The maximum number of marks available for your answers in SECTION B is 50 marks.

UNIVERSITY OF EAST ANGLIA School of Environmental Sciences Main Series UG Examination 017-018 ADVANCED QUANTITATIVE SKILLS ENV-4014Y Time allowed: 1.5 hours Answer ALL 6 questions Write answers to EACH SECTION in a SEPARATE booklet. The maximum number of marks available for your answers in SECTION A is 50 marks. The maximum number of marks available for your answers in SECTION B is 50 marks. The TOTAL number of marks available for the paper is 100. Numbers in square brackets [ ] indicate the relevant mark applied to each question. PROVIDED: Appendix 1: Formulae Sheet (8 pages) Calculators are permitted. Notes are not permitted in this examination. Do not turn over until you are told to do so by the Invigilator ENV-4014Y Module Contact: Dorothee Bakker, ENV Copyright of the University of East Anglia Version 1

SECTION A Answer ALL questions [50 marks] 1. Convert 0.3 psi (pound-force per square inch) into SI base units and give your answer in scientific notation and to decimal places. [10 marks]. The amount A(t) of the radioactive substance carbon-14 (or 14 C) decays according to the exponential function: A(t) = A 0 e rt where A 0 is the initial amount (at t = 0) and t is the time in years (t 0). The half-life of carbon-14 is 5,730 years. In other words, the amount A of this radioactive substance in 5,730 years is equal to half of its initial amount A 0. Find r (round to 3 significant figures). [0 marks] 3. The Aral Sea used to be one of the biggest lakes in the world, but it has dramatically lost water since the middle of the last century. This loss of water can be described - from 1973 to 010 - by the empirically derived quadratic function: f(x) = y = 0.550 x,1.7 x +,3,810 where y is the water volume in km 3 and x is the year. Determine the water volume of the Aral Sea in 1973 and in 010 and calculate in which year the volume would return to 1973 levels if the above function is used for prediction purposes. [0 marks] END OF SECTION A ENV-4014Y Version 1

3 SECTION B Answer ALL questions [50 marks] 4. Ocean uptake of carbon dioxide (CO) leads to ocean acidification as the extra dissolved carbon dioxide reacts with water to form carbonic acid. The ph, a measure of acidity, is defined as follows: ph = log 10 ([H + ]) where [H + ] is the hydrogen ion concentration (in mol l -1 or M). The global mean surface ocean ph was 8.1 in the year 000. Global mean surface ocean ph has been predicted to decrease by 0.3 ph units by the year 100 relative to the year 000, if CO emissions from human activity continue unabated (in a business as usual scenario ). What was the hydrogen ion concentration in the year 000 and what would it become in 100? Use scientific notation and significant figures. Is this an increase or a decrease in the hydrogen ion concentration over time? Express the change in the hydrogen ion concentration from the year 000 to the year 100 as a relative change in percent. Report this percentage to 3 significant figures. [0 marks] 5. An astronaut tosses a tennis ball vertically upwards, while walking on another planet. The height of the tennis ball, z(t) (in m), over time t (in s) can be described as: z(t) =.6 t + 15.6 t + 1 Use differentiation to determine the maximum height that the ball reaches and when it reaches this height. Report the height to 3 significant figures and the time to significant figures. [15 marks] 6. This empirical fit describes the annual emissions, E, of carbon dioxide (CO) by fossil fuel burning and cement use (in Pg a -1 in carbon equivalents): E = 0.147 X +.688 Here X are the years since 1958, with X = 0 for 1958 and X = 57 for 015. By integration calculate the cumulative CO emissions over this 57-year period. Report your answer to 4 significant figures. [15 marks] END OF PAPER ENV-4014Y Version 1

Appendix A Appendix A 8 Pages ENV-4014Y Formulae Sheet SI BASE QUANTITIES Base Quantity Name Quantity Symbol Base Unit Name Base Unit Symbol Length l metre m Mass m kilogram kg Time t second s Electric current I ampere A Thermodynamic temperature T kelvin K Amount of substance n mole mol Luminous intensity I v candela cd SI PREFIXES Multiple Prefix Name Prefix Symbol Multiple Prefix Name Prefix Symbol 10 18 exa E 10-1 deci d 10 15 peta P 10 - centi c 10 1 tera T 10-3 milli m 10 9 giga G 10-6 micro µ 10 6 mega M 10-9 nano n 10 3 kilo k 10-1 pico p 10 hecto h 10-15 femto f 10 1 deca da 10-18 atto a The following units are no longer correct. Unit Symbol SI Equivalent micron 1 micrometre (1 m) Ångström Å 0.1 nanometres (0.1 nm) 1 of 8

