IB Mathematics: Analysis and Approaches Quiz: Integration Basics
20 questions · exam conditions
0:00
Integration BasicsQuestion 1 of 20

Let f′′(x)=6xf''(x) = 6x. The graph of f(x)f(x) has a tangent line with equation y=2x+3y=2x+3 at the point where x=1x=1. Find f(x)f(x).

x3−x+5x^3 - x + 5
x3+2x+2x^3 + 2x + 2
x3−x+3x^3 - x + 3
3x2−13x^2 - 1
← Back to quizzes

IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Integration Basics

Practice Integration Basics in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Integration Basics, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let f′′(x)=6xf''(x) = 6x. The graph of f(x)f(x) has a tangent line with equation y=2x+3y=2x+3 at the point where x=1x=1. Find f(x)f(x).

  1. x3−x+5x^3 - x + 5 (correct answer)
  2. x3+2x+2x^3 + 2x + 2
  3. x3−x+3x^3 - x + 3
  4. 3x2−13x^2 - 1
Explanation: First, integrate f′′(x)=6xf''(x)=6x to find f′(x)f'(x): f′(x)=∫6xdx=3x2+C1f'(x) = \int 6x dx = 3x^2 + C_1. The gradient of the tangent line y=2x+3y=2x+3 is 2. This means that at x=1x=1, the gradient of f(x)f(x) is 2. So, f′(1)=2f'(1)=2. Substitute this into the expression for f′(x)f'(x): 3(1)2+C1=2⇒3+C1=2⇒C1=−13(1)^2 + C_1 = 2 \Rightarrow 3+C_1=2 \Rightarrow C_1=-1. Thus, f′(x)=3x2−1f'(x) = 3x^2-1. Now integrate f′(x)f'(x) to find f(x)f(x): f(x)=∫(3x2−1)dx=x3−x+C2f(x) = \int (3x^2-1) dx = x^3 - x + C_2. To find C2C_2, we need a point on the curve f(x)f(x). At x=1x=1, the point is on the tangent line, so y=2(1)+3=5y=2(1)+3=5. The point is (1,5)(1,5). Substitute this into the expression for f(x)f(x): f(1)=13−1+C2=5⇒0+C2=5⇒C2=5f(1) = 1^3 - 1 + C_2 = 5 \Rightarrow 0+C_2=5 \Rightarrow C_2=5. Therefore, f(x)=x3−x+5f(x) = x^3 - x + 5.

Question 2

A particle moves along a straight line with velocity v(t)=8−2tv(t) = 8-2t m/s for t≥0t \ge 0. What is the total distance travelled by the particle in the first 6 seconds?

  1. 12 m
  2. 16 m
  3. 20 m (correct answer)
  4. 24 m
Explanation: Total distance travelled is the integral of the speed, ∣v(t)∣|v(t)|. First, find when the velocity changes sign: v(t)=8−2t=0v(t) = 8-2t=0 gives t=4t=4. For 0≤t<40 \le t < 4, v(t)>0v(t)>0. For t>4t > 4, v(t)<0v(t)<0. The total distance is ∫06∣8−2t∣dt=∫04(8−2t)dt+∫46−(8−2t)dt\int_0^6 |8-2t| dt = \int_0^4 (8-2t) dt + \int_4^6 -(8-2t) dt. First integral: [8t−t2]04=(32−16)−0=16[8t-t^2]_0^4 = (32-16)-0 = 16. Second integral: ∫46(2t−8)dt=[t2−8t]46=(36−48)−(16−32)=−12−(−16)=4\int_4^6 (2t-8) dt = [t^2-8t]_4^6 = (36-48)-(16-32) = -12 - (-16) = 4. Total distance = 16+4=2016+4=20 m. Distractor A (12m) is the displacement ∫06v(t)dt\int_0^6 v(t) dt.

Question 3

The integral ∫02(x2−3x+2)dx\int_0^2 (x^2-3x+2) dx represents the net signed area between the curve y=x2−3x+2y=x^2-3x+2 and the x-axis from x=0x=0 to x=2x=2. What is the total area of the regions bounded by the curve and the x-axis over this interval?

