Calculus 2 Quiz: Properties Of Definite Integrals
20 questions · exam conditions
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Properties Of Definite IntegralsQuestion 1 of 20

Let I1=∫01(1−x2)3dxI_1 = \int_0^1 (1-x^2)^3 dx and I2=∫01(1−x2)4dxI_2 = \int_0^1 (1-x^2)^4 dx. Which statement correctly compares the two integrals?

I1<I2I_1 < I_2
I1>I2I_1 > I_2
I1=I2I_1 = I_2
The relationship cannot be determined without evaluating the integrals.
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Calculus 2 Quiz

Calculus 2 Quiz: Properties Of Definite Integrals

Practice Properties Of Definite Integrals in Calculus 2 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 Properties Of Definite Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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.

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Question 1

Let I1=∫01(1−x2)3dxI_1 = \int_0^1 (1-x^2)^3 dx and I2=∫01(1−x2)4dxI_2 = \int_0^1 (1-x^2)^4 dx. Which statement correctly compares the two integrals?

  1. I1<I2I_1 < I_2
  2. I1>I2I_1 > I_2 (correct answer)
  3. I1=I2I_1 = I_2
  4. The relationship cannot be determined without evaluating the integrals.
Explanation: This question uses the comparison property of integrals. On the interval (0,1)(0, 1), the term 1−x21-x^2 is between 0 and 1. For any number yy such that 0<y<10 < y < 1, we have y3>y4y^3 > y^4. Therefore, for x∈(0,1)x \in (0, 1), we have (1−x2)3>(1−x2)4(1-x^2)^3 > (1-x^2)^4. At the endpoints x=0 and x=1, the two integrands are equal. Since one integrand is strictly greater than the other on the interior of the interval, its integral must also be greater. Thus, ∫01(1−x2)3dx>∫01(1−x2)4dx\int_0^1 (1-x^2)^3 dx > \int_0^1 (1-x^2)^4 dx, which means I1>I2I_1 > I_2.

Question 2

If ∫04h(x)dx=12\int_0^4 h(x) dx = 12 and ∫04h(x)dx=∫02h(x)dx+∫24h(x)dx\int_0^4 h(x) dx = \int_0^2 h(x) dx + \int_2^4 h(x) dx, and it's known that h(x)=h(4−x)h(x) = h(4-x) for all x∈[0,4]x \in [0,4], what is ∫02h(x)dx\int_0^2 h(x) dx?

  1. 66 (correct answer)
  2. 44
  3. 88
  4. 33
Explanation: The condition h(x)=h(4−x)h(x) = h(4-x) means hh is symmetric about x=2x = 2. Using substitution u=4−xu = 4-x in ∫24h(x)dx\int_2^4 h(x) dx gives ∫02h(4−u)du=∫02h(u)du=∫02h(x)dx\int_0^2 h(4-u) du = \int_0^2 h(u) du = \int_0^2 h(x) dx. Therefore, ∫02h(x)dx=∫24h(x)dx\int_0^2 h(x) dx = \int_2^4 h(x) dx, and since their sum is 12, each equals 6. Choice B uses incorrect interval length. Choice C assumes one interval contains all the area. Choice D incorrectly divides by 4 instead of 2.

Question 3

If f(x)f(x) is continuous on [a,b][a,b] and ∫abf(x)dx=8\int_a^b f(x) dx = 8, ∫acf(x)dx=3\int_a^c f(x) dx = 3 where a<c<ba < c < b, and ∫cdf(x)dx=−2\int_c^d f(x) dx = -2 where c<d<bc < d < b, what is the value of ∫dbf(x)dx\int_d^b f(x) dx?

  1. 77 (correct answer)
  2. 33
  3. 1313
  4. −3-3
Explanation: Using the additive property of definite integrals: ∫abf(x)dx=∫acf(x)dx+∫cdf(x)dx+∫dbf(x)dx\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^d f(x) dx + \int_d^b f(x) dx. Substituting known values: 8=3+(−2)+∫dbf(x)dx8 = 3 + (-2) + \int_d^b f(x) dx, so ∫dbf(x)dx=8−3+2=7\int_d^b f(x) dx = 8 - 3 + 2 = 7. Choice B incorrectly uses only ∫acf(x)dx\int_a^c f(x) dx. Choice C adds all given integrals without considering the additive property correctly. Choice D incorrectly assumes ∫dbf(x)dx=−∫acf(x)dx\int_d^b f(x) dx = -\int_a^c f(x) dx.

