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AP Calculus BC Quiz

AP Calculus BC Quiz: Applying Properties Of Definite Integrals

Practice Applying Properties Of Definite Integrals in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

Given ∫14f(x) dx=7\int_{1}^{4} f(x)\,dx=7∫14​f(x)dx=7 and ∫12f(x) dx=3\int_{1}^{2} f(x)\,dx=3∫12​f(x)dx=3, what is ∫24f(x) dx\int_{2}^{4} f(x)\,dx∫24​f(x)dx?

Select an answer to continue

What this quiz covers

This quiz focuses on Applying Properties Of Definite Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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

Given ∫14f(x) dx=7\int_{1}^{4} f(x)\,dx=7∫14​f(x)dx=7 and ∫12f(x) dx=3\int_{1}^{2} f(x)\,dx=3∫12​f(x)dx=3, what is ∫24f(x) dx\int_{2}^{4} f(x)\,dx∫24​f(x)dx?

  1. 101010
  2. 444 (correct answer)
  3. −4-4−4
  4. 777
  5. −10-10−10

Explanation: This problem tests the properties of definite integrals, specifically the additivity over adjacent intervals. The ∫14f(x) dx\int_{1}^{4} f(x) \, dx∫14​f(x)dx equals the ∫12f(x) dx\int_{1}^{2} f(x) \, dx∫12​f(x)dx plus the ∫24f(x) dx\int_{2}^{4} f(x) \, dx∫24​f(x)dx. Given the total from 1 to 4 is 7 and from 1 to 2 is 3, subtract to find from 2 to 4: 7−3=47 - 3 = 47−3=4. Therefore, the value is 4. A tempting distractor is 10, which might result from adding the given integrals instead of subtracting. Remember this transferable checklist for definite integral properties: verify interval additivity, apply linearity for constants and sums, reverse limits with a negative sign, and consider even or odd function symmetries when applicable.

Question 2

Let ppp be odd and ∫05p(x) dx=8\int_{0}^{5} p(x)\,dx=8∫05​p(x)dx=8. Find ∫−55p(x) dx\int_{-5}^{5} p(x)\,dx∫−55​p(x)dx.

  1. 161616
  2. 888
  3. 000 (correct answer)
  4. −8-8−8
  5. −16-16−16

Explanation: This problem involves the property of odd functions in definite integrals. For an odd function where p(−x)=−p(x)p(-x) = -p(x)p(−x)=−p(x), the integral over a symmetric interval [−a,a][-a,a][−a,a] equals zero: ∫−aap(x) dx=0\int_{-a}^{a} p(x)\,dx = 0∫−aa​p(x)dx=0. We can verify this by splitting: ∫−55p(x) dx=∫−50p(x) dx+∫05p(x) dx\int_{-5}^{5} p(x)\,dx = \int_{-5}^{0} p(x)\,dx + \int_{0}^{5} p(x)\,dx∫−55​p(x)dx=∫−50​p(x)dx+∫05​p(x)dx. For odd functions, ∫−50p(x) dx=−∫05p(x) dx=−8\int_{-5}^{0} p(x)\,dx = -\int_{0}^{5} p(x)\,dx = -8∫−50​p(x)dx=−∫05​p(x)dx=−8. Therefore: −8+8=0-8 + 8 = 0−8+8=0. Students often mistakenly double the given integral, getting 16, without recognizing the odd function property. Key properties checklist: odd functions integrate to zero over symmetric intervals, even functions double the positive half.

Question 3

Given ∫−25f(x) dx=7\int_{-2}^{5} f(x)\,dx=7∫−25​f(x)dx=7 and ∫−21f(x) dx=−3\int_{-2}^{1} f(x)\,dx=-3∫−21​f(x)dx=−3, find ∫15f(x) dx\int_{1}^{5} f(x)\,dx∫15​f(x)dx.