Appendix A 8 Pages SI DERIVED AND SUPPLEMENTARY QUANTITIES SI Derived Quantity SI Derived SI Derived Unit SI Equivalent Unit Name Name Plane angle radian rad m m -1 Solid angle steradian sr m m - Frequency hertz Hz s -1 Force newton N kg m s - Pressure, stress pascal Pa N m - Energy, work, heat joule J N m Power, radiant flux watt W J s -1 Electric charge coulomb C A s Electric potential volt V J C -1 Electric resistance ohm V A -1 Electric conductance siemens S -1 Electric capacitance farad F C V -1 Magnetic flux weber Wb V s Magnetic flux density tesla T Wb m - (= kg A -1 s - ) Luminous flux lumen lm cd sr Illuminance lux lx lm m - Area m Volume m 3 Velocity m s -1 Acceleration m s - Angular velocity rad s -1 Angular acceleration rad s - Wave number m -1 Density kg m -3 Concentration (amount) mol m -3 Specific volume m 3 kg -1 Weight N (= kg m s - ) Momentum kg m s -1 Moment of force, torque kg m s - Moment of inertia kg m Surface tension N m -1 Viscosity (dynamic) N s m - ( = Pa s) Viscosity (kinematic) m s -1 Heat flux density W m - Heat capacity, entropy J K -1 Specific heat capacity J kg -1 K -1 Thermal conductivity W m -1 K -1 Specific energy J kg -1 Energy density J m -3 Electric field strength V m -1 Conductivity S m -1 Magnetic field strength A m -1 Magnetic susceptibility m 3 Luminance cd m - Radiance W m - sr Irradiance W m - Radiant intensity W sr -1 of 8

Appendix A 8 Pages MENSURATION AND GEOMETRY Circle Radius r, diameter d Area, Circumference, Length of arc, A = πr = π d 4 C = πr = πd L = rθ (in radians) Sphere Where r is the radius of the sphere, a and b are the radii of the two sections, and h is the perpendicular distance between the sections. Volume, V = 4 3 πr3 Surface area, S = 4 πr Volume of the zone of a sphere, Surface area of the zone of a sphere, V z = πh 6 (3a + 3b + h ) S z = πa + πb + πrh ERROR CALCULATION Errors of Sums and Differences For P P P R S R R S S and P ΔP R ΔR S ΔS Errors of Products and Quotients For P P R R S S and P ΔP R ΔR S ΔS P P R R S S Errors of Powers For P P R R a S S b P P R a R S b S 3 of 8

ALGEBRA Appendix A 8 Pages Quadratic Equation If ax bx c 0, b x ( b 4ac) a Indices a m a n = a (m+n) m a mn a n a m n mn ( a ) a m 1 a m a a a 1 m n / m a 0 1 m m a a n Binomial Expansion a b = (a + b)(a b) a 3 b 3 = (a b)(a + ab + b ) a 3 + b 3 = (a + b)(a ab + b ) (a + b) = a + ab + b (a + b) 3 = a 3 + 3a b + 3ab + b 3 Three laws of logarithms log a X + log a Y = log a XY log a X log a Y = log a X Y nlog a X = log a X n Exponential growth/decay X(t) = X 0 e rt e =.718 (exponential constant) The quantity X(t) depends exponentially on time t for growth rate r with X0 the initial amount and base e. 4 of 8

Appendix A 8 Pages TRIGONOMETRICAL IDENTITIES AND EQUATIONS sin θ = cos θ = opposite hypotenuse = BC AC adjacent hypotenuse = AB AC tan θ = opposite adjacent = BC AB tan θ = sin θ cos θ 360 o = π radians 1 o = π/180 radians 1 radian = 180/π degrees sin(α + β) = sin α cos β + sin β cos α cos(α + β) = cos α cos β sin α sin β tan(α + β) = tan α + tan β 1 tan α tan β sin(α β) = sin α cos β sin β cos α cos(α β) = cos α cos β sin α sin β tan(α β) = tan α tan β 1 + tan α tan β α + β β β + β sin α + sin β = sin ( ) cos (α ) sin α sin β = sin (α ) cos (α ) α + β β β + β cos α + cos β = cos ( ) cos (α ) cos α cos β = sin (α ) sin (α ) sin α = sin( α) tan α = tan( α) sin α + cos α = 1 cos α = cos( α) tan α = tan α 1 tan α sin α = sin α cos α cos α = (cos α) (cos β) = cos α sin α = cos α 1 = 1 sin α EQUATION OF A STRAIGHT LINE Gradient m, making intercept c on y-axis: y = mx + c Two lines, with m and m 1 as gradients, are perpendicular, if m = 1 m 1 5 of 8

