Numerical Integration - IB Mathematics Applications and Interpretation

1. The rate at which a social media post gains likes is modelled by $L'(t) = 50t^2 e^{-0.5t}$ likes per hour, for $t \ge 0$. To the nearest hour, how long does it take for the post to accumulate its first 500 likes?
   Answer: 8 hours We need to find the value of $T$ for which the total number of likes, $\int_{0}^{T} L'(t) \,dt$, is equal to 500. We need to solve the equation $\int_{0}^{T} 50t^2 e^{-0.5t} \,dt = 500$. Using a graphing calculator's solver or by finding the intersection of $y = \int_{0}^{x} 50t^2 e^{-0.5t} \,dt$ and $y = 500$, we find $T \approx 8.16$ hours. To the nearest hour, this is 8 hours.
2. A company's daily profit is changing at a rate of $P'(t) = 50 - 3t^2$ hundreds of dollars per day, where $t$ is the number of days from the start of a new campaign. What is the net change in profit over the first 5 days of the campaign?
   Answer: $12,500 When you see a rate of change function like $P'(t)$, you're dealing with a derivative that tells you how fast profit is changing at each moment. To find the total change over an interval, you need to integrate this rate function - this is a fundamental application of the Fundamental Theorem of Calculus. To find the net change in profit over the first 5 days, integrate $P'(t) = 50 - 3t^2$ from $t = 0$ to $t = 5$: $\int_0^5 (50 - 3t^2) dt = [50t - t^3]_0^5 = (50 \cdot 5 - 5^3) - (0) = 250 - 125 = 125$ Since the rate is given in hundreds of dollars per day, the result is 125 hundreds of dollars, which equals $12,500. Looking at the wrong answers: Answer B (-$2500) likely comes from a sign error or incorrectly calculating the definite integral. Answer C ($16,667) might result from using an incorrect antiderivative or evaluation error. Answer D ($20,833) could come from forgetting to subtract $t^3$ when finding the antiderivative, using only the $50t$ term. The correct answer is A ($12,500). Remember this key principle: when given a rate function, always integrate to find total change over an interval. Also, pay careful attention to units - here the rate was in "hundreds of dollars," so don't forget to convert your final numerical answer to actual dollars.
3. Find the volume of wood required to make the bowl, in cm³.
   Answer: 686 The volume of a solid of revolution about the x-axis is given by the formula $V = \pi \int_a^b y^2 \,dx$. In this case, we must calculate $V = \pi \int_{1}^{6} (0.2(x-1)^2 + 3)^2 \,dx$. Using a calculator's numerical integration function, the integral evaluates to approximately 218.3. Therefore, $V \approx \pi \times 218.3 \approx 685.8$ cm³.
4. Two particles, A and B, start from the same position and move along the same line. Their velocities are given by $v_A(t) = 10 + 2\cos(t)$ and $v_B(t) = 11$, for $t \ge 0$. What is the distance between the two particles at time $t=4\pi$?
   Answer: $4\pi$ The position of each particle is the integral of its velocity. The distance between them is the absolute difference in their positions. Position of A is $x_A(t) = \int_0^{4\pi} (10+2\cos(t)) \,dt = [10t+2\sin(t)]_0^{4\pi} = 40\pi$. Position of B is $x_B(t) = \int_0^{4\pi} 11 \,dt = [11t]_0^{4\pi} = 44\pi$. The distance between them is $|44\pi - 40\pi| = 4\pi$.
5. The rate at which water leaks from a storage tank is modelled by the function $R(t) = 15e^{-0.2t}$ litres per hour, where $t$ is the number of hours since the leak started. Find the total amount of water, in litres, that has leaked from the tank between $t=3$ and $t=7$.
   Answer: 25.9 The total amount of water leaked between $t=3$ and $t=7$ is the definite integral of the rate function $R(t)$ over this interval. The calculation is $\int_{3}^{7} 15e^{-0.2t} \,dt$. Using a graphing calculator, this integral evaluates to approximately 25.9 litres.
6. The marginal cost of producing $x$ computer chips is given by $C'(x) = 40 - 0.1x + 0.0015x^2$ dollars per chip. Calculate the increase in cost when production is increased from 200 chips to 300 chips.
   Answer: $5500 The increase in cost is the definite integral of the marginal cost function from $x=200$ to $x=300$. The integral to calculate is $\int_{200}^{300} (40 - 0.1x + 0.0015x^2) \,dx$. Using a calculator's numerical integration feature, the result is $5500.
7. Given that $f(x)$ is a continuous and positive function for all $x$, and $g(x) = \int_a^x f(t) \,dt$ for some constant $a$. Which statement about the function $g(x)$ must be true?
   Answer: The function $g(x)$ is always increasing. When you encounter questions about functions defined by integrals, focus on the Fundamental Theorem of Calculus and what it tells you about the derivative of the accumulation function. Since $g(x) = \int_a^x f(t) \,dt$, the Fundamental Theorem of Calculus gives us $g'(x) = f(x)$. Because we're told that $f(x)$ is positive for all $x$, this means $g'(x) > 0$ everywhere. When a function has a positive derivative throughout its domain, the function must be strictly increasing. This confirms that option A is correct. Let's examine why the other options fail. Option B claims $g(x)$ is concave up, but concavity depends on the second derivative $g''(x) = f'(x)$. Since we know nothing about whether $f(x)$ is increasing or decreasing, we can't determine the sign of $f'(x)$, so $g(x)$ could be concave up, concave down, or neither. Option C suggests $g(x)$ is always positive, but $g(a) = \int_a^a f(t) \,dt = 0$. While $g(x)$ increases from this point, it starts at zero, not above it. Option D claims there's a local minimum at $x = a$. While $g(a) = 0$ and $g(x)$ increases from there, this isn't technically a local minimum since the function doesn't decrease before $x = a$ and then increase—it simply starts increasing from that point. Study tip: Always differentiate accumulation functions using the Fundamental Theorem of Calculus. The sign of the integrand directly tells you whether the accumulation function is increasing or decreasing.