  1. −10-10−10
  2. 444
  3. 101010 (correct answer)
  4. −4-4−4
  5. −1-1−1

Explanation: This problem tests your understanding of the additive property of definite integrals. We know that ∫−25f(x) dx=∫−21f(x) dx+∫15f(x) dx\int_{-2}^{5} f(x)\,dx = \int_{-2}^{1} f(x)\,dx + \int_{1}^{5} f(x)\,dx∫−25​f(x)dx=∫−21​f(x)dx+∫15​f(x)dx when we split the interval at x=1x=1x=1. Substituting the given values: 7=−3+∫15f(x) dx7 = -3 + \int_{1}^{5} f(x)\,dx7=−3+∫15​f(x)dx, which gives us ∫15f(x) dx=10\int_{1}^{5} f(x)\,dx = 10∫15​f(x)dx=10. A common error would be to subtract the integrals directly without considering the interval relationship, yielding 7−(−3)=107-(-3)=107−(−3)=10 by coincidence but with flawed reasoning. When applying definite integral properties, always check: interval additivity, constant factor rules, and sign changes when reversing limits.

Question 4

Given ∫−44h(x) dx=6\int_{-4}^{4} h(x)\,dx=6∫−44​h(x)dx=6 and hhh is even, find ∫04h(x) dx\int_{0}^{4} h(x)\,dx∫04​h(x)dx.

  1. 000
  2. 121212
  3. −6-6−6
  4. 333 (correct answer)
  5. 666

Explanation: This problem tests your knowledge of even function properties in definite integrals. For an even function where h(−x)=h(x)h(-x) = h(x)h(−x)=h(x), we have the property ∫−aah(x) dx=2∫0ah(x) dx\int_{-a}^{a} h(x)\,dx = 2\int_{0}^{a} h(x)\,dx∫−aa​h(x)dx=2∫0a​h(x)dx. Given ∫−44h(x) dx=6\int_{-4}^{4} h(x)\,dx = 6∫−44​h(x)dx=6, we can write 6=2∫04h(x) dx6 = 2\int_{0}^{4} h(x)\,dx6=2∫04​h(x)dx. Solving for the desired integral: ∫04h(x) dx=3\int_{0}^{4} h(x)\,dx = 3∫04​h(x)dx=3. A common mistake is thinking that half the interval gives half the integral value, yielding 3 by coincidence but missing the even function property. When working with symmetric integrals, always identify: even functions double the half-interval integral, odd functions give zero over symmetric intervals.

Question 5

Given ∫26f(x) dx=−9\int_{2}^{6} f(x)\,dx=-9∫26​f(x)dx=−9, evaluate ∫26(−f(x)) dx\int_{2}^{6} \big(-f(x)\big)\,dx∫26​(−f(x))dx.

  1. −9-9−9
  2. 999 (correct answer)
  3. 000
  4. −18-18−18
  5. 181818

Explanation: This problem tests the constant multiple property of definite integrals. The integral of −f(x)-f(x)−f(x) equals the negative of the integral of f(x)f(x)f(x): ∫26(−f(x)) dx=−∫26f(x) dx\int_{2}^{6} \big(-f(x)\big)\,dx = -\int_{2}^{6} f(x)\,dx∫26​(−f(x))dx=−∫26​f(x)dx. Given ∫26f(x) dx=−9\int_{2}^{6} f(x)\,dx = -9∫26​f(x)dx=−9, we have: ∫26(−f(x)) dx=−(−9)=9\int_{2}^{6} \big(-f(x)\big)\,dx = -(-9) = 9∫26​(−f(x))dx=−(−9)=9. A tempting mistake is to think that negating the function makes the integral more negative, yielding −18-18−18, but the negative sign actually reverses the sign of the integral. Property checklist: constants factor out of integrals, negative signs flip the integral's sign, and this applies regardless of the original integral's sign.

Question 6

Given ∫−25f(x) dx=7\int_{-2}^{5} f(x)\,dx=7∫−25​f(x)dx=7 and ∫−21f(x) dx=−3\int_{-2}^{1} f(x)\,dx=-3∫−21​f(x)dx=−3, what is ∫15f(x) dx\int_{1}^{5} f(x)\,dx∫15​f(x)dx?

  1. 444
  2. 101010 (correct answer)
  3. −10-10−10
  4. −4-4−4
  5. 111

Explanation: This problem tests your understanding of the additive property of definite integrals. We know that ∫−25f(x) dx=∫−21f(x) dx+∫15f(x) dx\int_{-2}^{5} f(x)\,dx = \int_{-2}^{1} f(x)\,dx + \int_{1}^{5} f(x)\,dx∫−25​f(x)dx=∫−21​f(x)dx+∫15​f(x)dx because we can split an integral at any intermediate point. Substituting the given values: 7=−3+∫15f(x) dx7 = -3 + \int_{1}^{5} f(x)\,dx7=−3+∫15​f(x)dx. Solving for the unknown integral: ∫15f(x) dx=7−(−3)=10\int_{1}^{5} f(x)\,dx = 7 - (-3) = 10∫15​f(x)dx=7−(−3)=10. A common error would be to subtract instead of add, getting 7−3=47 - 3 = 47−3=4 (choice A), which ignores that we're adding a negative value. Remember: when splitting integrals, the sum of the parts equals the whole, and pay attention to signs.