Question 4

Let r(x)r(x) be a continuous function such that ∫08r(x)dx=24\int_0^8 r(x) dx = 24. If s(x)=r(x)+r(8−x)s(x) = r(x) + r(8-x), what is ∫04s(x)dx\int_0^4 s(x) dx?

  1. 1212
  2. 2424 (correct answer)
  3. 4848
  4. 66
Explanation: ∫04s(x)dx=∫04[r(x)+r(8−x)]dx=∫04r(x)dx+∫04r(8−x)dx\int_0^4 s(x) dx = \int_0^4 [r(x) + r(8-x)] dx = \int_0^4 r(x) dx + \int_0^4 r(8-x) dx. For the second integral, let u=8−xu = 8-x, so du=−dxdu = -dx. When x=0x = 0, u=8u = 8. When x=4x = 4, u=4u = 4. Therefore: ∫04r(8−x)dx=∫84r(u)(−du)=∫48r(u)du\int_0^4 r(8-x) dx = \int_8^4 r(u)(-du) = \int_4^8 r(u) du. So ∫04s(x)dx=∫04r(x)dx+∫48r(x)dx=∫08r(x)dx=24\int_0^4 s(x) dx = \int_0^4 r(x) dx + \int_4^8 r(x) dx = \int_0^8 r(x) dx = 24. Choice A assumes we get half the original integral. Choice C doubles the result incorrectly. Choice D uses quarter of the original value.

Question 5

If f(x)f(x) is continuous on [0,6][0,6] and ∫02f(x)dx=A\int_0^2 f(x) dx = A, ∫24f(x)dx=B\int_2^4 f(x) dx = B, and ∫46f(x)dx=C\int_4^6 f(x) dx = C, which expression correctly represents ∫15f(x)dx\int_1^5 f(x) dx in terms of AA, BB, and CC?

  1. A+B+C−12A−12CA + B + C - \frac{1}{2}A - \frac{1}{2}C
  2. A+B+C−∫01f(x)dx−∫56f(x)dxA + B + C - \int_0^1 f(x) dx - \int_5^6 f(x) dx
  3. ∫12f(x)dx+B+∫45f(x)dx\int_1^2 f(x) dx + B + \int_4^5 f(x) dx (correct answer)
  4. B+12A+12CB + \frac{1}{2}A + \frac{1}{2}C
Explanation: When you encounter problems involving definite integrals over different intervals, the key principle is that integrals can be split and combined using interval additivity: ∫acf(x)dx=∫abf(x)dx+∫bcf(x)dx\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx. To find ∫15f(x)dx\int_1^5 f(x) dx, you need to break this interval into pieces that relate to your given information. The interval [1,5] spans from 1 to 5, so you can write: ∫15f(x)dx=∫12f(x)dx+∫24f(x)dx+∫45f(x)dx\int_1^5 f(x) dx = \int_1^2 f(x) dx + \int_2^4 f(x) dx + \int_4^5 f(x) dx Since ∫24f(x)dx=B\int_2^4 f(x) dx = B, this becomes ∫12f(x)dx+B+∫45f(x)dx\int_1^2 f(x) dx + B + \int_4^5 f(x) dx, which matches answer choice C exactly. Answer A incorrectly assumes that ∫01f(x)dx=12A\int_0^1 f(x) dx = \frac{1}{2}A and ∫56f(x)dx=12C\int_5^6 f(x) dx = \frac{1}{2}C. This would only be true if the function were constant over each interval, which isn't given. Answer B is actually correct in principle—it expresses the idea that you take the total integral and subtract the unwanted pieces—but it doesn't express the result "in terms of A, B, and C" as the question requires. Answer D makes the same incorrect assumption as A, presuming that half of each interval contributes proportionally to the integral value. Remember: when working with definite integrals over intervals, always use the additivity property to break complex intervals into manageable pieces that match your given information. Avoid assuming the function has special properties (like being constant) unless explicitly stated.

Question 6

Given that ∫05f(x)dx=10\int_0^5 f(x) dx = 10 and ∫05∣f(x)∣dx=16\int_0^5 |f(x)| dx = 16. What is the value of the integral of f(x) over the subset of [0,5][0,5] where f(x) is negative?

