)/2+g(1.5)+g(2)+g(2.5)+g(3)/2]$.

   Why: Half-weight the endpoints and fully weight the three interior heights.

3. The numerical result is about 3.2395055, or 3.24 to two decimal places.

   Why: Use the printed functions and interval without changing the source record.

Reviewed answer: $3.2395055\dots$ square units.

Takeaway: Accurate table heights matter as much as choosing the right rule.

Authored worked explanation for the original source problem.

Source review note: The source gives 2.656. Neither the printed eight-rectangle right signed sum (0.8125) nor its absolute-height geometric sum (2.46875) equals that value.

1. Width is 1/4, with right sample inputs k/4 for k=1,…,8.

   Why: The curve crosses the axis at √2, so some sampled heights are negative.

2. The signed sum is $(1/4){{\Sigma }_{k=1}}^8[2-(k/4{)}^2]=(1/4\left)\right(16-204/16)=13/16$.

   Why: Adding negative heights estimates net signed area.

3. For geometric area, use the absolute heights; the last three heights are −1/4, −17/16, and −2. Flipping them adds $2(1/4)(1/4+17/16+2)$, giving 79/32.

   Why: Geometric area cannot subtract below-axis regions.

Reviewed answer: Right signed sum $13/16=0.8125$; right geometric-area sum $79/32=2.46875$.

Takeaway: State whether the requested area is signed or geometric when the curve crosses the axis.

Authored worked explanation for the original source problem.

Source review note: The source gives 1.27. The specified four midpoint rectangles yield exactly 75/64=1.171875.

1. The gap is $g\left(x\right)=2-x-x^2$ on [0,1], with width 1/4.

   Why: The line is above the parabola throughout the interval.

2. Midpoints 1/8, 3/8, 5/8, 7/8 have gap heights 119/64, 95/64, 63/64, 23/64.

   Why: Evaluate both curves at the same midpoint.

3. Multiply their sum by 1/4 to obtain 300/256=75/64.

   Why: Every rectangle has the common width.

Reviewed answer: $75/64=1.171875$ square units.

Takeaway: Gap estimates need a midpoint in both functions.

Authored worked explanation for the original source problem.

1. Unit widths and right inputs 1,2,… produce $1/2+1/4+1/8+\dots$.

   Why: The interval starts at zero, so the first right sample is 1.

2. The sum through N rectangles is $1-2^{-N}$.

   Why: Multiply the finite geometric sum by 1/2 and subtract to verify the formula.

3. As N grows, the leftover 2^(−N) tends to 0, giving 1.

   Why: This is the limit of these fixed-width rectangles, not the exact area under the curve.

Reviewed answer: $1$ square unit as a rectangle-sum estimate.

Takeaway: An infinite count of fixed-width rectangles can still underestimate the exact area.

Authored worked explanation for the original source problem.

Source review note: The source gives 1.6, a coarse right-sum estimate. The fixed-width infinite sum is about 1.6449; the improper integral from 0 is divergent, so the sum must not be presented as that exact area.

1. Right samples give $S_N={{\Sigma }_{k=1}}^{N}1/k^2$.

   Why: The infinite height at x=0 is not a right sample, so the rectangle sum itself can be finite.

2. For k>N, $1/\left[k\right(k+1\left)\right]<1/k^2<1/\left[k\right(k-1\left)\right]$; the two comparison tails telescope to 1/(N+1) and 1/N.

   Why: A telescoping comparison bounds the unknown tail without using integration from the next chapter.

3. For N=100, S_N≈1.6349839, so $1.6448849<S_{\infty }<1.6449839$.

   Why: Adding the tail bounds gives a controlled numerical estimate near 1.6449.

4. The actual curve is unbounded at zero, so this sampled estimate alone cannot establish a finite geometric area there.

   Why: A numerical sampling scheme can miss a singular endpoint.

Reviewed answer: The unit-width right sum is about 1.6449; it does not represent a finite exact area from zero.

Takeaway: A finite sampled sum does not settle the behavior of the original area at a singular endpoint.

Authored solution to an added practice problem.

1. Width is 1; left samples for [1,4] with n=3 are 1,2,3.

   Why: Left sampling starts at the interval's initial endpoint.

2. $L_3=1(1+2+3)=6$.

   Why: The function height equals its input.

Reviewed answer: $6$ square units.

Takeaway: Indexing distinguishes a left sum from a right sum.

Authored solution to an added practice problem.

1. For f=2^(−x) on [0,3], width is 1 and the right heights are 1/2,1/4,1/8.

   Why: The right sum uses inputs 1,2,3.

2. R₃=7/8; the left heights 1,1/2,1/4 give L₃=7/4.

   Why: The same partition selects the opposite endpoints.