Appendix A 8 Pages CONVERSION FACTORS FOR NON-SI UNITS Quantity Unit Name Unit Symbol SI Equivalent Length 1 mil 0.054 mm 1 inch 1 in 5.4 mm 1 foot 1 ft 0.3048 m 1 yard 1 yd 0.9144 m 1 mile 1 mi 1.609344 km 1 nautical mile 1 nmi, M 1.85 km Area 1 square inch 1 in 6.4516 x 10 mm 1 square foot 1 ft 9.9030 x 10 - m 1 square yard 1 yd 0.83617 m 1 square mile 1 mi.58999 km 1 acre 1 ac 4046.8564 m 1 hectare 1 ha 10 4 m Volume 1 cubic inch 1 in 3 1.63871 x 10 4 mm 3 1 cubic foot 1 ft 3 0.083168 m 3 1 cubic yard 1 yd 3 0.764555 m 3 1 UK pint 1 pt 0.568615 dm 3 1 UK gallon 1 imp gal 4.54609 x 10-3 m 3 1 US gallon 1 US gal 3.784118 x 10-3 m 3 Mass 1 ounce 1 oz 8.395 g 1 pound 1 lb 0.4535937 kg 1 slug 14.5939 kg 1 tonne 1 t 10 3 kg Density 1 pound per cubic foot 1 lb ft -3 16.0185 kg m -3 Velocity 1 foot per second 1 ft s -1 0.3048 m s -1 1 kilometre per hour 1 km h -1 0.778 m s -1 1 mile per hour 1 mph, mi h -1 0.44704 m s -1 1 knot 1 kn 0.51444 m s -1 Angular measure 1 1 degree 0.01745 rad 1 1 minute.90889 x 10-4 rad 1 1 second 4.84814 x 10-6 rad Force 1 dyne 1 dyn 10-5 N 1 poundal 1 pdl 0.13855 N 1 pound-force 1 lbf 4.448 N Torque 1 pound-force inch 1 lbf in 0.11985 N m 1 pound-force foot 1 lbf ft 1.355818 N m Pressure 1 pound-force per square foot 1 lbf ft - 0.0478803 kpa 1 foot of water 1 ft HO.98898 kpa 1 millimetre of mercury 1 mm Hg 133.3 Pa 1 torr 1 Torr 133.3 Pa 1 kilogram-force per square 1 kgf cm - 98.0665 kpa centimetre 1 bar 1 bar 100 kpa = 10 5 Pa 1 atmosphere 1 atm 101.35 kpa 6 of 8

Appendix A 8 Pages Quantity Unit Name Unit Symbol SI Equivalent Viscosity 1 poise 1 P 0.1 N s m - (dynamic) 1 pound per foot per second 1 lb ft -1 s -1 1.48816 N s m - Viscosity 1 stokes 1 St 10-4 m s -1 (kinematic) Energy 1 kilowatt hour 1 kw h 3.6 MJ 1 British Thermal Unit 1 Btu 1.05506 kj 1 therm 1 thm 105.506 MJ 1 kilocalorie 1 kcal 4.1868 kj 1 British Thermal Unit per 1 Btu lb -1.36 kj kg -1 pound 1 foot poundal ft-pdl 1 ft-pdl 0.04140 J Power 1 horse power 1 hp 0.745700 kw 1 British Thermal Unit per 1 Btu h -1 0.93071 W hour 1 British Thermal Unit per 1 Btu ft - 11.35657 kj m - square foot 1 British Thermal Unit per 1 Btu h -1 ft -1 3.15460 W m - hour per foot 1 calorie per 1 cal cm - 41.8680 kj m - Magnetic field 1 gauss 1 G 10-4 T 1 oersted 1 Oe (10 3 /4π) A m -1 Discharge 1 cubic foot per second 1 ft 3 s -1.83 x 10 - m 3 s -1 1 US gallon per minute 1 US gal min -1 6.309 x 10-5 m 3 s -1 Intrinsic 1 square foot 1 ft 9.90 x 10 - m permeability 1 darcy 1 d 9.870 x 10-13 m Hydraulic 1 foot per second 1 ft s -1 3.048 x 10-1 m s -1 conductivity 1 US gallon per day per 1 US gal d -1 ft - 4.70 x 10-7 m s -1 square foot Transmissivity 1 square foot per second 1 ft s -1 9.90 x 10 - m s -1 1 US gallon per day per foot 1 US gal d -1 ft -1 1.438 x 10-7 m s -1 7 of 8