  1. -6
  2. 3
  3. -3 (correct answer)
  4. 13
Explanation: Let P be the region in [0,5][0,5] where f(x)≥0f(x) \ge 0 and N be the region where f(x)<0f(x) < 0. We have ∫05f(x)dx=∫Pf(x)dx+∫Nf(x)dx=10\int_0^5 f(x) dx = \int_P f(x) dx + \int_N f(x) dx = 10. Also, ∫05∣f(x)∣dx=∫Pf(x)dx+∫N(−f(x))dx=∫Pf(x)dx−∫Nf(x)dx=16\int_0^5 |f(x)| dx = \int_P f(x) dx + \int_N (-f(x)) dx = \int_P f(x) dx - \int_N f(x) dx = 16. Let A=∫Pf(x)dxA = \int_P f(x) dx and B=∫Nf(x)dxB = \int_N f(x) dx. We have the system of equations: A+B=10A+B=10 and A−B=16A-B=16. Subtracting the second equation from the first gives (A+B)−(A−B)=10−16(A+B)-(A-B) = 10-16, which simplifies to 2B=−62B = -6, so B=−3B = -3. The question asks for the integral over the region where f(x) is negative, which is B.

Question 7

Let f(x) be a continuous periodic function with period 4. If ∫04f(x)dx=10\int_0^4 f(x) dx = 10, what is the value of ∫210f(x)dx\int_2^{10} f(x) dx?

  1. 10
  2. 20 (correct answer)
  3. 25
  4. 30
Explanation: The interval of integration [2,10][2, 10] has a length of 10−2=810-2=8. Since the function has a period of 4, this interval represents two full periods of the function. For a periodic function f with period P, the integral over any interval of length P is the same. Therefore, the integral over an interval of length nP is n times the integral over one period. Here, the interval length is 8=2×48 = 2 \times 4, so ∫210f(x)dx=2×∫04f(x)dx=2×10=20\int_2^{10} f(x) dx = 2 \times \int_0^4 f(x) dx = 2 \times 10 = 20.

Question 8

If ∫abk(x)dx=15\int_a^b k(x) dx = 15 where k(x)k(x) is continuous, and we define m(x)=3k(2x+1)m(x) = 3k(2x+1), what is ∫a−12b−12m(x)dx\int_{\frac{a-1}{2}}^{\frac{b-1}{2}} m(x) dx?

  1. 152\frac{15}{2}
  2. 4545
  3. 3030
  4. 452\frac{45}{2} (correct answer)
Explanation: When you encounter integrals involving function transformations like m(x)=3k(2x+1)m(x) = 3k(2x+1), you need to carefully handle the substitution and scaling effects. To evaluate ∫a−12b−12m(x)dx=∫a−12b−123k(2x+1)dx\int_{\frac{a-1}{2}}^{\frac{b-1}{2}} m(x) dx = \int_{\frac{a-1}{2}}^{\frac{b-1}{2}} 3k(2x+1) dx, use substitution. Let u=2x+1u = 2x + 1, so du=2dxdu = 2dx or dx=12dudx = \frac{1}{2}du. When x=a−12x = \frac{a-1}{2}, we get u=2⋅a−12+1=a−1+1=au = 2 \cdot \frac{a-1}{2} + 1 = a-1+1 = a. When x=b−12x = \frac{b-1}{2}, we get u=2⋅b−12+1=b−1+1=bu = 2 \cdot \frac{b-1}{2} + 1 = b-1+1 = b. Substituting: ∫a−12b−123k(2x+1)dx=∫ab3k(u)⋅12du=32∫abk(u)du=32⋅15=452\int_{\frac{a-1}{2}}^{\frac{b-1}{2}} 3k(2x+1) dx = \int_a^b 3k(u) \cdot \frac{1}{2} du = \frac{3}{2} \int_a^b k(u) du = \frac{3}{2} \cdot 15 = \frac{45}{2} Answer choice (A) 152\frac{15}{2} incorrectly applies only the 12\frac{1}{2} factor from the substitution while ignoring the coefficient 3. (B) 45 mistakenly ignores the 12\frac{1}{2} factor from dx=12dudx = \frac{1}{2}du, keeping only 3×153 \times 15. (C) 30 likely comes from incorrectly using 2×152 \times 15, perhaps confusing the transformation coefficients. Study tip: When dealing with function transformations f(ax+b)f(ax+b), always use substitution and track both the coefficient scaling and the dxdx adjustment factor. The final result involves both the outer coefficient and the reciprocal of the inner coefficient.