Question 7

If ∫−31f(x) dx=7\int_{-3}^{1} f(x)\,dx=7∫−31​f(x)dx=7, what is ∫1−3f(x) dx\int_{1}^{-3} f(x)\,dx∫1−3​f(x)dx?

  1. 777
  2. −7-7−7 (correct answer)
  3. 000
  4. 141414
  5. −14-14−14

Explanation: This problem assesses the skill of applying properties of definite integrals, particularly the reversal of limits property. The integral from 1 to -3 is the negative of the integral from -3 to 1. Since the given integral is 7, the desired one is -7. This stems from the definition where swapping limits introduces a negative sign: ∫_b^a f(x) dx = -∫_a^b f(x) dx. A tempting distractor might be choice A, 7, which ignores the sign change from reversing the limits. Remember, key properties of definite integrals include additivity over intervals, linearity with constants and sums, reversal of limits negating the value, and symmetry for even or odd functions.

Question 8

Given ∫−12f(x) dx=4\int_{-1}^{2} f(x)\,dx=4∫−12​f(x)dx=4 and ∫−123f(x) dx=k\int_{-1}^{2} 3f(x)\,dx=k∫−12​3f(x)dx=k, what is kkk?

  1. 43\tfrac{4}{3}34​
  2. 111
  3. 777
  4. 121212 (correct answer)
  5. −12-12−12

Explanation: This problem assesses the skill of applying properties of definite integrals, specifically the scalar multiple property. The integral of 3f(x) over [-1,2] is 3 times the integral of f(x) over the same interval. Given that integral is 4, k = 3*4 = 12. This is a direct application of ∫ c f(x) dx = c ∫ f(x) dx for constant c. A tempting distractor might be choice A, 4/3, which could result from dividing instead of multiplying by 3. Remember, key properties of definite integrals include additivity over intervals, linearity with constants and sums, reversal of limits negating the value, and symmetry for even or odd functions.

Question 9

Given ∫17f(x) dx=12\int_{1}^{7} f(x)\,dx=12∫17​f(x)dx=12 and ∫37f(x) dx=5\int_{3}^{7} f(x)\,dx=5∫37​f(x)dx=5, what is ∫13f(x) dx\int_{1}^{3} f(x)\,dx∫13​f(x)dx?

  1. 171717
  2. 777 (correct answer)
  3. −7-7−7
  4. 555
  5. −17-17−17

Explanation: This problem tests the properties of definite integrals, specifically additivity over adjacent intervals. The integral from 1 to 7 of f(x) dx equals the integral from 1 to 3 plus from 3 to 7. Given from 1 to 7 is 12 and from 3 to 7 is 5, subtract: 12 - 5 = 7 for from 1 to 3. Thus, the value is 7. A tempting distractor is 17, which might come from adding the given integrals instead of subtracting. Remember this transferable checklist for definite integral properties: verify interval additivity, apply linearity for constants and sums, reverse limits with a negative sign, and consider even or odd function symmetries when applicable.

Question 10

If ∫28q(x) dx=11\int_{2}^{8} q(x)\,dx=11∫28​q(x)dx=11, what is ∫28q(x) dx+∫82q(x) dx\int_{2}^{8} q(x)\,dx+\int_{8}^{2} q(x)\,dx∫28​q(x)dx+∫82​q(x)dx?

  1. 222222
  2. −22-22−22
  3. 111111
  4. 000 (correct answer)
  5. 111

Explanation: This problem combines two key properties of definite integrals. First, we recognize that ∫₈² q(x)dx = -∫₂⁸ q(x)dx by the reversal property. Since ∫₂⁸ q(x)dx = 11, we have ∫₈² q(x)dx = -11. Therefore, ∫₂⁸ q(x)dx + ∫₈² q(x)dx = 11 + (-11) = 0. This result illustrates that integrating from a to b and then from b back to a always yields zero, representing a "round trip" with no net accumulation. Students might incorrectly add 11 + 11 = 22, forgetting the sign reversal. Always check: when limits are reversed, negate the value, and remember that ∫ₐᵇ f(x)dx + ∫ᵇᵃ f(x)dx = 0.