  1. 16\frac{1}{6}
  2. 13\frac{1}{3}
  3. 56\frac{5}{6}
  4. 1 (correct answer)
Explanation: The curve is a parabola y=(x−1)(x−2)y=(x-1)(x-2) with roots at x=1x=1 and x=2x=2. Over the interval [0,2][0, 2], the curve is above the x-axis for x∈[0,1)x \in [0, 1) and below the x-axis for x∈(1,2]x \in (1, 2]. The total area is the sum of the absolute values of the integrals over these sub-intervals: Area = ∫01(x2−3x+2)dx+∣∫12(x2−3x+2)dx∣\int_0^1 (x^2-3x+2) dx + |\int_1^2 (x^2-3x+2) dx|. Let F(x)=x33−3x22+2xF(x) = \frac{x^3}{3} - \frac{3x^2}{2} + 2x. F(0)=0F(0)=0, F(1)=13−32+2=2−9+126=56F(1) = \frac{1}{3} - \frac{3}{2} + 2 = \frac{2-9+12}{6} = \frac{5}{6}, F(2)=83−122+4=83−6+4=83−2=23F(2) = \frac{8}{3} - \frac{12}{2} + 4 = \frac{8}{3} - 6 + 4 = \frac{8}{3} - 2 = \frac{2}{3}. First integral: F(1)−F(0)=56F(1)-F(0) = \frac{5}{6}. Second integral: F(2)−F(1)=23−56=4−56=−16F(2)-F(1) = \frac{2}{3} - \frac{5}{6} = \frac{4-5}{6} = -\frac{1}{6}. Total area = 56+∣−16∣=56+16=66=1\frac{5}{6} + |-\frac{1}{6}| = \frac{5}{6} + \frac{1}{6} = \frac{6}{6} = 1. The value of the single integral from 0 to 2 would be F(2)−F(0)=2/3F(2)-F(0) = 2/3, which is a possible distractor.

Question 4

The gradient of a curve y=f(x)y=f(x) is given by dydx=3x2+k\frac{dy}{dx} = 3x^2+k. The curve has a local minimum at the point (1,−4)(1, -4). Find the value of f(2)f(2).

  1. -3
  2. 0 (correct answer)
  3. 2
  4. 4
Explanation: At a local minimum, the gradient dydx\frac{dy}{dx} is 0. So, at x=1x=1, 3(1)2+k=0⇒3+k=0⇒k=−33(1)^2+k = 0 \Rightarrow 3+k=0 \Rightarrow k=-3. The gradient function is dydx=3x2−3\frac{dy}{dx} = 3x^2-3. To find f(x)f(x), we integrate: f(x)=∫(3x2−3)dx=x3−3x+Cf(x) = \int (3x^2-3) dx = x^3 - 3x + C. The curve passes through (1,−4)(1, -4), so we can find CC: −4=13−3(1)+C⇒−4=1−3+C⇒−4=−2+C⇒C=−2-4 = 1^3 - 3(1) + C \Rightarrow -4 = 1-3+C \Rightarrow -4 = -2+C \Rightarrow C=-2. So the equation of the curve is f(x)=x3−3x−2f(x) = x^3-3x-2. We need to find f(2)f(2): f(2)=23−3(2)−2=8−6−2=0f(2) = 2^3 - 3(2) - 2 = 8-6-2 = 0.

Question 5

An anti-derivative of f(x)=(2x+3)4f(x) = (2x+3)^4 is F(x)F(x). Which of the following could be F(x)F(x)?

  1. 8(2x+3)38(2x+3)^3
  2. (2x+3)55\frac{(2x+3)^5}{5}
  3. (2x+3)510\frac{(2x+3)^5}{10} (correct answer)
  4. 5(2x+3)55(2x+3)^5
Explanation: This requires using the reverse chain rule for integration. We are looking for ∫(2x+3)4dx\int (2x+3)^4 dx. Let u=2x+3u = 2x+3, then du=2dxdu = 2 dx or dx=12dudx = \frac{1}{2}du. The integral becomes ∫u4(12du)=12∫u4du=12u55+C=u510+C\int u^4 (\frac{1}{2} du) = \frac{1}{2} \int u^4 du = \frac{1}{2} \frac{u^5}{5} + C = \frac{u^5}{10} + C. Substituting back u=2x+3u=2x+3, we get (2x+3)510+C\frac{(2x+3)^5}{10} + C. Option C is a possible anti-derivative (with C=0C=0). Distractor B is a common error where the 1a\frac{1}{a} factor from the inner function's derivative is forgotten. Distractor A is the derivative, not the anti-derivative.