3. Since the function is nonnegative and decreasing, the right sum underestimates and the left sum overestimates.

   Why: Every slice's heights lie between its endpoint heights.

Reviewed answer: $R_3=7/8$, $L_3=7/4$; the area lies between them.

Takeaway: Use monotonicity to turn endpoint sums into bounds.

Authored solution to an added practice problem.

1. On [−1,1] with n=2, width is 1 and the midpoints are −1/2 and 1/2.

   Why: Midpoints are centered within the two intervals.

2. Each height is $1+(1/2{)}^2=5/4$, so M₂=5/2.

   Why: The even quadratic has matching heights at opposite samples.

Reviewed answer: $5/2=2.5$ square units.

Takeaway: Symmetry can simplify a midpoint sum.

Authored solution to an added practice problem.

1. For f=2x+3 on [−1,2], three trapezoids have width 1 and boundary heights 1,3,5,7.

   Why: The vertical shift keeps all heights positive.

2. T₃=1(1/2+3+5+7/2)=12.

   Why: Endpoints have half weight.

3. A line's graph is the exact sloping top of each trapezoid.

   Why: No curved gap remains above or below those tops.

Reviewed answer: $12$ square units, exactly.

Takeaway: Trapezoids exactly fill a region under a linear function.

Authored solution to an added practice problem.

1. On each subinterval of a decreasing nonnegative graph, the left height is the largest and the right height the smallest.

   Why: Every other height lies between those boundary values.

2. Each left rectangle contains that slice's region, while each right rectangle fits inside it.

   Why: Adding the slice comparisons preserves the inequality.

Reviewed answer: Left is an overestimate and right an underestimate.

Takeaway: Monotonicity determines the direction of endpoint-sampling error.

Authored solution to an added practice problem.

1. The width is 1. Left heights are 2,4,8; right heights are 4,8,10.

   Why: Use the table's ordered boundary values.

2. The sums are 14 and 22, and monotonic increase makes them lower and upper bounds.

   Why: The added monotonicity assumption bounds unsampled heights.

Reviewed answer: $L_3=14$, $R_3=22$, and geometric area lies between 14 and 22.

Takeaway: A table needs a shape assumption to turn samples into guaranteed bounds.

Authored solution to an added practice problem.

1. On [0,1], the trapezoid contributes $1(2+4)/2=3$.

   Why: Its width is 1.

2. On [1,3], it contributes $2(4+8)/2=12$, so the total is 15.

   Why: Use each interval's own width rather than a single uniform Δx.

Reviewed answer: $15$ square units.

Takeaway: Nonuniform data require local widths.

Authored solution to an added practice problem.

1. The upper curve is y=3 and the lower is y=x², so the gap is 3−x² on [0,1].

   Why: Use a common input for both boundary curves.

2. With two rectangles, width 1/2 and midpoints 1/4,3/4, the gap heights are 47/16 and 39/16.

   Why: The samples stay halfway inside each interval.

3. The estimate is $(1/2)(47/16+39/16)=43/16$.

   Why: Multiply by width only after adding both gap heights.

Reviewed answer: $43/16=2.6875$ square units.

Takeaway: Build the gap before sampling it.

Authored solution to an added practice problem.

1. For a concave-up graph, each chord lies above or on the graph.

   Why: The slope increases, so the graph bends below its joining secant line.

2. Each trapezoid fills up to the chord and therefore contains the curve's slice.

   Why: Sum the slice inequalities across the interval.

Reviewed answer: The trapezoidal estimate is an overestimate.

Takeaway: Concavity predicts a trapezoid error's sign.

Authored solution to an added practice problem.

1. For $f\left(x\right)=4-x^2$ on [0,2], the graph is concave down.

   Why: Its second derivative is −2.

2. With two rectangles, width 1 and midpoints 1/2,3/2 give $M_2=15/4+7/4=11/2$.

   Why: The midpoint samples sit high relative to each interval's average height.

3. The tangent-line symmetry around each midpoint bounds the curve below those midpoint rectangles in total.

   Why: A concave-down curve lies below its midpoint tangent, whose odd linear change cancels over a symmetric interval.

Reviewed answer: $M_2=11/2$ square units; an overestimate for this concave-down graph.

Takeaway: Concavity can determine the error direction without evaluating an integral.

Authored solution to an added practice problem.

1. Width is 1 and the right heights are −2,1,1, giving −2+1+1=0.

   Why: Negative height contributes negatively to net signed area.

2. For a geometric-area estimate use 2,1,1, yielding 4.

   Why: Replace sampled negative heights by their absolute values.

