📝 Integrating rates of change net change (25 MCQs)

📖 From Calculus • 6. Integration • 25 questions available

←Previous Topic:Mean value theorem for integrals📝 Integrating rates of change net changeRectilinear motion revisted using integration:Next Topic→

What is Integrating rates of change net change?

Definition:
The Net Change Theorem states that the integral of a rate of change gives the net change in the quantity: $\int_a^b F'(x) \, dx = F(b) - F(a)$ . It applies to velocity, growth rates, and flow rates.

Example:
If water flows into a tank at $r(t) = 2t$ L/min, net change in volume from t=1 to t=3 is $\int_1^3 2t \, dt = [t^2]_1^3 = 9 - 1 = 8$ Liters.

Reason:
It connects derivatives (rates) to integrals (totals), allowing calculation of total accumulation from known instantaneous rates, fundamental in dynamics and fluid mechanics.

Easy

Medium

Hard

📝 All Integrating rates of change net change MCQs

Q1. A population of bacteria grows at a rate of $r(t) = 500e^{0.2t}$ cells per hour. What is the net change in the population from $t=0$ to $t=5$ hours?

\int_0^5 500e^{0.2t} dt

✅

\int_0^5 500e^{0.2t} dt + C

500e^{0.2(5)} - 500e^{0.2(0)}

500e^{0.2(5)} + C

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The net change in a quantity over an interval is given by the definite integral of its rate of change over that interval. Options C and D represent the antiderivative evaluated at a single point, not the change over the interval. Option B incorrectly includes a constant of integration, which cancels out in a definite integral. The correct expression is the definite integral from 0 to 5, as stated in option A.

Q2. A particle moves along a line with velocity $v(t) = 4t - 12$ m/s. The displacement from $t=1$ to $t=5$ is 0, but the distance traveled is 16 m. What is the most plausible explanation?

A.The particle’s velocity changes sign, so displacement and distance differ. ✅

B.The velocity is constant, so displacement and distance are equal.

C.The particle’s position at

t=1

and

t=5

is the same, so the distance must also be 0.

D.The acceleration is zero, so displacement and distance are the same.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Displacement is the integral of velocity (net signed area), while distance traveled is the integral of speed (total area). If the velocity changes sign, these two values can differ. Option C is a common misconception: if displacement is zero, the distance traveled can still be positive. Options B and D ignore the possibility of sign changes in velocity.

Q3. Water flows into a tank at a rate of $r(t) = 10 + 2t$ L/min. At $t=0$ , the tank contains 100 L. How much water is in the tank at $t=10$ minutes?

100 + \int_0^{10} (10 + 2t) dt

✅

\int_0^{10} (10 + 2t) dt

100 + 10 + 2(10)

100 + \int_0^{10} (10 + 2t) dt + C

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The total amount at time $T$ is the initial amount plus the integral of the rate of change from 0 to $T$ . Option B gives only the net change, not the final amount. Option C is an incorrect calculation. Option D includes a constant of integration, which is unnecessary since the initial condition is specified. Option A correctly adds the initial amount to the definite integral.

Q4. The rate of change of the volume of a balloon is $V'(t) = 8\pi t$ cm³/s. Which statement is true?

A.The volume function is

V(t) = 4\pi t^2 + C

. ✅

B.The volume function is

V(t) = 4\pi t^2

C.The volume function is

V(t) = 8\pi

D.The volume function is

V(t) = 8\pi t^2 + C

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since $V'(t) = 8\pi t$ , the volume function is the antiderivative: $V(t) = 4\pi t^2 + C$ . Option B omits the constant of integration, which is necessary unless an initial condition is given. Option C is the derivative of the volume function, and option D incorrectly integrates the constant factor. Option A correctly represents the general antiderivative.

Q5. A student claims that if the rate of change of a function is always positive on an interval, then the function itself must be increasing on that interval. Is this always true, and why?

A.Yes, because a positive derivative implies an increasing function. ✅

B.No, because the function could be negative but increasing.

C.Yes, because the function’s values will be positive.

