This problem tests the properties of definite integrals, specifically the additivity over adjacent intervals. The $\int_{1}^{4} f(x) , dx$ equals the $\int_{1}^{2} f(x) , dx$ plus the $\int_{2}^{4} f(x) , dx$. Given the total from 1 to 4 is 7 and from 1 to 2 is 3, subtract to find from 2 to 4: $7 - 3 = 4$. 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.

This problem involves the property of odd functions in definite integrals. For an odd function where $p(-x) = -p(x)$, the integral over a symmetric interval $[-a,a]$ equals zero: $\int_{-a}^{a} p(x),dx = 0$. We can verify this by splitting: $\int_{-5}^{5} p(x),dx = \int_{-5}^{0} p(x),dx + \int_{0}^{5} p(x),dx$. For odd functions, $\int_{-5}^{0} p(x),dx = -\int_{0}^{5} p(x),dx = -8$. Therefore: $-8 + 8 = 0$. 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.

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.

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.

This problem tests the constant multiple property of definite integrals. The integral of $-f(x)$ equals the negative of the integral of $f(x)$: $\int_{2}^{6} \big(-f(x)\big),dx = -\int_{2}^{6} f(x),dx$. Given $\int_{2}^{6} f(x),dx = -9$, we have: $\int_{2}^{6} \big(-f(x)\big),dx = -(-9) = 9$. A tempting mistake is to think that negating the function makes the integral more negative, yielding $-18$, 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.

This problem tests your understanding of the additive property of definite integrals. We know that $\int_{-2}^{5} f(x),dx = \int_{-2}^{1} f(x),dx + \int_{1}^{5} f(x),dx$ because we can split an integral at any intermediate point. Substituting the given values: $7 = -3 + \int_{1}^{5} f(x),dx$. Solving for the unknown integral: $\int_{1}^{5} f(x),dx = 7 - (-3) = 10$. A common error would be to subtract instead of add, getting $7 - 3 = 4$ (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.

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.

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.

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.

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.