Reviewed answer: Signed right sum 0; absolute-height right sum 4.

Takeaway: Cancellation can hide a nonzero geometric region.

Authored solution to an added practice problem.

1. For f=x³ on [0,2], n=2 gives width 1 and heights 0,1, so L₂=1.

   Why: The function is increasing and nonnegative.

2. With n=4, width 1/2 and left heights 0,1/8,1,27/8 give L₄=9/4.

   Why: Narrower left rectangles fill more of each increasing slice.

Reviewed answer: $L_2=1$, $L_4=9/4$; both underestimate.

Takeaway: Refining this increasing function raises its lower estimate.

Authored solution to an added practice problem.

1. Width is 1/n, and the kth right height is k/n, giving $R_n=(1/n^2){{\Sigma }_{k=1}}^{n}k$.

   Why: Separate the sample height from the rectangle width.

2. Use $\Sigma k=n(n+1)/2$ to get the formula, which tends to 1/2.

   Why: The limiting rectangle sum gives the area as widths approach zero.

Reviewed answer: $R_n=(n+1)/\left(2n\right)$; limiting area 1/2.

Takeaway: A finite arithmetic sum can describe every resolution at once.

Authored solution to an added practice problem.

1. For f=1/(1+x) on [0,2], width is 1. Boundary heights are 1,1/2,1/3, so L₂=3/2 and R₂=5/6.

   Why: The decreasing shape makes the left sum larger.

2. Midpoint heights at 1/2 and 3/2 are 2/3 and 2/5, giving M₂=16/15.

   Why: Midpoints differ from the boundary samples.

3. T₂=1/2+1/2+1/6=7/6.

   Why: Half-weight endpoints and fully weight the interior boundary.

Reviewed answer: $L_2=3/2$, $R_2=5/6$, $M_2=16/15$, $T_2=7/6$.

Takeaway: The same partition compares sampling rules across a curved function.

Authored solution to an added practice problem.

1. Increase the exponent in 6x² to 3 and divide its coefficient by 3, giving 2x³.

   Why: Differentiating reverses the inverse power step.

2. The constant −4 has primitive −4x, and an arbitrary C contributes zero derivative.

   Why: Every constant shift has the same derivative.

Reviewed answer: $F\left(x\right)=2x^3-4x+C$.

Takeaway: An antiderivative is a family until extra data fix its constant.

Authored solution to an added practice problem.

1. Differentiating cos(3x) gives −3sin(3x).

   Why: The inside derivative supplies a factor 3.

2. Multiply by −1/3 so the derivative becomes exactly sin(3x).

   Why: Undo the chain multiplier rather than multiplying by it again.

Reviewed answer: $-(1/3)\cos \left(3x\right)+C$.

Takeaway: Check a proposed primitive by differentiation.

Authored solution to an added practice problem.

1. All primitives of 2x are x²+C.

   Why: The derivative of any constant is zero.

2. F(1)=6 gives 1+C=6, so C=5.

   Why: The extra point selects one member of the family.

Reviewed answer: $F\left(x\right)=x^2+5$.

Takeaway: One initial value determines the additive constant.

Authored solution to an added practice problem.

1. Differentiate (1/2)sin(2x): the chain factor 2 cancels the 1/2, leaving cos(2x).

   Why: The derivative check exposes the missing coefficient.

2. For the stated integrand, sin(2x)+C differentiates to 2cos(2x).

   Why: Match the entire integrand, not just its function type.

Reviewed answer: The proposed function differentiates to $\cos \left(2x\right)$; a primitive of 2cos(2x) is sin(2x)+C.

Takeaway: Differentiation is the quickest primitive check.

Authored solution to an added practice problem.

1. For times 0,1,2 seconds and velocities 2,6,14 m/s, each interval has width 1 s.

   Why: Multiplying time by velocity gives displacement units.

2. $T_2=1\left[\right(2+6)/2+(6+14)/2]=4+10=14$ m.

   Why: Average neighboring velocities on each interval.

Reviewed answer: Estimated displacement 14 m.

Takeaway: Area estimates can estimate accumulated change, with units explaining why.

Authored solution to an added practice problem.

1. For f=2^(−x) with unit-width right rectangles from zero, the sum through N is 1−2^(−N), so the remaining tail is 2^(−N).

   Why: The same geometric ratio controls every leftover block.

2. Require $2^{-N}<0.01$; N=6 gives 1/64>0.01, while N=7 gives 1/128<0.01.

   Why: Check neighboring integers to find the first sufficient rectangle count.

Reviewed answer: After 7 right rectangles; the remaining tail is $1/128<0.01$.

Takeaway: A known tail formula makes an approximation tolerance concrete.