D.No, because the integral of a positive function is always zero.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If the derivative $f'(x) > 0$ for all $x$ in an interval, then by the Mean Value Theorem, $f$ is strictly increasing on that interval. Option B is a misconception: a function can be increasing even if its values are negative. Option C confuses the sign of the function with its rate of change. Option D is false, as the integral of a positive function is positive, not zero. Option A correctly applies the relationship between the sign of the derivative and the monotonicity of the function.

Q6. Given the graph of $f'(x)$ below, which integral correctly represents the change in $f(x)$ from $x=0$ to $x=4$ ? (Assume the graph shows a positive area of 10 units² and a negative area of 6 units².)

\int_0^4 f&#039;(x) dx = 4

✅

\int_0^4 f&#039;(x) dx = 16

\int_0^4 f&#039;(x) dx = 10

\int_0^4 f&#039;(x) dx = 6

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The change in $f(x)$ is given by the net signed area under $f'(x)$ . The net signed area is the positive area minus the negative area: $10 - 6 = 4$ . Option B incorrectly adds the areas. Option C ignores the negative area. Option D ignores the positive area. Option A correctly computes the net change.

Q7. A car’s velocity is given by $v(t) = 20 - 0.5t$ m/s for $0 \le t \le 40$ . The displacement is 400 m, but the distance traveled is 400 m as well. What does this imply?

A.The car never changes direction. ✅

B.The car changes direction exactly once.

C.The car’s acceleration is zero.

D.The car’s speed is always positive.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: If displacement equals distance traveled, the velocity must not change sign; the car moves in one direction. Options B and D are incorrect because if the velocity changes sign, displacement and distance would differ. Option C is false because the acceleration is constant and non-zero. Option A correctly identifies that the car never changes direction.

Q8. The rate of growth of a tree is $r(t) = 2t + 1$ cm/year. The tree’s height at $t=0$ is 5 cm. What is the height at $t=3$ years?

5 + \int_0^3 (2t + 1) dt = 17

cm ✅

\int_0^3 (2t + 1) dt = 12

5 + 2(3) + 1 = 12

5 + \int_0^3 (2t + 1) dt = 12

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The height is $h(3) = h(0) + \int_0^3 r(t) dt = 5 + [t^2 + t]_0^3 = 5 + (9 + 3) = 17$ cm. Option B gives only the net change. Option C misapplies the rate. Option D miscalculates the integral. Option A correctly computes the final height.

Q9. The marginal cost of producing $x$ units is $C'(x) = 3x^2 - 2x + 5$ dollars per unit. If the fixed cost is $C(0) = 100$ , what is the total cost to produce 10 units?

100 + \int_0^{10} (3x^2 - 2x + 5) dx

✅

\int_0^{10} (3x^2 - 2x + 5) dx

100 + 3(10)^2 - 2(10) + 5 = 385

100 + \int_0^{10} (3x^2 - 2x + 5) dx + C

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The total cost is $C(10) = C(0) + \int_0^{10} C'(x) dx$ . Option B omits the fixed cost. Option C incorrectly applies the rate. Option D includes an unnecessary constant. Option A correctly incorporates the fixed cost and the integral.

Q10. A function $F$ has $F'(x) = x^2 - 4$ . If $F(0) = 3$ , what is the change in $F$ from $x=0$ to $x=3$ ?

A.3

B.-3 ✅

C.6

D.0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The change is $F(3) - F(0) = \int_0^3 (x^2 - 4) dx = [x^3/3 - 4x]_0^3 = 9 - 12 = -3$ . Option A is $F(0)$ . Option C is the integral of the positive part. Option D is a common error. Option B is the correct net change.

Q11. The temperature of a cup of coffee changes at a rate of $T'(t) = -10e^{-0.2t}$ °C/min. If the initial temperature is $90^\circ$ C, which expression gives the temperature at $t=5$ ?

