Applying Properties of Definite Integrals - AP Calculus BC

1. If $\int_{1}^{6} p(x)\,dx=11$ and $\int_{1}^{6} q(x)\,dx=-4$, what is $\int_{1}^{6} \big(3p(x)-2q(x)\big)\,dx$?
   Answer: $41$ This problem tests the linearity property of definite integrals, specifically how integrals distribute over linear combinations. We can split the integral: $\int_{1}^{6} (3p(x)-2q(x))\,dx = \int_{1}^{6} 3p(x)\,dx - \int_{1}^{6} 2q(x)\,dx$. Factoring out constants: $= 3\int_{1}^{6} p(x)\,dx - 2\int_{1}^{6} q(x)\,dx = 3(11) - 2(-4) = 33 + 8 = 41$. A common mistake is to forget the negative sign when subtracting a negative, getting $33 - 8 = 25$ (choice A). Remember the linearity checklist: distribute integrals over sums/differences, factor out constants, and carefully track all signs.
2. If $\int_{-4}^{4} f(x)\,dx=0$ and $f$ is odd, what is $\int_{0}^{4} f(x)\,dx$?
   Answer: $-\int_{-4}^{0} f(x)\,dx$ This problem tests the properties of definite integrals, specifically the symmetry for odd functions. For an odd function f, the integral from -a to a is 0, and the integral from 0 to a equals the negative of the integral from -a to 0. Given the integral from -4 to 4 is 0, which is consistent, the integral from 0 to 4 is -integral from -4 to 0. Therefore, the expression is -∫_{-4}^{0} f(x) dx. A tempting distractor is 0, which might result from confusing the full symmetric integral with the half-interval. 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.
3. If $\int_{0}^{4} g(x)\,dx=9$, what is $\int_{4}^{0} \big(2g(x)\big)\,dx$?
   Answer: $-18$ 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: $\int_{4}^{0} 2g(x)\,dx = 2\int_{4}^{0} g(x)\,dx$. Next, reversing the limits of integration changes the sign: $\int_{4}^{0} g(x)\,dx = -\int_{0}^{4} g(x)\,dx = -9$. Therefore, $2 \times (-9) = -18$. A tempting error is to forget the sign change when reversing limits and get $2 \times 9 = 18$ (choice A). Always check: constant factors come out unchanged, but flipping limits introduces a negative sign.
4. If $\int_{-2}^{3} f(x)\,dx=9$, evaluate $\int_{-2}^{3} \big(\tfrac12 f(x)\big)\,dx$.
   Answer: $\tfrac{9}{2}$ This problem demonstrates the constant multiple property of definite integrals. When a function is multiplied by a constant, the integral is multiplied by the same constant: $\int_{-2}^{3} \big(\tfrac{1}{2} f(x)\big)\,dx = \tfrac{1}{2} \int_{-2}^{3} f(x)\,dx$. Given $\int_{-2}^{3} f(x)\,dx = 9$, we have: $\tfrac{1}{2} \times 9 = \tfrac{9}{2}$. A common error is to divide the limits by 2 or apply the factor incorrectly, perhaps getting $\tfrac{9}{4}$. Remember these properties: constants factor out completely from integrals, the factor applies to the entire integral value, and this works for any constant including fractions.
5. Given $\int_{0}^{2} f(x)\,dx=5$ and $\int_{2}^{6} f(x)\,dx=-1$, what is $\int_{0}^{6} f(x)\,dx$?​
   Answer: $4$ This problem tests your ability to apply the additive property of definite integrals. When we have adjacent intervals, we can combine them: ∫₀⁶ f(x)dx = ∫₀² f(x)dx + ∫₂⁶ f(x)dx. Substituting the given values, we get ∫₀⁶ f(x)dx = 5 + (-1) = 4. The key insight is recognizing that the point x = 2 connects the two intervals [0,2] and [2,6] to form [0,6]. A common error would be subtracting the integrals (getting 6) instead of adding them, forgetting that definite integrals can be negative. Remember the checklist: identify interval endpoints, check if intervals are adjacent, and add (not subtract) the integral values.
6. Given $\int_{0}^{2} f(x)\,dx=5$ and $\int_{2}^{5} f(x)\,dx=-1$, what is $\int_{0}^{5} f(x)\,dx$?
   Answer: $4$ This problem assesses the skill of applying properties of definite integrals, specifically the additivity property over adjacent intervals. The integral from 0 to 5 can be split into the sum of integrals from 0 to 2 and 2 to 5. Given that the first is 5 and the second is -1, adding them yields 5 + (-1) = 4. This directly applies the property that ∫_a^c f(x) dx = ∫_a^b f(x) dx + ∫_b^c f(x) dx when b is between a and c. A tempting distractor might be choice B, 6, which could result from mistakenly adding the absolute values instead of the signed values. 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.