Question 9

Let f(x) be an integrable function. Suppose ∫27f(x)dx=10\int_2^7 f(x) dx = 10 and ∫52f(x)dx=−4\int_5^2 f(x) dx = -4. What is the value of ∫57f(x)dx\int_5^7 f(x) dx?

  1. 6 (correct answer)
  2. 14
  3. -6
  4. -14
Explanation: First, use the property ∫baf(x)dx=−∫abf(x)dx\int_b^a f(x) dx = -\int_a^b f(x) dx. This means ∫25f(x)dx=−∫52f(x)dx=−(−4)=4\int_2^5 f(x) dx = -\int_5^2 f(x) dx = -(-4) = 4. Next, use the additivity property ∫27f(x)dx=∫25f(x)dx+∫57f(x)dx\int_2^7 f(x) dx = \int_2^5 f(x) dx + \int_5^7 f(x) dx. Substituting the known values gives 10=4+∫57f(x)dx10 = 4 + \int_5^7 f(x) dx. Solving for the unknown integral, we get ∫57f(x)dx=10−4=6\int_5^7 f(x) dx = 10 - 4 = 6.

Question 10

If ∫14f(x)dx=6\int_1^4 f(x) dx = 6, which of the following expressions represents the value of ∫12xf(x2)dx\int_1^2 x f(x^2) dx?

  1. 3 (correct answer)
  2. 6
  3. 12
  4. 2
Explanation: This problem can be solved with a u-substitution. Let u=x2u = x^2. Then du=2xdxdu = 2x dx, which means xdx=12dux dx = \frac{1}{2} du. We must also change the limits of integration. When x=1x=1, u=12=1u = 1^2 = 1. When x=2x=2, u=22=4u = 2^2 = 4. Substituting these into the integral gives: ∫12f(x2)(xdx)=∫14f(u)(12du)=12∫14f(u)du\int_1^2 f(x^2) (x dx) = \int_1^4 f(u) (\frac{1}{2} du) = \frac{1}{2} \int_1^4 f(u) du. Since ∫14f(x)dx=6\int_1^4 f(x) dx = 6, and the variable of integration is a dummy variable, ∫14f(u)du=6\int_1^4 f(u) du = 6. Therefore, the value is 12(6)=3\frac{1}{2} (6) = 3.

Question 11

If f(x) is a continuous function and ∫08f(x)dx=20\int_0^8 f(x) dx = 20, what is the value of ∫02f(4x)dx\int_0^2 f(4x) dx?

  1. 5 (correct answer)
  2. 10
  3. 20
  4. 80
Explanation: Use a u-substitution to evaluate ∫02f(4x)dx\int_0^2 f(4x) dx. Let u=4xu = 4x, so du=4dxdu = 4 dx, or dx=14dudx = \frac{1}{4} du. The limits of integration must be converted as well: when x=0x=0, u=4(0)=0u=4(0)=0; when x=2x=2, u=4(2)=8u=4(2)=8. The integral becomes ∫08f(u)14du=14∫08f(u)du\int_0^8 f(u) \frac{1}{4} du = \frac{1}{4} \int_0^8 f(u) du. Since we are given ∫08f(x)dx=20\int_0^8 f(x) dx = 20, and the variable of integration does not matter, ∫08f(u)du=20\int_0^8 f(u) du = 20. The result is 14(20)=5\frac{1}{4} (20) = 5.

Question 12

Let f(x) be a continuous function. If ∫09f(x)dx=4\int_0^9 f(x) dx = 4, what is the value of ∫03xf(x2)dx\int_0^3 x f(x^2) dx?

  1. 1
  2. 2 (correct answer)
  3. 4
  4. 8
Explanation: To evaluate ∫03xf(x2)dx\int_0^3 x f(x^2) dx, use a u-substitution. Let u=x2u = x^2, which means du=2xdxdu = 2x dx, or xdx=12dux dx = \frac{1}{2} du. The limits of integration must be converted: when x=0x=0, u=02=0u=0^2=0; when x=3x=3, u=32=9u=3^2=9. Substituting these into the integral gives ∫09f(u)(12du)=12∫09f(u)du\int_0^9 f(u) (\frac{1}{2} du) = \frac{1}{2} \int_0^9 f(u) du. We are given that ∫09f(x)dx=4\int_0^9 f(x) dx = 4. Since the variable of integration is a dummy variable, ∫09f(u)du=4\int_0^9 f(u) du = 4. Therefore, the value of the expression is 12(4)=2\frac{1}{2}(4) = 2.