90 + \int_0^5 -10e^{-0.2t} dt

✅

\int_0^5 -10e^{-0.2t} dt

90 - 10e^{-0.2(5)}

90 + \int_0^5 -10e^{-0.2t} dt + C

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The temperature at time $T$ is $T(0) + \int_0^T T'(t) dt$ . Option B gives the change, not the final temperature. Option C is the value of the rate, not the temperature. Option D includes an unnecessary constant. Option A correctly adds the initial temperature to the integral.

Q12. A student evaluates $\int_0^2 v(t) dt$ to find displacement, but the problem asks for distance traveled. What mistake has the student made?

A.The student used velocity instead of speed. ✅

B.The student used speed instead of velocity.

C.The student forgot to add the initial position.

D.The student used acceleration instead of velocity.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Displacement is the integral of velocity (which accounts for direction), while distance traveled is the integral of speed (absolute value of velocity). Option B would be correct for displacement. Options C and D are unrelated to the core error. Option A correctly identifies the mistake.

Q13. The rate of water loss from a tank is $r(t) = 5t$ L/min. If the tank initially has 200 L, what is the volume after 4 minutes?

200 - \int_0^4 5t dt = 160

L ✅

\int_0^4 5t dt = 40

200 + \int_0^4 5t dt = 240

200 - 5(4) = 180

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The rate is a loss, so $V(t) = V(0) - \int_0^t r(u) du$ . Option C incorrectly adds the integral. Option B gives only the loss. Option D misapplies the rate. Option A correctly subtracts the integral from the initial volume.

Q14. If $f'(x) = 2x$ and $f(0) = 1$ , what is $f(2)$ ?

A.5 ✅

B.4

C.3

D.6

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: $f(x) = \int 2x dx = x^2 + C$ . Using $f(0) = 1$ , $C = 1$ . Thus, $f(2) = 4 + 1 = 5$ . Option B is $x^2$ without the constant. Option C is a common error. Option D is incorrect. Option A correctly applies the initial condition.

Q15. A particle’s acceleration is $a(t) = 6t$ m/s². If $v(0) = 2$ m/s, what is the velocity at $t=3$ ?

v(3) = 2 + \int_0^3 6t dt = 29

m/s ✅

v(3) = \int_0^3 6t dt = 27

m/s

v(3) = 2 + 6(3) = 20

m/s

v(3) = 2 + 3(3)^2 = 29

m/s

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Velocity is the integral of acceleration: $v(t) = v(0) + \int_0^t a(u) du$ . Option B omits the initial velocity. Option C misapplies the formula. Option D is a misEasy. Option A correctly integrates and adds the initial velocity.

Q16. The rate of change of the area of a circle is $A'(r) = 2\pi r$ . If $A(0) = 0$ , what is $A(5)$ ?

25\pi

✅

10\pi

5\pi

0

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: $A(r) = \int 2\pi r dr = \pi r^2 + C$ . With $A(0) = 0$ , $C = 0$ , so $A(5) = 25\pi$ . Option B is the derivative. Option C is a common error. Option D is incorrect. Option A correctly integrates the rate of change.

Q17. A population grows at a rate of $r(t) = 100e^{0.1t}$ per year. The initial population is 500. What is the population after 10 years?

500 + \int_0^{10} 100e^{0.1t} dt

✅

\int_0^{10} 100e^{0.1t} dt

500 + 100e^{0.1(10)}

500 + \int_0^{10} 100e^{0.1t} dt + C

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The population is $P(10) = P(0) + \int_0^{10} r(t) dt$ . Option B gives only the net change. Option C incorrectly evaluates the rate. Option D includes an unnecessary constant. Option A correctly adds the initial population.

Q18. If $f'(x) = 3x^2 - 6x + 2$ , and $f(1) = 0$ , what is the change in $f$ from $x=1$ to $x=3$ ?

A.8 ✅

B.4

C.12

D.0

💡 Difficulty: medium | ✅ Correct: A

Q19. A student argues that since the integral of a rate of change gives the net change, the constant of integration is always needed. Is the student correct?

A.No, the constant cancels when evaluating a definite integral. ✅

B.Yes, the constant is always part of the general antiderivative.