Appendix A 8 Pages CALCULUS Differentiation y = f(x) y = f (x) Notes constant 0 x 1 x x x n nx n 1 e x e x e kx ke kx k is constant sin x cos x cos x sin x sin kx k cos kx k is constant cos kx k sin kx k is constant ln kx 1 x k is constant Integration f(x) f(x)dx Notes k, constant x n kx + c 1 n + 1 xn+1 + c 1 x ln x + c e x e kx sin x cos x sin kx cos kx e x + c e kx k + c cos x + c sin x + c cos kx + c k sin kx k + c k is constant k is constant k is constant Note: For technical reasons, when finding the gradient/integral functions of trigonometrical functions the angle x must always be measured in radians. 8 of 8

Exam feedback and model answers ENV-4014Y Advanced Quantitative Skills Dorothee Bakker and Johannes Laube Maths exam of 7 June 018 Most common feedback: Confusion between significant figures and decimal places: For example, 5.888 has 3 decimal places and 5 significant figures. Significant figures refers to all the numbers, except zeros at the beginning and the end. Decimal places refers to the numbers after the decimal point. Rounding errors were common: 5.567 rounds to 6 (1 s.f.), 5.6 ( s. f.) and 5.57 (3 s. f.). 5.45 rounds to 5 (1 s. f.), 5.5 ( s. f.) and 5.45 (3 s. f.). Scientific notation In scientific notation 0.0000000079 M is written as 7.9 x10-9 M. Most answers required a unit. On Q1: - Correct conversion for the force, but not the area (both the number and the unit need to be squared). - Result not converted to SI base units (needed here are metres, kilograms and seconds). - The answer should have been given to two decimal places. - It is poundal-force PER square inch, so the two converted quantities should have been divided by each other. - Kilogram is the unit of mass, not force. - Well done for answering in a sentence. - Two attempts on the question, but none crossed out. - Individual steps not explained. - Answer not given in Scientific Notation. - Pressure was mistaken for torque. On Q: - Two minor comments are that a) the unit of r is missing and b)the answer could have been expressed as a sentence. - A logarithm needs to be taken for rearranging an equation to solve for a term in an exponent. - A(5730) should have been replaced with A(0)/. - Exponents in a numerator cannot be cancelled out with numbers in a denominator. - A(5,730) is not the same as A times 5,730 - A(0) is not zero at t = 0. - Two attempts on the question, but none crossed out. - Multiple errors in rearranging an equation. - A(t) and A(0) are amounts not times. 1

On Q3: - The formula to solve a quadratic equation has been recognised as important but is only needed for the second part of the question. - The first part of the question is calculated correctly but incorrectly rounded. - The second part of the question requires the formula to solve a quadratic equation. On Q4: Many students struggled with the calculation of the hydrogen ion concentration, notably with the negative sign before the logarithm. In scientific notation 0.0000000079 M is written as 7.9 x10-9 M. The scientific notation was required for reporting the hydrogen ion concentration. This is an increase in the hydrogen ion concentration. In fact, the hydrogen ion concentration almost doubles. Hence the name ocean acidification. The percentage relative change is calculated as the difference of the new value and the original value, divided by the original value, times 100 percent. Full marks were also awarded to calculation of the percentage relative change as the new value divided by the original value, times 100 percent. Many students reversed the order of the new value and the original value in the calculation. On Q5: Many students did well on this question. This question requires taking the first derivative, setting it to zero and solving for time. Entry of this time in the original equation for the height as a function of time then gives the maximum height of the tennis ball. Mistakes included errors in the first derivative and in solving the equation for time. A few students correctly calculated the time, but then confused time with height. On Q6: This question required definite integration for the period of 0 to 57 years. Some students struggled to integrate the equation. Other students did not clearly carry out definite integration. Cumulative CO emissions over a period of time have the unit Pg, not Pg a -1 (or Pg yr -1 ). The unit 'loses' the 'per year' as a result of the integration over time (here years).

