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 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.

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 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.

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.

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

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.