C.Yes, because every function has infinitely many antiderivatives.

D.No, because the definite integral does not exist.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For a definite integral, the constant of integration cancels out: $F(b) + C - (F(a) + C) = F(b) - F(a)$ . Option B is true for indefinite integrals. Option C is a misconception. Option D is false. Option A correctly explains why the constant is unnecessary for definite integrals.

Q20. The rate of change of a function $f$ is given by $f'(x) = 2x - 3$ . If $f(2) = 4$ , which of the following is the correct expression for $f(5)$ ?

4 + \int_2^5 (2x - 3) dx

✅

\int_2^5 (2x - 3) dx

4 + 2(5) - 3 = 11

4 + \int_0^5 (2x - 3) dx - \int_0^2 (2x - 3) dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The Fundamental Theorem of Calculus states $f(b) = f(a) + \int_a^b f'(x) dx$ . Option B gives only the change. Option C misapplies the function. Option D is equivalent to A, but more complex. Option A is the direct and correct Easy.

Q21. A particle’s velocity is given by $v(t) = t^2 - 4t + 3$ . Over the interval [0, 4], what is the distance traveled? (Hint: Find where velocity changes sign.)

\int_0^1 (t^2 - 4t + 3) dt - \int_1^3 (t^2 - 4t + 3) dt + \int_3^4 (t^2 - 4t + 3) dt

✅

\int_0^1 (t^2 - 4t + 3) dt + \int_1^3 (t^2 - 4t + 3) dt + \int_3^4 (t^2 - 4t + 3) dt

\int_0^4 (t^2 - 4t + 3) dt

\int_1^3 (t^2 - 4t + 3) dt

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Distance traveled is the integral of speed, $|v(t)|$ . The velocity changes sign at $t = 1$ and $t = 3$ . Option B and C do not account for sign changes correctly. Option D only considers part of the interval. Option A correctly splits the integral into subintervals where the sign is constant.

Q22. The rate of change of a function $f$ is $f'(x) = \sin x$ . If $f(0) = 2$ , what is $f(\pi)$ ?

A.2 ✅

B.0

\pi

2 + \int_0^\pi \sin x dx = 4

💡 Difficulty: medium | ✅ Correct: A

Q23. The velocity of a particle is $v(t) = 6t - 2t^2$ . The particle’s displacement from $t=0$ to $t=3$ is 9. What is the distance traveled?

A.9

B.More than 9 ✅

C.Less than 9

D.Cannot be determined

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: If velocity changes sign, displacement and distance differ. Here, $v(t) = 2t(3 - t)$ , which is positive for $0 < t < 3$ . Since velocity is non-negative, displacement equals distance, so distance is also 9. Option B is incorrect because the distance is exactly 9. Options C and D are incorrect. Option A correctly states that displacement equals distance when velocity does not change sign.

Q24. A student uses $\int_0^5 f'(x) dx$ to find the final value of a function. What is the student missing?

A.The initial value of the function. ✅

B.The constant of integration.

C.The derivative of the function.

D.The variable of integration.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The integral alone gives the net change, not the final value. The final value is $f(5) = f(0) + \int_0^5 f'(x) dx$ . Option B is a common misconception. Options C and D are not the main issue. Option A correctly identifies the missing initial condition.

Q25. The rate of change of the radius of a spherical balloon is $r'(t) = 2$ cm/s. If the initial radius is 5 cm, what is the radius after 3 seconds?

r(t) = 5 + 2t

;

r(3) = 11

r(t) = 2t + C

;

r(3) = 6

r(t) = 5 + \int_0^t 2 du = 5 + 2t

;

r(3) = 11

cm ✅

r(t) = 2t

;

r(3) = 6

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: If $r'(t) = 2$ , then $r(t) = 2t + C$ . Using $r(0) = 5$ , $C = 5$ , so $r(3) = 11$ . Option A is correct but lacks the integral notation. Option B is incomplete. Option D omits the initial radius. Option C correctly expresses the solution with the integral.

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