📝 Properties of definite integrals (26 MCQs)

📖 Explanation: The integral of a product is NOT the product of integrals; this is a common but incorrect generalization from the sum property. Integration is a linear operator, meaning it preserves addition and scalar multiplication, but it does not preserve multiplication. The product of integrals would be valid only in special cases, not as a general property. This is a critical distinction between integration and differentiation, where the product rule does exist.

📖 Explanation: The integral over a union of adjacent intervals equals the sum of integrals over each interval: ∫[0,5] f(x) dx = ∫[0,3] f(x) dx + ∫[3,5] f(x) dx = 8 + (-2) = 6. This additivity property is fundamental and allows breaking complex integration regions into simpler parts. Note that even when the function is negative on some subintervals, the property still holds as the integral represents signed area.

📖 Explanation: The student is correct! The additivity property states ∫[a,b] f(x) dx = ∫[a,c] f(x) dx + ∫[c,b] f(x) dx for any c between a and b. Here, 8/3 = 1/3 + ∫[1,2] x² dx, so the latter integral equals 7/3. This property is one of the most useful for splitting difficult integrals or finding unknown integrals from known values. The student's reasoning is valid and demonstrates proper Easy of interval additivity.

📖 Explanation: If f(x) ≤ g(x) pointwise on an interval, then the integral of f is less than or equal to the integral of g over that interval, provided both functions are integrable. This is the comparison property of definite integrals, which is a direct consequence of the fact that integration preserves inequalities (when the integrand is nonnegative). Since ∫ f = 12, the integral of g must be at least 12. This property is crucial for bounding integrals without computing them exactly.

📖 Explanation: By linearity, ∫[1,4] [3f(x) - 2g(x)] dx = 3∫[1,4] f(x) dx - 2∫[1,4] g(x) dx = 3(10) - 2(3) = 30 - 6 = 24. This combines both the constant multiple and difference properties. This type of linear combination is common in Easys where we need to compute integrals of expressions built from known integrals. The coefficients can be any real numbers, and the property extends to any finite linear combination of functions.

📖 Explanation: If f is bounded between m and M on [a,b], then the integral of f is bounded between the areas of the rectangles of heights m and M: m(b-a) ≤ ∫ f(x) dx ≤ M(b-a). This is the boundedness property of definite integrals. It follows from applying the comparison property to the inequalities m ≤ f(x) ≤ M. This is particularly useful when exact integration is difficult but bounds are needed, such as in estimating errors or proving convergence.

📖 Explanation: This is the reversal of limits property: switching the upper and lower limits of integration changes the sign of the integral. This property is a convention that ensures consistency when we want to integrate 'backwards' or when deriving other formulas. It is essential for handling situations where the upper limit might be less than the lower limit, such as in Easys involving velocity and displacement over time intervals that are traversed in either direction.

📖 Explanation: The integral over a single point (when the upper and lower limits are equal) is always 0, regardless of the integrand, provided the function is defined at that point. This is a fundamental property: there is no area (signed or otherwise) when the interval has zero width. This property is particularly important when dealing with initial conditions in differential equations or when using the Fundamental Theorem of Calculus, as it allows constant terms to be determined.

📖 Explanation: For odd functions, the integral over a symmetric interval [-a,a] is 0. This follows from ∫[-a,a] f(x) dx = ∫[-a,0] f(x) dx + ∫[0,a] f(x) dx, and substituting u=-x in the first integral yields ∫[-a,0] f(x) dx = -∫[0,a] f(u) du. Since f is odd, ∫[-a,0] f(x) dx = -∫[0,a] f(x) dx, so the sum is 0. This property has geometric significance: the signed area on one side cancels with that on the other. It is widely used to simplify integrals involving sine, tangent, or polynomials with only odd powers.

📖 Explanation: The comparison property states that if f(x) ≤ g(x) and ∫ f = ∫ g, then f(x) = g(x) for all x in [a,b]. This is a stronger conclusion than just equality at a point. It follows because if f(x) < g(x) on any subinterval of positive length, the integral of g - f would be positive. The function h(x) = g(x) - f(x) is nonnegative and has integral 0; for continuous nonnegative functions, this forces h to be identically zero. This property is useful in proving uniqueness of solutions to certain integral equations.

📖 Explanation: The student made a fundamental error: ∫ f² is NOT the same as (∫ f)² in general. For f(x)=x, ∫[0,2] x² dx = 8/3, while [∫[0,2] x dx]² = (2)² = 4. This misconception arises from incorrectly applying algebraic rules to integrals. The only algebraic operations that distribute over integration are addition/subtraction and scalar multiplication—not multiplication, division, exponentiation, or function composition. This error can lead to drastically incorrect results in probability (variance calculations), physics (root-mean-square values), and statistics (where integrals of squared functions represent energy or second moments).

📖 Explanation: Let u = 2x. Then du = 2 dx, and when x=0, u=0; when x=2, u=4. Thus ∫[0,2] 2f(2x) dx = ∫[0,4] f(u) du = 6. This tests substitution in definite integrals combined with the constant multiple property. The 2 outside the function and the factor from du exactly cancel. This type of transformation is common in change of variables and scaling arguments, where the factor can be 'absorbed' by adjusting limits. Understanding this also connects to the area interpretation: stretching the graph horizontally compresses it, but the integral remains invariant under such combined scaling.