Question 11

Given ∫13q(x) dx=2\int_{1}^{3} q(x)\,dx=2∫13​q(x)dx=2 and ∫13r(x) dx=−5\int_{1}^{3} r(x)\,dx=-5∫13​r(x)dx=−5, what is ∫13(q(x)−r(x)) dx\int_{1}^{3} \big(q(x)-r(x)\big)\,dx∫13​(q(x)−r(x))dx?

  1. −7-7−7
  2. 777 (correct answer)
  3. −3-3−3
  4. 333
  5. −10-10−10

Explanation: This problem tests the linearity property of definite integrals, specifically how to handle differences. By linearity, ∫13(q(x)−r(x)) dx=∫13q(x) dx−∫13r(x) dx\int_{1}^{3} \big(q(x)-r(x)\big)\,dx = \int_{1}^{3} q(x)\,dx - \int_{1}^{3} r(x)\,dx∫13​(q(x)−r(x))dx=∫13​q(x)dx−∫13​r(x)dx. Substituting the given values: 2−(−5)=2+5=72 - (-5) = 2 + 5 = 72−(−5)=2+5=7. The key is recognizing that subtracting a negative integral adds its absolute value. A common mistake is computing 2−5=−32 - 5 = -32−5=−3, forgetting that r(x)r(x)r(x)'s integral is negative. Remember the linearity checklist: split sums and differences, maintain signs carefully, and combine results algebraically.

Question 12

Given ∫16f(x) dx=4\int_{1}^{6} f(x)\,dx=4∫16​f(x)dx=4 and ∫16g(x) dx=−3\int_{1}^{6} g(x)\,dx=-3∫16​g(x)dx=−3, find ∫16(f(x)−2g(x)) dx\int_{1}^{6} \left(f(x)-2g(x)\right)\,dx∫16​(f(x)−2g(x))dx.​

  1. −2-2−2
  2. 101010 (correct answer)
  3. −10-10−10
  4. 222
  5. 111

Explanation: This problem requires applying the linearity property of definite integrals to a linear combination of functions. We need to evaluate ∫16(f(x)−2g(x)) dx\int_{1}^{6} (f(x)-2g(x))\,dx∫16​(f(x)−2g(x))dx using the given values. By linearity, this equals ∫16f(x) dx−2∫16g(x) dx\int_{1}^{6} f(x)\,dx - 2\int_{1}^{6} g(x)\,dx∫16​f(x)dx−2∫16​g(x)dx. Substituting the known values: 4−2(−3)=4+6=104 - 2(-3) = 4 + 6 = 104−2(−3)=4+6=10. A common error is to forget that subtracting a negative gives a positive, incorrectly computing 4−2(3)=−24 - 2(3) = -24−2(3)=−2. Remember the linearity checklist: split linear combinations, factor out constants, then substitute known integral values carefully with their signs.

Question 13

If ∫−22f(x) dx=6\int_{-2}^{2} f(x)\,dx=6∫−22​f(x)dx=6 and fff is even, what is ∫02f(x) dx\int_{0}^{2} f(x)\,dx∫02​f(x)dx?

  1. 121212
  2. 000
  3. 333 (correct answer)
  4. 666
  5. −3-3−3

Explanation: This problem tests the properties of definite integrals, specifically the symmetry for even functions. For an even function f, the integral from -a to a is twice the integral from 0 to a. Given the integral from -2 to 2 is 6, divide by 2 to find from 0 to 2: 6 / 2 = 3. Thus, the value is 3. A tempting distractor is 6, which might come from mistakenly thinking the full integral applies directly without symmetry adjustment. Remember this transferable checklist for definite integral properties: verify interval additivity, apply linearity for constants and sums, reverse limits with a negative sign, and consider even or odd function symmetries when applicable.

Question 14

If ∫16p(x) dx=11\int_{1}^{6} p(x)\,dx=11∫16​p(x)dx=11 and ∫16q(x) dx=−4\int_{1}^{6} q(x)\,dx=-4∫16​q(x)dx=−4, what is ∫16(3p(x)−2q(x)) dx\int_{1}^{6} \big(3p(x)-2q(x)\big)\,dx∫16​(3p(x)−2q(x))dx?