8. The rate of ice melting on a glacier is given by $M(t) = 0.5 + \sin^2(\frac{\pi t}{12})$ billion tonnes per month, where $t$ is the number of months from the start of the year. Approximately how much ice melts in total during the first 6 months of the year?
   Answer: 4.50 billion tonnes The total amount of ice melted is the definite integral of the melting rate from $t=0$ to $t=6$. We need to calculate $\int_0^6 (0.5 + \sin^2(\frac{\pi t}{12})) \,dt$. Using a calculator's numerical integration feature, this integral evaluates to approximately 4.50.
9. The velocity of a particle moving along a line is given by $v(t) = t^2 - 7t + 10$ metres per second, for $t \ge 0$. Find the particle's total displacement during the first 6 seconds.
   Answer: 6.00 m Displacement is the definite integral of the velocity function over the given time interval. We need to calculate $\int_{0}^{6} (t^2 - 7t + 10) \,dt$. Using a calculator, this evaluates to $[\frac{t^3}{3} - \frac{7t^2}{2} + 10t]_0^6 = (72 - 126 + 60) - 0 = 6$. The displacement is 6 metres.
10. The area of a healing wound is decreasing at a rate of $A'(t) = -15(t+1)^{-2}$ square millimetres per day. What is the total decrease in the area of the wound from the end of day 1 ($t=1$) to the end of day 4 ($t=4$)?
    Answer: 4.50 mm² When you encounter a rate of change function and need to find the total change over an interval, you're looking at a definite integral problem. The rate $A'(t) = -15(t+1)^{-2}$ tells you how fast the wound area is changing at any time $t$, so integrating this rate gives you the total change. To find the total decrease from day 1 to day 4, you need to evaluate $\int_1^4 A'(t) \, dt = \int_1^4 -15(t+1)^{-2} \, dt$. First, find the antiderivative: $\int -15(t+1)^{-2} \, dt = -15 \cdot \frac{(t+1)^{-1}}{-1} = 15(t+1)^{-1} = \frac{15}{t+1}$ Now evaluate the definite integral: $\left[\frac{15}{t+1}\right]_1^4 = \frac{15}{4+1} - \frac{15}{1+1} = \frac{15}{5} - \frac{15}{2} = 3 - 7.5 = -4.5$ Since we want the decrease (positive value), the total decrease is 4.50 mm². Answer A (4.50 mm²) is correct. Answer B (3.75 mm²) likely comes from an integration error or incorrect bounds. Answer C (6.00 mm²) might result from forgetting the negative sign in the rate function. Answer D (9.00 mm²) could come from doubling the correct answer or making multiple calculation errors. Remember: when given a rate function, always integrate to find total change. Pay careful attention to signs—negative rates mean decreasing quantities, but the question asks for the amount of decrease, which should be positive.
11. A company's initial investment of $50,000 is expected to generate a continuous income stream at a rate of $R(t) = 15000e^{0.04t}$ dollars per year, where $t$ is in years. Find the total income generated during the first five years.
    Answer: $83,214 The total income generated over a period is the definite integral of the income rate function. We need to calculate $\int_{0}^{5} 15000e^{0.04t} \,dt$. Using a calculator, this integral is approximately $83,214. The initial investment is extra information not needed for this calculation.
12. The rate of change of a country's population is modelled by $P'(t)$, where $P$ is the population in millions and $t$ is the number of years since 2020. What is the correct interpretation of the statement $\int_{5}^{10} P'(t) \,dt = 2.5$?
    Answer: The population increased by a total of 2.5 million between 2025 and 2030. When you encounter an integral of a derivative like $\int_{5}^{10} P'(t) \,dt$, you're looking at one of the most fundamental relationships in calculus: the Fundamental Theorem of Calculus. This theorem tells us that integrating a rate of change function gives us the total change over that interval. Since $P'(t)$ represents the rate of population change, $\int_{5}^{10} P'(t) \,dt = P(10) - P(5)$. This equals 2.5, meaning the population at $t = 10$ (year 2030) minus the population at $t = 5$ (year 2025) is 2.5 million. In other words, the population increased by a total of 2.5 million between 2025 and 2030, making D correct. Option A confuses the change in population with the actual population value. The integral doesn't tell us what the population was in 2030, only how much it changed. Option B misinterprets the integral as an average population. To find average population, you'd need $\frac{1}{5}\int_{5}^{10} P(t) \,dt$, not the integral of the derivative. Option C confuses total change with average rate of change. The average rate would be $\frac{\int_{5}^{10} P'(t) \,dt}{10-5} = \frac{2.5}{5} = 0.5$ million per year. Remember this key pattern: whenever you see $\int_{a}^{b} f'(x) \,dx$, it always equals $f(b) - f(a)$—the total change in the original function. This appears frequently on IB exams in contexts involving rates and accumulation.
13. The rate of profit for two products, A and B, are modelled by $P_A(t) = 60e^{-0.1t}$ and $P_B(t) = 20 + 2t$ respectively, in thousands of dollars per year. Find the total excess profit of product A over product B during the time interval $t=0$ to $t=5$ years.
    Answer: $81,400 The excess profit is the integral of the difference between the two profit rates. We need to calculate $\int_{0}^{5} (P_A(t) - P_B(t)) \,dt = \int_{0}^{5} (60e^{-0.1t} - (20 + 2t)) \,dt$. Using a calculator, this integral is approximately 81.4. Since the units are in thousands of dollars, the total excess profit is $81,400.

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