📖 Explanation: If f is nonnegative on an interval, its integral must be nonnegative. Moreover, if the integral is 0, then f must be 0 almost everywhere (for continuous functions, this means everywhere). The other options are false: the integral of a nonnegative function cannot be negative, and it is zero only in trivial cases. This property is fundamental to the interpretation of the definite integral as area: area cannot be negative. It also underpins the proof of the comparison property and is crucial in establishing the Mean Value Theorem for Integrals.

📖 Explanation: For even functions, the integral over a symmetric interval is twice the integral from 0 to a. This follows from ∫[-a,a] f(x) dx = ∫[-a,0] f(x) dx + ∫[0,a] f(x) dx, and substituting u=-x in the first integral yields ∫[0,a] f(u) du. Since f is even, f(-u)=f(u). This property simplifies many calculations and is often used when the integrand contains cosines or even powers of x. It also has a geometric interpretation: the area from -a to 0 equals the area from 0 to a.

📖 Explanation: By the reversal of limits property, ∫[1,0] f(x) dx = -∫[0,1] f(x) dx = -3. This property is often used when we need to change the direction of integration, such as when applying substitutions that reverse the order of integration limits. It is also crucial in the context of the Fundamental Theorem of Calculus, where we often encounter expressions like F(b) - F(a) = -[F(a) - F(b)], which corresponds to reversing limits. The sign change reflects the orientation of the interval when summing infinitesimal contributions.

📖 Explanation: The student's reasoning is flawed because sin x is indeed an odd function, but the property ∫[-a,a] f(x) dx = 0 applies only when the interval is symmetric about the origin. Here, the interval is [0,π], not symmetric, so the oddness of the function does not imply the integral is zero. In fact, ∫[0,π] sin x dx = 2. This common mistake highlights the importance of understanding the domain when applying function symmetry properties. The symmetry property is about canceling areas on opposite sides of the y-axis over symmetric intervals.

📖 Explanation: This is the triangle inequality for integrals: the absolute value of the integral is less than or equal to the integral of the absolute value. It follows from -|f(x)| ≤ f(x) ≤ |f(x)| and then integrating both sides: -∫|f| ≤ ∫f ≤ ∫|f|. This property is fundamental in bounding integrals and in establishing convergence theorems. It also has a geometric interpretation: the net signed area (which can have cancellations) is always less than or equal to the total area (where cancellations are removed). This is crucial in probability (bounding expectations) and analysis (dominated convergence).

📖 Explanation: For a continuous nonnegative function, if its integral over an interval is zero, then the function must be identically zero on that interval. This is because if f(x0) > 0 at some point, by continuity there would be a neighborhood around x0 where f(x) > 0, giving a positive integral over that subinterval. The integral of a nonnegative function can be zero only if the function is zero almost everywhere, and for continuous functions this means everywhere. This property is used to prove uniqueness of solutions in differential equations and to show that certain integral equations have only trivial solutions.

📖 Explanation: Distance traveled is the integral of speed, which is the absolute value of velocity. This is distinct from displacement, which is the integral of velocity (signed). The absolute value ensures that all motion contributes positively to distance traveled, regardless of direction. This distinction is crucial in physics and engineering: while displacement might be zero (if the particle returns to start), distance traveled would be positive. The integral of |v(t)| is also the total area under the speed curve. This concept also appears in economics as 'total variation' of a function.

📖 Explanation: Using the additivity property on intervals: ∫[0,2] f = ∫[0,1] f + ∫[1,2] f. Thus 10 = 3 + ∫[1,2] f, so ∫[1,2] f = 7. This is a straightforward Easy of interval additivity, but it's easy to make sign errors or to forget that the property works in either direction. This type of reasoning is common when we only have partial information and need to compute the integral over a subinterval. It's also used when splitting integrals for integration by parts or when dealing with piecewise-defined functions.

📖 Explanation: The average value of a function h on [a,b] is (1/(b-a))∫[a,b] h(x) dx. Here h = f+g, so the average is (1/(1-0))∫[0,1] (f+g) dx = ∫[0,1] f + ∫[0,1] g = 2+5 = 7. Since the interval length is 1, the average equals the integral. This combines the concept of average value with linearity of the integral. Students often forget the division by the interval length or incorrectly average the integrals. The average value is a key concept in physics (average velocity, average current) and statistics (mean of a continuous distribution).

📖 Explanation: Using linearity, ∫[0,5] [2f - g] dx = 2∫ f dx - ∫ g dx = 2(10) - (-3) = 20 + 3 = 23. This problem combines constant multiple and difference properties with careful attention to signs. The minus sign before g requires subtracting a negative, which results in addition. Many students make arithmetic errors with signs in such combined expressions. This type of calculation is common in engineering where we have known integrated quantities and need to compute derived quantities like total energy (sum of energies) or net force (difference of forces).

📖 Explanation: For a continuous nonnegative function with zero integral, the function must be identically zero on the interval. This is a key result that follows from continuity: if f(c)>0 for some c, then by continuity there is a neighborhood around c where f(x) > f(c)/2 > 0, making the integral over that neighborhood positive and hence the total integral positive. This property is used to prove that if two continuous functions have the same integral over all intervals, they must be equal. It also underlies the uniqueness of solutions to certain differential equations and the completeness of function spaces in analysis.

📖 Explanation: The integral is linear, so it distributes over addition of any finite number of functions. Thus both A (directly applying the sum property) and B (applying it in two steps) are correct. The sum property extends by induction to any number of terms. This extensibility is often used when functions are expressed as series or when we need to integrate piecewise-defined functions by splitting them into simpler parts. The linearity of integration is one of its most powerful and frequently used properties.