Question 13

If ∫13f(x)dx=5\int_1^3 f(x) dx = 5, what is the value of ∫24[f(x−1)+3]dx\int_2^4 [f(x-1) + 3] dx?

  1. 5
  2. 8
  3. 11 (correct answer)
  4. 14
Explanation: First, use linearity to split the integral: ∫24f(x−1)dx+∫243dx\int_2^4 f(x-1) dx + \int_2^4 3 dx. For the first term, use a u-substitution with u=x−1u = x-1, so du=dxdu = dx. The limits of integration change from x=2,x=4x=2, x=4 to u=2−1=1,u=4−1=3u=2-1=1, u=4-1=3. The integral becomes ∫13f(u)du\int_1^3 f(u) du, which is given to be 5. The second term is ∫243dx=[3x]24=3(4)−3(2)=12−6=6\int_2^4 3 dx = [3x]_2^4 = 3(4) - 3(2) = 12 - 6 = 6. The total value is the sum of the two parts: 5+6=115 + 6 = 11.

Question 14

Given that f(x)=e−x2+1f(x) = e^{-x^2} + 1. Which of the following is a possible value for ∫02f(x)dx\int_0^2 f(x) dx?

  1. 2.0
  2. 4.1
  3. 1.9
  4. 3.5 (correct answer)
Explanation: When you encounter a definite integral that can't be solved analytically, you need to estimate its value using properties of the integrand and geometric reasoning. For f(x)=e−x2+1f(x) = e^{-x^2} + 1, notice that e−x2e^{-x^2} is always positive and decreases as xx moves away from zero. At x=0x = 0, e0=1e^{0} = 1, so f(0)=2f(0) = 2. At x=2x = 2, e−4≈0.018e^{-4} \approx 0.018, so f(2)≈1.018f(2) \approx 1.018. The function starts at 2, peaks there, and gradually decreases toward 1 as xx increases. To estimate ∫02f(x)dx\int_0^2 f(x) dx, consider bounds. The minimum value occurs near x=2x = 2 where f(x)≈1f(x) \approx 1, and the maximum occurs at x=0x = 0 where f(x)=2f(x) = 2. Since the function decreases smoothly, the integral should be between 2×1=22 \times 1 = 2 (if the function were constantly at its minimum) and 2×2=42 \times 2 = 4 (if constantly at its maximum). A reasonable estimate falls closer to the middle-upper range since the function stays relatively high over most of the interval. Answer choice A (2.0) represents the absolute minimum bound. Answer choice C (1.9) is impossibly low since even the minimum rectangle estimate gives 2.0. Answer choice B (4.1) exceeds the maximum possible bound of 4.0. Answer choice D (3.5) falls reasonably within our expected range. Study tip: When estimating definite integrals, always establish upper and lower bounds first by identifying the function's maximum and minimum values over the interval.

Question 15

Let f(x) be an odd function and g(x) be an even function. If ∫03f(x)dx=4\int_0^3 f(x) dx = 4 and ∫03g(x)dx=5\int_0^3 g(x) dx = 5, what is the value of ∫−33[f(x)+g(x)+2]dx\int_{-3}^3 [f(x) + g(x) + 2] dx?

  1. 10
  2. 12
  3. 22 (correct answer)
  4. 30
Explanation: Using the linearity property, the integral can be split: ∫−33f(x)dx+∫−33g(x)dx+∫−332dx\int_{-3}^3 f(x) dx + \int_{-3}^3 g(x) dx + \int_{-3}^3 2 dx. For an odd function f(x), ∫−aaf(x)dx=0\int_{-a}^a f(x) dx = 0. For an even function g(x), ∫−aag(x)dx=2∫0ag(x)dx=2(5)=10\int_{-a}^a g(x) dx = 2\int_0^a g(x) dx = 2(5) = 10. The integral of the constant is ∫−332dx=[2x]−33=2(3)−2(−3)=6−(−6)=12\int_{-3}^3 2 dx = [2x]_{-3}^3 = 2(3) - 2(-3) = 6 - (-6) = 12. Summing these results gives 0+10+12=220 + 10 + 12 = 22.