7. Given $\int_{1}^{6} f(x)\,dx=-4$ and $\int_{1}^{3} f(x)\,dx=2$, find $\int_{3}^{6} f(x)\,dx$.
   Answer: $-6$ This problem assesses the skill of applying properties of definite integrals, specifically the additivity property to find a subinterval. The integral from 1 to 6 equals the sum from 1 to 3 and 3 to 6, so ∫$3^6$ = ∫$1^6$ - ∫$1^3$ = -4 - 2 = -6. This rearranges the additivity rule to isolate the desired integral. The property holds because the intervals are adjacent and cover the total range. A tempting distractor might be choice A, -2, which could result from subtracting in the wrong order. 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.
8. If $\int_{-3}^{1} f(x)\,dx=-6$ and $\int_{1}^{4} f(x)\,dx=2$, what is $\int_{-3}^{4} f(x)\,dx$?​
   Answer: $-4$ This problem tests the additivity property of definite integrals over adjacent intervals. We need to find $\int_{-3}^{4} f(x)\,dx$ using the given information about integrals over sub-intervals. By additivity, $\int_{-3}^{4} f(x)\,dx = \int_{-3}^{1} f(x)\,dx + \int_{1}^{4} f(x)\,dx$. Substituting the known values: $\int_{-3}^{4} f(x)\,dx = -6 + 2 = -4$. A tempting error is to add the absolute values, getting 8, or to subtract incorrectly. Remember the additivity property: when intervals share an endpoint, add the integrals directly with their given signs.
9. If $f$ is odd and $\int_{0}^{4} f(x)\,dx=9$, what is $\int_{-4}^{4} f(x)\,dx$?
   Answer: $0$ This problem assesses the skill of applying properties of definite integrals, particularly the symmetry property for odd functions. For an odd function f, ∫_{-a}^a f(x) dx = 0, as the areas cancel out symmetrically. Given ∫$0^4$ f = 9, the integral from -4 to 0 would be -9, summing to 0 overall. This directly uses the odd function property where f(-x) = -f(x), leading to symmetric cancellation. A tempting distractor might be choice A, 18, which might come from doubling the positive integral without considering the odd 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.
10. If $\int_{-3}^{1} f(x)\,dx=2$ and $\int_{-3}^{1} g(x)\,dx=-5$, what is $\int_{-3}^{1} \big(3f(x)-2g(x)\big)\,dx$?
    Answer: $16$ This problem tests the properties of definite integrals, specifically the linearity property for sums and scalar multiples. The integral of 3f(x) - 2g(x) from -3 to 1 is 3 times the integral of f minus 2 times the integral of g over the same interval. Given integral of f is 2 and g is -5, this computes to $3 \times 2 - 2 \times (-5) = 6 + 10 = 16$. Thus, the value is 16. A tempting distractor is 4, which might come from mistakenly adding instead of properly applying the coefficients. 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.
11. Given $\int_{-2}^{5} f(x)\,dx=7$ and $\int_{-2}^{1} f(x)\,dx=-3$, find $\int_{1}^{5} f(x)\,dx$.
    Answer: $10$ This problem tests your understanding of the additive property of definite integrals. We know that $\int_{-2}^{5} f(x)\,dx = \int_{-2}^{1} f(x)\,dx + \int_{1}^{5} f(x)\,dx$ when we split the interval at $x=1$. Substituting the given values: $7 = -3 + \int_{1}^{5} f(x)\,dx$, which gives us $\int_{1}^{5} f(x)\,dx = 10$. A common error would be to subtract the integrals directly without considering the interval relationship, yielding $7-(-3)=10$ by coincidence but with flawed reasoning. When applying definite integral properties, always check: interval additivity, constant factor rules, and sign changes when reversing limits.
12. If $\int_{0}^{3} w(x)\,dx=-2$ and $\int_{3}^{7} w(x)\,dx=5$, what is $\int_{0}^{7} -w(x)\,dx$?
    Answer: $-3$ This problem combines additivity and the constant multiple property of definite integrals. First, using additivity: $\int_{0}^{7} w(x)\,dx = \int_{0}^{3} w(x)\,dx + \int_{3}^{7} w(x)\,dx = -2 + 5 = 3$. Then, by the constant multiple property: $\int_{0}^{7} -w(x)\,dx = -\int_{0}^{7} w(x)\,dx = -(3) = -3$. A common error is to negate each integral separately before adding, giving $-(-2) + (-5) = 2 - 5 = -3$, which happens to match but uses incorrect reasoning. Always combine integrals first using additivity, then apply the constant multiple rule to the result.
13. If $\int_{2}^{8} q(x)\,dx=11$, what is $\int_{2}^{8} q(x)\,dx+\int_{8}^{2} q(x)\,dx$?
    Answer: $0$ 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.
14. If $\int_{-2}^{3} p(x)\,dx=-4$ and $\int_{3}^{5} p(x)\,dx=10$, find $\int_{-2}^{5} p(x)\,dx$.
    Answer: $6$ This problem applies the additive property of definite integrals across adjacent intervals. Since the intervals [-2,3] and [3,5] share the endpoint x = 3, we can combine them: ∫₋₂⁵ p(x)dx = ∫₋₂³ p(x)dx + ∫₃⁵ p(x)dx. Substituting the given values: ∫₋₂⁵ p(x)dx = (-4) + 10 = 6. The key is recognizing that x = 3 connects the two intervals to form the complete interval [-2,5]. A common mistake is subtracting the integrals (getting -14) instead of adding them. When applying integral properties, always check: are intervals adjacent (sharing an endpoint), then add the integral values algebraically.