Question 6

The integral ∫02(x2−3x+2)dx\int_0^2 (x^2-3x+2) dx represents the net signed area between the curve y=x2−3x+2y=x^2-3x+2 and the x-axis from x=0x=0 to x=2x=2. What is the total area of the regions bounded by the curve and the x-axis over this interval?

  1. 16\frac{1}{6}
  2. 13\frac{1}{3}
  3. 56\frac{5}{6}
  4. 1 (correct answer)
Explanation: The curve is a parabola y=(x−1)(x−2)y=(x-1)(x-2) with roots at x=1x=1 and x=2x=2. Over the interval [0,2][0, 2], the curve is above the x-axis for x∈[0,1)x \in [0, 1) and below the x-axis for x∈(1,2]x \in (1, 2]. The total area is the sum of the absolute values of the integrals over these sub-intervals: Area = ∫01(x2−3x+2)dx+∣∫12(x2−3x+2)dx∣\int_0^1 (x^2-3x+2) dx + |\int_1^2 (x^2-3x+2) dx|. Let F(x)=x33−3x22+2xF(x) = \frac{x^3}{3} - \frac{3x^2}{2} + 2x. F(0)=0F(0)=0, F(1)=13−32+2=2−9+126=56F(1) = \frac{1}{3} - \frac{3}{2} + 2 = \frac{2-9+12}{6} = \frac{5}{6}, F(2)=83−122+4=83−6+4=83−2=23F(2) = \frac{8}{3} - \frac{12}{2} + 4 = \frac{8}{3} - 6 + 4 = \frac{8}{3} - 2 = \frac{2}{3}. First integral: F(1)−F(0)=56F(1)-F(0) = \frac{5}{6}. Second integral: F(2)−F(1)=23−56=4−56=−16F(2)-F(1) = \frac{2}{3} - \frac{5}{6} = \frac{4-5}{6} = -\frac{1}{6}. Total area = 56+∣−16∣=56+16=66=1\frac{5}{6} + |-\frac{1}{6}| = \frac{5}{6} + \frac{1}{6} = \frac{6}{6} = 1. The value of the single integral from 0 to 2 would be F(2)−F(0)=2/3F(2)-F(0) = 2/3, which is a possible distractor.

Question 7

The area of the region bounded by the curves y=x2+1y=x^2+1 and y=x+3y=x+3 is given by which integral?

  1. ∫−12(x2−x−2)dx\int_{-1}^2 (x^2-x-2) dx
  2. ∫−12(−x2+x+2)dx\int_{-1}^2 (-x^2+x+2) dx (correct answer)
  3. ∫−21(x2−x−2)dx\int_{-2}^1 (x^2-x-2) dx
  4. ∫−21(−x2+x+2)dx\int_{-2}^1 (-x^2+x+2) dx
Explanation: To find the area between two curves, we first need to find their points of intersection by setting the equations equal to each other: x2+1=x+3x^2+1 = x+3. This simplifies to x2−x−2=0x^2-x-2=0, which factors as (x−2)(x+1)=0(x-2)(x+1)=0. The points of intersection are at x=−1x=-1 and x=2x=2. These will be the limits of integration. To determine which function is greater on the interval [−1,2][-1, 2], we can test a point, for example x=0x=0. At x=0x=0, y=x2+1=1y=x^2+1=1 and y=x+3=3y=x+3=3. Since 3>13>1, the line y=x+3y=x+3 is the upper curve. The area is given by ∫−12(upper curve−lower curve)dx=∫−12((x+3)−(x2+1))dx=∫−12(−x2+x+2)dx\int_{-1}^2 (\text{upper curve} - \text{lower curve}) dx = \int_{-1}^2 ((x+3) - (x^2+1)) dx = \int_{-1}^2 (-x^2+x+2) dx.

Question 8

Evaluate ∫π/6π/3sin⁡(2x)dx\int_{\pi/6}^{\pi/3} \sin(2x) dx.