Question 16

Given that ∫−11f(x)dx=0\int_{-1}^1 f(x) dx = 0 and ∫01f(x)dx=3\int_0^1 f(x) dx = 3. What is the value of ∫−10(2f(x)+5)dx\int_{-1}^0 (2f(x)+5) dx?

  1. -1 (correct answer)
  2. 1
  3. 11
  4. -6
Explanation: First, use the additivity property: ∫−11f(x)dx=∫−10f(x)dx+∫01f(x)dx\int_{-1}^1 f(x) dx = \int_{-1}^0 f(x) dx + \int_0^1 f(x) dx. We are given ∫−11f(x)dx=0\int_{-1}^1 f(x) dx = 0 and ∫01f(x)dx=3\int_0^1 f(x) dx = 3. So, 0=∫−10f(x)dx+30 = \int_{-1}^0 f(x) dx + 3, which implies ∫−10f(x)dx=−3\int_{-1}^0 f(x) dx = -3. Now, we evaluate the required integral using linearity: ∫−10(2f(x)+5)dx=2∫−10f(x)dx+∫−105dx\int_{-1}^0 (2f(x)+5) dx = 2\int_{-1}^0 f(x) dx + \int_{-1}^0 5 dx. Substitute the value we found: 2(−3)+[5x]−10=−6+(5(0)−5(−1))=−6+5=−12(-3) + [5x]_{-1}^0 = -6 + (5(0) - 5(-1)) = -6 + 5 = -1.

Question 17

Let f(x) be an even function such that ∫02f(x)dx=5\int_0^2 f(x) dx = 5 and ∫04f(x)dx=12\int_0^4 f(x) dx = 12. What is the value of ∫−24f(x)dx\int_{-2}^4 f(x) dx?

  1. 7
  2. 17 (correct answer)
  3. 22
  4. 24
Explanation: We can split the integral using the additivity property: ∫−24f(x)dx=∫−20f(x)dx+∫04f(x)dx\int_{-2}^4 f(x) dx = \int_{-2}^0 f(x) dx + \int_0^4 f(x) dx. Since f(x) is an even function, we know that ∫−a0f(x)dx=∫0af(x)dx\int_{-a}^0 f(x) dx = \int_0^a f(x) dx. Therefore, ∫−20f(x)dx=∫02f(x)dx=5\int_{-2}^0 f(x) dx = \int_0^2 f(x) dx = 5. Substituting the known values, we get 5+12=175 + 12 = 17. An alternative correct method is ∫−24f(x)dx=∫−22f(x)dx+∫24f(x)dx=2∫02f(x)dx+(∫04f(x)dx−∫02f(x)dx)=2(5)+(12−5)=10+7=17\int_{-2}^4 f(x) dx = \int_{-2}^2 f(x) dx + \int_2^4 f(x) dx = 2\int_0^2 f(x) dx + (\int_0^4 f(x) dx - \int_0^2 f(x) dx) = 2(5) + (12-5) = 10+7=17.

Question 18

Let f(x) be a continuous function on [0,6][0, 6] such that 1≤f(x)≤31 \le f(x) \le 3 for all x. If ∫03f(x)dx=5\int_0^3 f(x) dx = 5, what is the tightest possible range for the value of I=∫06f(x)dxI = \int_0^6 f(x) dx?

  1. 6≤I≤186 \le I \le 18
  2. 3≤I≤93 \le I \le 9
  3. 8≤I≤148 \le I \le 14 (correct answer)
  4. 11≤I≤2311 \le I \le 23
Explanation: We can write I=∫06f(x)dx=∫03f(x)dx+∫36f(x)dxI = \int_0^6 f(x) dx = \int_0^3 f(x) dx + \int_3^6 f(x) dx. We are given ∫03f(x)dx=5\int_0^3 f(x) dx = 5. For the second integral, we use the comparison property on the interval [3,6][3, 6]. The width of this interval is 6−3=36-3=3. Since 1≤f(x)≤31 \le f(x) \le 3, we have 1(3)≤∫36f(x)dx≤3(3)1(3) \le \int_3^6 f(x) dx \le 3(3), which simplifies to 3≤∫36f(x)dx≤93 \le \int_3^6 f(x) dx \le 9. Adding 5 to all parts of this inequality gives 5+3≤5+∫36f(x)dx≤5+95+3 \le 5 + \int_3^6 f(x) dx \le 5+9, which results in 8≤I≤148 \le I \le 14.