Exam questions with answers 1. Convert 0.3 psi (pound-force per square inch) into SI base units and give your answer in scientific notation and to decimal places. [10 marks] 1 pound-force = 4.448 N (from the ENV-4014Y Formulae Sheet) 1 pound-force = 4.448 kg m s - 1 inch = 5.4 mm (from the ENV-4014Y Formulae Sheet) 1 inch = 5.4 10-3 m 1 square inch = (5.4 10-3 m) 1 square inch = (5.4 10-3 ) m 1 square inch = 6.4516 10-4 m 0.3 (pound-force) / (1 square inch) = 0.3 (4.448 kg m s - ) / (6.4516 10-4 m ) 0.3 (pound-force) / (1 square inch) =.1 10 3 kg m -1 s -. The amount A(t) of the radioactive substance carbon-14 (or 14 C) decays according to the exponential function A(t) = A 0 e rt where A 0 is the initial amount (at t = 0) and t is the time in years (t 0). The half-life of carbon-14 is 5,730 years. In other words, the amount A of this radioactive substance in 5,730 years is equal to half of its initial amount A 0. Find r (round to 3 significant figures). [0 marks] List variables with values: A 0 =? A(5,730) = 1 A 0 t = 5,730 r =? Substitute values in the formula: A(t) = A 0 e rt 3

A(5,730) = A 0 e r 5,730 Make r the subject and solve for r: 1 A 0 = A 0 e 5,730r e 5,730r = 0.5 ln(e 5,730r ) = ln(0.5) 5,730r = ln(0.5) r = ln(0.5) 5,730 = 0.6931 5,730 = 1.1 10 4 a 1 (3 s. f. ) or, since scientific notation was not specifically asked for: r = 0.00011 a 1. 3. The Aral Sea used to be one of the biggest lakes in the world, but it has dramatically lost water since the middle of the last century. This loss of water can be described - from 1973 to 010 - by the empirically derived quadratic function f(x) = y = 0.550 x,1.7 x +,3,810 where y is the water volume in km 3 and x is the year. Determine the water volume of the Aral Sea in 1973 and in 010 and calculate in which year the volume would return to 1973 levels if the above function is used for prediction purposes. [0 marks] Firstly we need to find f(1973) and f(010): f(1973) = 0.550 (1973),1.7 1973 +,3,810 = 858.64038 km 3 = 859 km 3 (The number of s.f. of the result of Addition or Subtraction is determined by rightmost common decimal position.) f(010) = 0.550 (010),1.7 010 +,3,810 = 91.1 km 3 = 91 km 3 Secondly we are asked to find x for f(1973) which we just determined to be 859 km 3. First write the equation: 859 = 0.550 x,1.7 x +,3,810 Then rearrange to give the form ax + bx + c = 0: 4

0 = 0.550 x,1.7 x +,3,810 859 = 0.550 x,1.7 x +,,951 Then solve using the general formula given by: x = b± b 4ac a x = (,1.7) ± (,1.7) 4 0.550,,951 0.550 x = 1973.000 or x = 047.701 The question asks for the date of return to 1973 water volume levels, so only the second solution really makes sense. The correct answer sentence is: The water volume of the Aral Sea was 859 km 3 in 1973 and 91 km 3 in 010 and volumes would return to 1973 levels in 047 if the above function is used for prediction purposes. N.B. This is unfortunately unlikely to happen as water volumes do not appear to be recovering significantly to date. 4. Ocean uptake of carbon dioxide (CO) leads to ocean acidification as the extra dissolved carbon dioxide reacts with water to form carbonic acid. The ph, a measure of acidity, is defined as follows: ph = log 10 ([H + ]) where [H + ] is the hydrogen ion concentration (in mol l -1 or M). The global mean surface ocean ph was 8.1 in the year 000. Global mean surface ocean ph has been predicted to decrease by 0.3 ph units by the year 100 relative to the year 000, if CO emissions from human activity continue unabated (in a business as usual scenario ). What was the hydrogen ion concentration in the year 000 and what would it become in 100? Use scientific notation and significant figures. Is this an increase or a decrease in the hydrogen ion concentration over time? Express the change in the hydrogen ion concentration from the year 000 to the year 100 as a relative change in percent. Report this percentage to 3 significant figures. [0 marks] ph = log 10 ([H + ]) For 000 ph 000 = log 10 ([H + ] 000 ) = 8.1 Multiply both sides of the equation with -1. 5