  1. 252525
  2. 414141 (correct answer)
  3. −41-41−41
  4. −25-25−25
  5. 191919

Explanation: This problem tests the linearity property of definite integrals, specifically how integrals distribute over linear combinations. We can split the integral: ∫16(3p(x)−2q(x)) dx=∫163p(x) dx−∫162q(x) dx\int_{1}^{6} (3p(x)-2q(x))\,dx = \int_{1}^{6} 3p(x)\,dx - \int_{1}^{6} 2q(x)\,dx∫16​(3p(x)−2q(x))dx=∫16​3p(x)dx−∫16​2q(x)dx. Factoring out constants: =3∫16p(x) dx−2∫16q(x) dx=3(11)−2(−4)=33+8=41= 3\int_{1}^{6} p(x)\,dx - 2\int_{1}^{6} q(x)\,dx = 3(11) - 2(-4) = 33 + 8 = 41=3∫16​p(x)dx−2∫16​q(x)dx=3(11)−2(−4)=33+8=41. A common mistake is to forget the negative sign when subtracting a negative, getting 33−8=2533 - 8 = 2533−8=25 (choice A). Remember the linearity checklist: distribute integrals over sums/differences, factor out constants, and carefully track all signs.

Question 15

If ∫04g(x) dx=9\int_{0}^{4} g(x)\,dx=9∫04​g(x)dx=9, what is ∫40(2g(x)) dx\int_{4}^{0} \big(2g(x)\big)\,dx∫40​(2g(x))dx?

  1. 181818
  2. −9-9−9
  3. −18-18−18 (correct answer)
  4. 999
  5. 000

Explanation: This problem requires applying two key properties of definite integrals: the constant multiple rule and the reversal of limits rule. First, we can factor out constants: ∫402g(x) dx=2∫40g(x) dx\int_{4}^{0} 2g(x)\,dx = 2\int_{4}^{0} g(x)\,dx∫40​2g(x)dx=2∫40​g(x)dx. Next, reversing the limits of integration changes the sign: ∫40g(x) dx=−∫04g(x) dx=−9\int_{4}^{0} g(x)\,dx = -\int_{0}^{4} g(x)\,dx = -9∫40​g(x)dx=−∫04​g(x)dx=−9. Therefore, 2×(−9)=−182 \times (-9) = -182×(−9)=−18. A tempting error is to forget the sign change when reversing limits and get 2×9=182 \times 9 = 182×9=18 (choice A). Always check: constant factors come out unchanged, but flipping limits introduces a negative sign.

Question 16

If ∫03w(x) dx=2\int_{0}^{3} w(x)\,dx=2∫03​w(x)dx=2 and ∫03x dx=92\int_{0}^{3} x\,dx=\tfrac{9}{2}∫03​xdx=29​, what is ∫03(w(x)+2x) dx\int_{0}^{3} \big(w(x)+2x\big)\,dx∫03​(w(x)+2x)dx?

  1. 111111 (correct answer)
  2. 132\tfrac{13}{2}213​
  3. 112\tfrac{11}{2}211​
  4. 202020
  5. −11-11−11

Explanation: This problem demonstrates the linearity property when integrating a sum of functions. We can split the integral: ∫03(w(x)+2x) dx=∫03w(x) dx+∫032x dx\int_{0}^{3} (w(x)+2x)\,dx = \int_{0}^{3} w(x)\,dx + \int_{0}^{3} 2x\,dx∫03​(w(x)+2x)dx=∫03​w(x)dx+∫03​2xdx. The first integral is given as 2, and the second equals 2×∫03x dx=2×92=92 \times \int_{0}^{3} x\,dx = 2 \times \frac{9}{2} = 92×∫03​xdx=2×29​=9. Therefore, the sum is 2+9=112 + 9 = 112+9=11. A common error is to forget to multiply by 2, getting 2+92=1322 + \frac{9}{2} = \frac{13}{2}2+29​=213​ (choice B). Remember the linearity properties: split sums into separate integrals, factor out constants before using given values.

Question 17

Given ∫−12r(x) dx=5\int_{-1}^{2} r(x)\,dx=5∫−12​r(x)dx=5, what is ∫−1−1r(x) dx+∫2−1r(x) dx\int_{-1}^{-1} r(x)\,dx+\int_{2}^{-1} r(x)\,dx∫−1−1​r(x)dx+∫2−1​r(x)dx?