Question 19

Let f(x)f(x) be a continuous function. If g(x)=∫1xf(t)dtg(x) = \int_1^x f(t) dt and ∫14f(x)dx=10\int_1^4 f(x) dx = 10, what is the value of g(4)−g(1)g(4) - g(1)?

  1. 0
  2. 9
  3. Cannot be determined.
  4. 10 (correct answer)
Explanation: This question tests your understanding of definite integrals and the Fundamental Theorem of Calculus. When you see a function defined as an integral with a variable upper limit, think about how definite integrals relate to the area under a curve. Let's work through this step by step. We have g(x)=∫1xf(t)dtg(x) = \int_1^x f(t) dt, so:
  • g(4)=∫14f(t)dtg(4) = \int_1^4 f(t) dt
  • g(1)=∫11f(t)dtg(1) = \int_1^1 f(t) dt
The key insight is that any integral from a point to itself equals zero, so g(1)=∫11f(t)dt=0g(1) = \int_1^1 f(t) dt = 0. This makes sense because there's no "area" when the upper and lower limits are the same. Therefore: g(4)−g(1)=∫14f(t)dt−0=∫14f(t)dtg(4) - g(1) = \int_1^4 f(t) dt - 0 = \int_1^4 f(t) dt Since we're told that ∫14f(x)dx=10\int_1^4 f(x) dx = 10, we have g(4)−g(1)=10g(4) - g(1) = 10. Looking at the wrong answers: (A) 0 would result from incorrectly thinking both g(4)g(4) and g(1)g(1) equal zero. (B) 9 might come from subtracting 1 from 10, perhaps confusing the limits of integration. (C) suggests we need more information about f(x)f(x), but we actually have exactly what we need. Study tip: Remember that ∫aaf(x)dx=0\int_a^a f(x) dx = 0 always, regardless of the function. When you see integrals with variable limits, substitute the limits directly and use this property to simplify your work.

Question 20

Let F(x)=∫0x(t2−4)dtF(x) = \int_0^x (t^2 - 4) dt. If F(a)=F(b)F(a) = F(b) where 0<a<2<b0 < a < 2 < b, which property of definite integrals explains why this equality can occur?

  1. The integral represents a constant function, so all values of FF must be equal regardless of limits
  2. The function t2−4t^2 - 4 changes sign in the interval [a,b][a,b], allowing positive and negative areas to cancel (correct answer)
  3. The symmetry of t2−4t^2 - 4 about t=2t = 2 ensures that F(a)=F(b)F(a) = F(b) whenever a+b=4a + b = 4
  4. The fundamental theorem requires that antiderivatives have the same values at symmetric points around zeros
Explanation: When you encounter a function defined as an integral with a variable upper limit, you're looking at the Fundamental Theorem of Calculus in action. The key insight here is understanding what happens when the integrand changes sign over the interval. Let's analyze F(x)=∫0x(t2−4)dtF(x) = \int_0^x (t^2 - 4) dt. The function t2−4t^2 - 4 equals zero when t2=4t^2 = 4, so at t=±2t = ±2. Since we're integrating from 0, the relevant zero is at t=2t = 2. For 0<t<20 < t < 2, we have t2<4t^2 < 4, so t2−4<0t^2 - 4 < 0. For t>2t > 2, we have t2>4t^2 > 4, so t2−4>0t^2 - 4 > 0. This means F(a)F(a) represents the (negative) area under the curve from 0 to aa, while F(b)F(b) includes both this negative area and positive area from 2 to bb. When these positive and negative contributions exactly balance, F(a)=F(b)F(a) = F(b). Choice A is wrong because F(x)F(x) is clearly not constant—it's an accumulation function that changes as xx changes. Choice C incorrectly claims the function has symmetry about t=2t = 2, but t2−4t^2 - 4 is symmetric about t=0t = 0, not t=2t = 2. Choice D misapplies the Fundamental Theorem, which doesn't require equal values at symmetric points. The correct answer is B because the sign change allows positive and negative areas to cancel out, creating the equality. Study tip: When analyzing integrals where the integrand changes sign, always identify where it equals zero—these are the points where area contributions switch from positive to negative.