  1. −12-\frac{1}{2}
  2. 0
  3. 12\frac{1}{2} (correct answer)
  4. 1
Explanation: First, find the anti-derivative of sin⁡(2x)\sin(2x). Using the reverse chain rule, this is −12cos⁡(2x)-\frac{1}{2}\cos(2x). Now, apply the Fundamental Theorem of Calculus: [−12cos⁡(2x)]π/6π/3=(−12cos⁡(2⋅π3))−(−12cos⁡(2⋅π6))=−12cos⁡(2π3)+12cos⁡(π3)[-\frac{1}{2}\cos(2x)]_{\pi/6}^{\pi/3} = (-\frac{1}{2}\cos(2 \cdot \frac{\pi}{3})) - (-\frac{1}{2}\cos(2 \cdot \frac{\pi}{6})) = -\frac{1}{2}\cos(\frac{2\pi}{3}) + \frac{1}{2}\cos(\frac{\pi}{3}). We know cos⁡(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2} and cos⁡(2π3)=−12\cos(\frac{2\pi}{3}) = -\frac{1}{2}. Substituting these values gives: −12(−12))+12(12)=14+14=12-\frac{1}{2}(-\frac{1}{2})) + \frac{1}{2}(\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.

Question 9

The region R is bounded by the curve y=e−xy=e^{-x}, the x-axis, and the lines x=0x=0 and x=ln⁡(a)x=\ln(a), where a>1a>1. The area of R is 23\frac{2}{3}. Find the value of aa.

  1. e2/3e^{2/3}
  2. 53\frac{5}{3}
  3. 2
  4. 3 (correct answer)
Explanation: The area of the region R is given by the integral A=∫0ln⁡(a)e−xdxA = \int_0^{\ln(a)} e^{-x} dx. The anti-derivative of e−xe^{-x} is −e−x-e^{-x}. Evaluating the definite integral: [−e−x]0ln⁡(a)=(−e−ln⁡(a))−(−e−0)=−eln⁡(a−1)+e0=−a−1+1=1−1a[-e^{-x}]_0^{\ln(a)} = (-e^{-\ln(a)}) - (-e^{-0}) = -e^{\ln(a^{-1})} + e^0 = -a^{-1} + 1 = 1 - \frac{1}{a}. We are given that the area is 23\frac{2}{3}. So, 1−1a=231 - \frac{1}{a} = \frac{2}{3}. Solving for aa: 1a=1−23=13\frac{1}{a} = 1 - \frac{2}{3} = \frac{1}{3}. Therefore, a=3a=3.

Question 10

Find the area of the region enclosed by the graph of y=3x−x2y=3x-x^2 and the line y=xy=x.

  1. 23\frac{2}{3}
  2. 11
  3. 43\frac{4}{3} (correct answer)
  4. 22
Explanation: First, find the points of intersection: 3x−x2=x⇒2x−x2=0⇒x(2−x)=03x-x^2 = x \Rightarrow 2x-x^2 = 0 \Rightarrow x(2-x) = 0. The intersection points are at x=0x=0 and x=2x=2. In the interval [0,2][0, 2], we test a point, e.g., x=1x=1. The parabola gives y=3(1)−12=2y=3(1)-1^2=2 and the line gives y=1y=1. So, the parabola y=3x−x2y=3x-x^2 is the upper curve. The area is A=∫02((3x−x2)−x)dx=∫02(2x−x2)dxA = \int_0^2 ((3x-x^2) - x) dx = \int_0^2 (2x-x^2) dx. Now, integrate: [x2−x33]02=(22−233)−(0)=4−83=12−83=43[x^2 - \frac{x^3}{3}]_0^2 = (2^2 - \frac{2^3}{3}) - (0) = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3}.

Question 11

Find the total area of the regions enclosed by the curve y=x3−4xy=x^3-4x and the x-axis.