log 10 ([H + ] 000 ) = 8.1 [H + ] 000 = 10 8.1 M = 7.9438 x 10-9 = 7.9 x 10-9 M ( s.f.) For 100 ph 100 = ph 000 0.3 = 7.8 Repeat the calculation above for the ph value in 100. ph 100 = log 10 ([H + ] 100 ) = 7.8 Multiply both sides of the equation with -1. log 10 ([H + ] 100 ) = 7.8 [H + ] 100 = 10 7.8 M = 1.58489 x 10-8 = 1.6 x 10-8 M ( s.f.) This is an increase in the hydrogen ion concentration. In fact, the hydrogen ion concentration almost doubles. This is why the process is called ocean acidification. Relative change as a percentage = 100% [H+ ] 100 [H + ] 000 [H + ] 000 = 100% 10 7.8 10 8.1 10 8.1 = 100% (10 ( 7.8 ( 8.1)) 10 ( 8.1 ( 8.1)) ) Relative change (%) = 100% (10 0.3 10 0 ) = 100% (1.9953 1) = 99.53% = 99.5% The hydrogen ion concentration was 7.9 10-9 M in 000 and would become 1.6 10-8 M in 100 ( significant figures, scientific notation). The relative increase in the hydrogen ion concentration would be 99.5% by the year 100 (3 significant figures). Use of the rounded concentrations for the hydrogen-ion concentration gives 10.53% which rounds to 103% (3 significant figures), a result with a slightly lower precision. An alternative answer (awarded full marks) is: Percentage = 100% [H+ ] 100 [H + ] 000 6

= 100% 10 7.8 10 8.1 = 100% 10 ( 7.8+8.1) Percentage = 100% 10 0.3 = 100% 1.9953 = 00% The hydrogen ion concentration was 7.9 10-9 M in 000 and would become 1.6 10-8 M in 100 ( significant figures, scientific notation). The hydrogen ion concentration in 100 would be 00% of the concentration in 000 (3 significant figures). Thus, the hydrogen ion concentration would nearly double by 100. Reference: Caldeira, K. and M. E. Wickett (003) Anthropogenic carbon and ocean ph. Nature 45: 365. 5. An astronaut tosses a tennis ball vertically upwards, while walking on another planet. The height of the tennis ball, z(t) (in m), over time t (in s) can be described as: z(t) =.6 t + 15.6 t + 1 Use differentiation to determine the maximum height that the ball reaches and when it reaches this height. Report the height to 3 significant figures and the time to significant figures. [15 marks] By differentiation work out the vertical velocity of the tennis ball as a function of time. The vertical velocity is the first derivative of the height over time. v(t) = z (t) = 5. t + 15.6 (in m s -1 ) The velocity is zero at the highest point. z (t) = 5. t + 15.6 = 0 t = 15.6 5. = 3 s z (t) = 5. The second derivative is negative, confirming that this is a maximum. zmax = z(3) =.6 3 + 15.6 3 + 1 = 4.4 m t = 3.0 s ( s. f.) zmax = 4.4 m (3 s. f.) 7

The ball reaches a maximum height of 4.4 meters (3 significant figures) after 3.0 seconds ( significant figures). 6. This empirical fit describes the annual emissions, E, of carbon dioxide (CO) by fossil fuel burning and cement use (in Pg a -1 in carbon equivalents): E = 0.147 X +.688 Here X are the years since 1958, with X = 0 for 1958 and X = 57 for 015. By integration calculate the cumulative CO emissions over this 57-year period. Report your answer to 4 significant figures. [15 marks] Cumulative emissions = 57 0 (0.147 X +.688) dx = 0.147 X +.688 X + c 0 57 = ( 0.147 57 +.688 57 + c 0.147 c ) 0.688 0 = ( 0.147 57 +.688 57) = 331.8968 = 331.9 Pg The cumulative CO emissions were 331.9 Pg in carbon equivalents over this 57 year period. Note: Data from: Le Quéré et al. (016) Global Carbon Budget 016. Earth System Science Data 8: 605-649. doi:10.5194/essd-8-605-016. END OF PAPER 8