15. If $\int_{2}^{7} f(x)\,dx=3$, what is $\int_{2}^{7} \left(-\tfrac{1}{2}f(x)\right)\,dx$?​
    Answer: $-\dfrac{3}{2}$ This problem tests the scalar multiplication property of definite integrals. We need to find $\int_{2}^{7} (-\frac{1}{2}f(x))\,dx$ given that $\int_{2}^{7} f(x)\,dx = 3$. By the linearity property, constants factor out: $\int_{2}^{7} (-\frac{1}{2}f(x))\,dx = -\frac{1}{2}\int_{2}^{7} f(x)\,dx$. Substituting the known value: $-\frac{1}{2}(3) = -\frac{3}{2}$. A tempting error is to multiply incorrectly or forget the negative sign, perhaps getting $\frac{3}{2}$ instead. Remember the scalar multiplication rule: when a constant multiplies the integrand, it factors out completely, preserving its sign.
16. Given $\int_{1}^{3} q(x)\,dx=2$ and $\int_{1}^{3} r(x)\,dx=-5$, what is $\int_{1}^{3} \big(q(x)-r(x)\big)\,dx$?
    Answer: $7$ This problem tests the linearity property of definite integrals, specifically how to handle differences. By linearity, $\int_{1}^{3} \big(q(x)-r(x)\big)\,dx = \int_{1}^{3} q(x)\,dx - \int_{1}^{3} r(x)\,dx$. Substituting the given values: $2 - (-5) = 2 + 5 = 7$. The key is recognizing that subtracting a negative integral adds its absolute value. A common mistake is computing $2 - 5 = -3$, forgetting that $r(x)$'s integral is negative. Remember the linearity checklist: split sums and differences, maintain signs carefully, and combine results algebraically.
17. If $\int_{2}^{6} f(x)\,dx=-9$, what is $\int_{6}^{2} \big(-f(x)\big)\,dx$?
    Answer: $-9$ This problem tests the properties of definite integrals, specifically reversal of limits and scalar multiples. The integral from 6 to 2 of -f(x) dx is -1 times the integral from 6 to 2 of f(x) dx, and integral from 6 to 2 of f is -integral from 2 to 6 of f. Given from 2 to 6 is -9, from 6 to 2 of f is -(-9)=9, so -9 for -f. Therefore, the value is -9. A tempting distractor is 9, which might result from forgetting the negative sign from the scalar. 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.
18. Given $\int_{0}^{4} f(x)\,dx=9$ and $\int_{0}^{1} f(x)\,dx=2$, find $\int_{1}^{4} f(x)\,dx$.​
    Answer: $7$ This problem tests the additivity property of definite integrals over adjacent intervals. We're given $\int_{0}^{4} f(x)\,dx = 9$ and $\int_{0}^{1} f(x)\,dx = 2$, and need to find $\int_{1}^{4} f(x)\,dx$. By the additivity property, $\int_{0}^{4} f(x)\,dx = \int_{0}^{1} f(x)\,dx + \int_{1}^{4} f(x)\,dx$. Substituting the known values: $9 = 2 + \int_{1}^{4} f(x)\,dx$, which gives us $\int_{1}^{4} f(x)\,dx = 7$. A common mistake is subtracting in the wrong order, getting -7 instead of 7. Remember the additivity checklist: identify the full interval, find the sub-intervals, and ensure endpoints match when adding.
19. If $\int_{-1}^{2} f(x)\,dx=4$, what is $\int_{-1}^{2} \big(f(x)+3\big)\,dx$?
    Answer: $13$ This problem combines the linearity property with integrating a constant function. We can split the integral: $\int_{-1}^{2} \big(f(x)+3\big)\,dx = \int_{-1}^{2} f(x)\,dx + \int_{-1}^{2} 3\,dx$. The first integral is given as 4, and the constant integral equals $3 \times (2-(-1)) = 3 \times 3 = 9$. Therefore: $4 + 9 = 13$. Students might incorrectly add 3 to the result instead of integrating the constant over the interval, getting 7. Key properties: linearity allows splitting sums, constants integrate to (constant × interval length), and always calculate the interval length as (upper limit - lower limit).
20. Given $\int_{-4}^{4} h(x)\,dx=6$ and $h$ is even, find $\int_{0}^{4} h(x)\,dx$.
    Answer: $3$ This problem tests your knowledge of even function properties in definite integrals. For an even function where $h(-x) = h(x)$, we have the property $\int_{-a}^{a} h(x)\,dx = 2\int_{0}^{a} h(x)\,dx$. Given $\int_{-4}^{4} h(x)\,dx = 6$, we can write $6 = 2\int_{0}^{4} h(x)\,dx$. Solving for the desired integral: $\int_{0}^{4} h(x)\,dx = 3$. 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.

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