  1. 0
  2. 4
  3. 8 (correct answer)
  4. 16
Explanation: First, find the x-intercepts by setting y=0y=0: x3−4x=x(x2−4)=x(x−2)(x+2)=0x^3-4x = x(x^2-4) = x(x-2)(x+2) = 0. The intercepts are x=−2,0,2x=-2, 0, 2. The area is split into two regions. For x∈[−2,0]x \in [-2, 0], the curve is above the x-axis. For x∈[0,2]x \in [0, 2], the curve is below the x-axis. The total area is ∫−20(x3−4x)dx+∣∫02(x3−4x)dx∣\int_{-2}^0 (x^3-4x) dx + |\int_0^2 (x^3-4x) dx|. Calculate the first integral: [x44−2x2]−20=(0)−((−2)44−2(−2)2)=−(164−8)=−(4−8)=4[\frac{x^4}{4}-2x^2]_{-2}^0 = (0) - (\frac{(-2)^4}{4}-2(-2)^2) = -(\frac{16}{4}-8) = -(4-8) = 4. Calculate the second integral: [x44−2x2]02=(244−2(2)2)−(0)=(164−8)=4−8=−4[\frac{x^4}{4}-2x^2]_0^2 = (\frac{2^4}{4}-2(2)^2) - (0) = (\frac{16}{4}-8) = 4-8 = -4. The area for this part is ∣−4∣=4|-4|=4. The total area is 4+4=84+4=8.

Question 12

The area of the region bounded by the curves y=x2+1y=x^2+1 and y=x+3y=x+3 is given by which integral?

  1. ∫−12(x2−x−2)dx\int_{-1}^2 (x^2-x-2) dx
  2. ∫−12(−x2+x+2)dx\int_{-1}^2 (-x^2+x+2) dx (correct answer)
  3. ∫−21(x2−x−2)dx\int_{-2}^1 (x^2-x-2) dx
  4. ∫−21(−x2+x+2)dx\int_{-2}^1 (-x^2+x+2) dx
Explanation: To find the area between two curves, we first need to find their points of intersection by setting the equations equal to each other: x2+1=x+3x^2+1 = x+3. This simplifies to x2−x−2=0x^2-x-2=0, which factors as (x−2)(x+1)=0(x-2)(x+1)=0. The points of intersection are at x=−1x=-1 and x=2x=2. These will be the limits of integration. To determine which function is greater on the interval [−1,2][-1, 2], we can test a point, for example x=0x=0. At x=0x=0, y=x2+1=1y=x^2+1=1 and y=x+3=3y=x+3=3. Since 3>13>1, the line y=x+3y=x+3 is the upper curve. The area is given by ∫−12(upper curve−lower curve)dx=∫−12((x+3)−(x2+1))dx=∫−12(−x2+x+2)dx\int_{-1}^2 (\text{upper curve} - \text{lower curve}) dx = \int_{-1}^2 ((x+3) - (x^2+1)) dx = \int_{-1}^2 (-x^2+x+2) dx.

Question 13

The rate of change of a population PP is modelled by dPdt=kt\frac{dP}{dt} = k\sqrt{t}, where tt is the time in years. The initial population is P0P_0. Find the population P(t)P(t) at time tt.

  1. P0+k2tP_0 + \frac{k}{2\sqrt{t}}
  2. P0+kt3/2P_0 + k t^{3/2}
  3. P0+2k3t3/2P_0 + \frac{2k}{3}t^{3/2} (correct answer)
  4. P0+3k2t3/2P_0 + \frac{3k}{2}t^{3/2}
Explanation: To find the population P(t)P(t), we need to integrate the rate of change with respect to time: P(t)=∫ktdt=∫kt1/2dtP(t) = \int k\sqrt{t} dt = \int k t^{1/2} dt. Using the power rule for integration, this becomes kt1/2+11/2+1+C=kt3/23/2+C=2k3t3/2+Ck \frac{t^{1/2+1}}{1/2+1} + C = k \frac{t^{3/2}}{3/2} + C = \frac{2k}{3} t^{3/2} + C. We are given that the initial population is P0P_0, which means at t=0t=0, P(0)=P0P(0) = P_0. Substituting this in: P(0)=2k3(0)3/2+C=P0P(0) = \frac{2k}{3}(0)^{3/2} + C = P_0, which implies C=P0C=P_0. Therefore, the population at time tt is P(t)=P0+2k3t3/2P(t) = P_0 + \frac{2k}{3}t^{3/2}.

Question 14

The region R is bounded by the curve y=e−xy=e^{-x}, the x-axis, and the lines x=0x=0 and x=ln⁡(a)x=\ln(a), where a>1a>1. The area of R is 23\frac{2}{3}. Find the value of aa.