  1. 000
  2. 555
  3. −5-5−5 (correct answer)
  4. 101010
  5. −10-10−10

Explanation: This problem involves understanding integrals with identical limits and the reversal property. First, ∫−1−1r(x) dx=0\int_{-1}^{-1} r(x)\,dx = 0∫−1−1​r(x)dx=0 because any integral from a point to itself equals zero. Second, ∫2−1r(x) dx=−∫−12r(x) dx=−5\int_{2}^{-1} r(x)\,dx = -\int_{-1}^{2} r(x)\,dx = -5∫2−1​r(x)dx=−∫−12​r(x)dx=−5 by the reversal of limits property. Therefore, the sum is 0+(−5)=−50 + (-5) = -50+(−5)=−5. Students might mistakenly think both integrals are zero (choice A) or forget the sign change when reversing limits to get 0+5=50 + 5 = 50+5=5 (choice B). Key properties: same-point integrals are always zero, and reversing limits changes the sign.

Question 18

If fff is even and ∫03f(x) dx=−5\int_{0}^{3} f(x)\,dx=-5∫03​f(x)dx=−5, what is ∫−33f(x) dx\int_{-3}^{3} f(x)\,dx∫−33​f(x)dx?

  1. −5-5−5
  2. 555
  3. −10-10−10 (correct answer)
  4. 101010
  5. 000

Explanation: This problem assesses the skill of applying properties of definite integrals, focusing on the symmetry for even functions. For an even function f, ∫_{-a}^a f(x) dx = 2 ∫0^a f(x) dx, due to mirror symmetry. Given ∫0^3 f = -5, the total is 2*(-5) = -10. This property holds because f(-x) = f(x), doubling the integral over the positive side. A tempting distractor might be choice A, -5, which ignores the doubling effect of the even symmetry. Remember, key properties of definite integrals include additivity over intervals, linearity with constants and sums, reversal of limits negating the value, and symmetry for even or odd functions.

Question 19

If ∫−22f(x) dx=0\int_{-2}^{2} f(x)\,dx=0∫−22​f(x)dx=0, what is ∫−22(f(x)+5) dx\int_{-2}^{2} (f(x)+5)\,dx∫−22​(f(x)+5)dx?

  1. 000
  2. 555
  3. 101010
  4. 202020 (correct answer)
  5. −20-20−20

Explanation: This problem assesses the skill of applying properties of definite integrals, emphasizing linearity and the integral of a constant. The integral of f(x) + 5 from -2 to 2 is the sum of ∫f and ∫5 dx over that interval. Given ∫f = 0, and ∫5 dx = 5*(2 - (-2)) = 20, the total is 0 + 20 = 20. This applies the linearity property and the fact that the integral of a constant c over [a,b] is c*(b-a). A tempting distractor might be choice C, 10, possibly from halving the interval length mistakenly. Remember, key properties of definite integrals include additivity over intervals, linearity with constants and sums, reversal of limits negating the value, and symmetry for even or odd functions.

Question 20

Given ∫04f(x) dx=3\int_{0}^{4} f(x)\,dx=3∫04​f(x)dx=3 and ∫04g(x) dx=−2\int_{0}^{4} g(x)\,dx=-2∫04​g(x)dx=−2, find ∫04(2f(x)−3g(x)) dx\int_{0}^{4} (2f(x)-3g(x))\,dx∫04​(2f(x)−3g(x))dx.

  1. 000
  2. 121212 (correct answer)
  3. −12-12−12
  4. 666
  5. −6-6−6

Explanation: This problem assesses the skill of applying properties of definite integrals, focusing on linearity with scalar multiples and sums. The integral of 2f(x)−3g(x)2f(x) - 3g(x)2f(x)−3g(x) from 0 to 4 equals 2 times the integral of fff minus 3 times the integral of ggg over the same interval. Plugging in the values, that's 2∗3−3∗(−2)=6+6=122*3 - 3*(-2) = 6 + 6 = 122∗3−3∗(−2)=6+6=12. This uses the properties that ∫(af+bg)=a∫f+b∫g\int (a f + b g) = a \int f + b \int g∫(af+bg)=a∫f+b∫g for constants a and b. A tempting distractor might be choice D, 6, which could come from forgetting to apply the negative sign to the g integral properly. Remember, key properties of definite integrals include additivity over intervals, linearity with constants and sums, reversal of limits negating the value, and symmetry for even or odd functions.