  1. e2/3e^{2/3}
  2. 53\frac{5}{3}
  3. 2
  4. 3 (correct answer)
Explanation: The area of the region R is given by the integral A=∫0ln⁡(a)e−xdxA = \int_0^{\ln(a)} e^{-x} dx. The anti-derivative of e−xe^{-x} is −e−x-e^{-x}. Evaluating the definite integral: [−e−x]0ln⁡(a)=(−e−ln⁡(a))−(−e−0)=−eln⁡(a−1)+e0=−a−1+1=1−1a[-e^{-x}]_0^{\ln(a)} = (-e^{-\ln(a)}) - (-e^{-0}) = -e^{\ln(a^{-1})} + e^0 = -a^{-1} + 1 = 1 - \frac{1}{a}. We are given that the area is 23\frac{2}{3}. So, 1−1a=231 - \frac{1}{a} = \frac{2}{3}. Solving for aa: 1a=1−23=13\frac{1}{a} = 1 - \frac{2}{3} = \frac{1}{3}. Therefore, a=3a=3.

Question 15

The value of ∫−aa(x3cos⁡x+1)dx\int_{-a}^a (x^3 \cos x + 1) dx is 6. What is the value of aa?

  1. 3 (correct answer)
  2. 6
  3. 6\sqrt{6}
  4. Cannot be determined
Explanation: The integral can be split: ∫−aax3cos⁡xdx+∫−aa1dx\int_{-a}^a x^3 \cos x dx + \int_{-a}^a 1 dx. The function f(x)=x3cos⁡xf(x) = x^3 \cos x is an odd function because f(−x)=(−x)3cos⁡(−x)=−x3cos⁡x=−f(x)f(-x) = (-x)^3 \cos(-x) = -x^3 \cos x = -f(x). The integral of an odd function over a symmetric interval [−a,a][-a, a] is 0. Therefore, the expression simplifies to ∫−aa1dx\int_{-a}^a 1 dx. This evaluates to [x]−aa=a−(−a)=2a[x]_{-a}^a = a - (-a) = 2a. We are given that the value is 6, so 2a=62a=6, which means a=3a=3.

Question 16

Given that ∫14f(x)dx=6\int_1^4 f(x) dx = 6 and ∫14g(x)dx=−2\int_1^4 g(x) dx = -2, find the value of ∫41(f(x)−2g(x))dx\int_4^1 (f(x) - 2g(x)) dx.

  1. -10 (correct answer)
  2. -2
  3. 2
  4. 10
Explanation: Using the properties of definite integrals: ∫41(f(x)−2g(x))dx=−∫14(f(x)−2g(x))dx\int_4^1 (f(x) - 2g(x)) dx = - \int_1^4 (f(x) - 2g(x)) dx. Using the linearity of integrals: −[∫14f(x)dx−2∫14g(x)dx])-[\int_1^4 f(x) dx - 2\int_1^4 g(x) dx]). Substitute the given values: −[6−2(−2)])=−[6+4]=−10-[6 - 2(-2)]) = -[6+4] = -10. Distractor D (10) arises from ignoring the negative sign from swapping the limits of integration. Distractor B (-2) comes from an error like 6−2(2)=26 - 2(2) = 2 then flipping the sign to -2.

Question 17

A particle moves along a straight line with velocity v(t)=8−2tv(t) = 8-2t m/s for t≥0t \ge 0. What is the total distance travelled by the particle in the first 6 seconds?

  1. 12 m
  2. 16 m
  3. 20 m (correct answer)
  4. 24 m
Explanation: Total distance travelled is the integral of the speed, ∣v(t)∣|v(t)|. First, find when the velocity changes sign: v(t)=8−2t=0v(t) = 8-2t=0 gives t=4t=4. For 0≤t<40 \le t < 4, v(t)>0v(t)>0. For t>4t > 4, v(t)<0v(t)<0. The total distance is ∫06∣8−2t∣dt=∫04(8−2t)dt+∫46−(8−2t)dt\int_0^6 |8-2t| dt = \int_0^4 (8-2t) dt + \int_4^6 -(8-2t) dt. First integral: [8t−t2]04=(32−16)−0=16[8t-t^2]_0^4 = (32-16)-0 = 16. Second integral: ∫46(2t−8)dt=[t2−8t]46=(36−48)−(16−32)=−12−(−16)=4\int_4^6 (2t-8) dt = [t^2-8t]_4^6 = (36-48)-(16-32) = -12 - (-16) = 4. Total distance = 16+4=2016+4=20 m. Distractor A (12m) is the displacement ∫06v(t)dt\int_0^6 v(t) dt.

Question 18

Find the area of the region enclosed by the graph of y=3x−x2y=3x-x^2 and the line y=xy=x.

  1. 23\frac{2}{3}
  2. 11
  3. 43\frac{4}{3} (correct answer)
  4. 22
Explanation: First, find the points of intersection: 3x−x2=x⇒2x−x2=0⇒x(2−x)=03x-x^2 = x \Rightarrow 2x-x^2 = 0 \Rightarrow x(2-x) = 0. The intersection points are at x=0x=0 and x=2x=2. In the interval [0,2][0, 2], we test a point, e.g., x=1x=1. The parabola gives y=3(1)−12=2y=3(1)-1^2=2 and the line gives y=1y=1. So, the parabola y=3x−x2y=3x-x^2 is the upper curve. The area is A=∫02((3x−x2)−x)dx=∫02(2x−x2)dxA = \int_0^2 ((3x-x^2) - x) dx = \int_0^2 (2x-x^2) dx. Now, integrate: [x2−x33]02=(22−233)−(0)=4−83=12−83=43[x^2 - \frac{x^3}{3}]_0^2 = (2^2 - \frac{2^3}{3}) - (0) = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3}.

Question 19

The region bounded by the graph of y=xy=\sqrt{x}, the x-axis, and the line x=ax=a has an area of 18. Find the value of aa.

  1. 6
  2. 9 (correct answer)
  3. 12
  4. 36
Explanation: The area is given by the integral ∫0axdx=∫0ax1/2dx\int_0^a \sqrt{x} dx = \int_0^a x^{1/2} dx. Evaluating the integral: [x3/23/2]0a=[23x3/2]0a=23a3/2−0=23a3/2[\frac{x^{3/2}}{3/2}]_0^a = [\frac{2}{3}x^{3/2}]_0^a = \frac{2}{3}a^{3/2} - 0 = \frac{2}{3}a^{3/2}. We are given that this area is 18. So, 23a3/2=18\frac{2}{3}a^{3/2} = 18. Multiply by 32\frac{3}{2}: a3/2=18⋅32=27a^{3/2} = 18 \cdot \frac{3}{2} = 27. To find aa, we can write (a1/2)3=33(a^{1/2})^3 = 3^3, so a1/2=3a^{1/2} = 3. Squaring both sides gives a=9a=9.

Question 20

The area of the region enclosed by the graph of y=42x+1y = \frac{4}{2x+1}, the x-axis, and the lines x=0x=0 and x=kx=k is ln⁡(25)\ln(25). Find the value of kk, where k>0k > 0.

  1. 1
  2. 2 (correct answer)
  3. 4
  4. 12
Explanation: The area is given by the definite integral ∫0k42x+1dx\int_0^k \frac{4}{2x+1} dx. First, find the anti-derivative of 42x+1\frac{4}{2x+1}. This is 4⋅12ln⁡∣2x+1∣=2ln⁡∣2x+1∣4 \cdot \frac{1}{2} \ln|2x+1| = 2\ln|2x+1|. Now, evaluate the definite integral: [2ln⁡∣2x+1∣]0k=2ln⁡(2k+1)−2ln⁡(1)=2ln⁡(2k+1)[2\ln|2x+1|]_0^k = 2\ln(2k+1) - 2\ln(1) = 2\ln(2k+1). We are given that this area is ln⁡(25)\ln(25). So, 2ln⁡(2k+1)=ln⁡(25)2\ln(2k+1) = \ln(25). Using logarithm properties, this becomes ln⁡((2k+1)2)=ln⁡(25)\ln((2k+1)^2) = \ln(25). Therefore, (2k+1)2=25(2k+1)^2 = 25. Since k>0k>0, we take the positive root: 2k+1=52k+1 = 5, which gives 2k=42k=4 and k=2k=2.