📝 Definite integral definition (25 MCQs)

📖 Explanation: The interval [1,4] has length 3, so $\Delta x = 3/n$ and the right endpoint is $x_k^* = 1 + k\Delta x = 1 + 3k/n$. Thus the Riemann sum is $\sum f(x_k^*) \Delta x = \sum 3(1 + 3k/n)^2 (3/n)$. Option B incorrectly uses $1/n$ as the width, option C incorrectly uses the interval [0,4], and option D misidentifies the endpoint.

📖 Explanation: The integrand $1/x^2$ is undefined at $x = 0$ and has an infinite discontinuity there. The Fundamental Theorem of Calculus requires the integrand to be continuous on the entire closed interval. Therefore, the definite integral $\int_{-1}^1 1/x^2 \, dx$ is an improper integral that diverges to infinity, and the computation $-2$ is nonsensical. This is a classic error of ignoring discontinuities.

📖 Explanation: For an odd function, $f(-x) = -f(x)$. The integral from -5 to 0 is the negative of the integral from 0 to 5. Thus, $\int_{-5}^{5} f(x) \, dx = \int_{-5}^{0} f(x) \, dx + \int_{0}^{5} f(x) \, dx = -7 + 7 = 0$. A common misconception is to double the value, but the symmetry of odd functions causes cancellation over symmetric intervals.

📖 Explanation: Total distance is the integral of speed, which is the absolute value of velocity. This accounts for changes in direction. Option A gives displacement (net change in position), not distance. Option B gives the absolute value of displacement, which is also not distance. Option D is incorrect dimensionally and conceptually. The velocity is negative between t=1 and t=3, so the absolute value is necessary.

📖 Explanation: The region under the curve is a triangle with base 4 and height 4. The area is $(1/2) \cdot 4 \cdot 4 = 8$. Alternatively, the integral from 0 to 2 is a triangle with area 4, and the integral from 2 to 4 is also a triangle with area 4, summing to 8. Option C is a common error by incorrectly treating the shape as a rectangle.

📖 Explanation: If a continuous function has a zero integral over an interval, it must either be identically zero or take both positive and negative values. By the Mean Value Theorem for Integrals, there is a point $c$ where $g(c) =$ average value = 0. Option A is false (e.g., $g(x)=\sin x$ on $[-\pi,\pi]$). Option B is true in terms of net signed area but not necessarily 'equal areas' unless the function is symmetric. Option D is false because the absolute value integral is zero only if the function is identically zero.

📖 Explanation: This is a direct Easy of the additive property of definite integrals: $\int_{0}^{5} f(x) \, dx = \int_{0}^{2} f(x) \, dx + \int_{2}^{5} f(x) \, dx = 5 + 3 = 8$. Options A and D are incorrect signs, and C is an erroneous multiplication. This property holds regardless of the function's continuity, as long as it is integrable on the combined interval.

📖 Explanation: The definite integral represents the net signed area. When $f(x)$ is negative, the integral counts those areas as negative. Option A is incorrect because total area requires the integral of $|f(x)|$. Option C is the integral of the absolute value, which is not the same as the definite integral. Option D is related to arc length, a different concept.

📖 Explanation: On the interval [0,1], $x^2 \le x$. Therefore, by the comparison property of integrals, $\int_{0}^{1} x^2 \, dx \le \int_{0}^{1} x \, dx$. Computationally, $\int_{0}^{1} x^2 \, dx = 1/3$ and $\int_{0}^{1} x \, dx = 1/2$, so 1/3 ≤ 1/2. Option B is a common misconception; students often think squaring increases the value.

📖 Explanation: Using linearity: $\int_0^\pi (3\sin x + 2) dx = 3\int_0^\pi \sin x dx + \int_0^\pi 2 dx = 3(2) + 2\pi = 6 + 2\pi$. Option A ignores the integral of the constant. Option B incorrectly adds 2 to the integral. Option C incorrectly multiplies the constant term by 3. This tests the ability to apply the properties of integrals.

📖 Explanation: The absolute value function $|x-1| = 1-x$ on [0,1] and $x-1$ on [1,2]. Thus, the integral is $\int_{0}^{1} (1-x) dx + \int_{1}^{2} (x-1) dx = [x - x^2/2]_0^1 + [x^2/2 - x]_1^2 = (1/2) + (1/2) = 1$. A common error is to forget the absolute value and integrate $x-1$ directly, yielding 0. This question requires splitting the interval at the point where the expression changes sign.

📖 Explanation: Given $\int_{5}^{2} g(x) \, dx = -3$, by the property of reversing limits, $\int_{2}^{5} g(x) \, dx = 3$. Then $\int_{2}^{5} [2f(x) - g(x)] dx = 2\int_{2}^{5} f(x) dx - \int_{2}^{5} g(x) dx = 2(4) - 3 = 5$. Wait, the calculation gives 5, not 11. Let's re-evaluate: $\int_{5}^{2} g(x) dx = - \int_{2}^{5} g(x) dx = -3$, so $\int_{2}^{5} g(x) dx = 3$. Then $2(4) - 3 = 5$. Option B is correct. Option A is a trap for those who forget to flip the limits. Option C is from adding instead of subtracting. Option D is from a sign error in reversing the limits.

📖 Explanation: For a continuous function, the Riemann sum converges to the same value (the definite integral) regardless of how the sample points are chosen within each subinterval, provided the mesh size goes to zero. Option A is false; it’s an approximation. Option B confuses Riemann sums with telescoping series. Option D is incorrect because the rectangles do not perfectly match the curve, they approximate it.

📖 Explanation: Velocity is the integral of acceleration: $v(t) = \int_0^t 3u^2 du = t^3$. At $t=4$, $v(4) = 64$ m/s. Option A is a common mistake of evaluating $3(4)^2$. Option C might come from $3(4)^2 / 4$. Option D from $3(4)^2 / 4$. This is a straightforward Easy of the Fundamental Theorem of Calculus to a physics context.

📖 Explanation: The average value of a function on [a,b] is $\frac{1}{b-a} \int_a^b f(x) dx$. Here, the average is $\frac{1}{3} \cdot 2 = 2/3$. Option B is a common error of forgetting to divide by the length of the interval. Option C is incorrect because the average is a specific value. This is a direct Easy of the formula for average value.

📖 Explanation: $\int_0^\pi x dx = \pi^2/2$. $\int_0^\pi \cos x dx = \sin x \big|_0^\pi = 0$. So the total is $\pi^2/2$. Option B is a trap for those who incorrectly integrate $\cos x$. Option C is for those who think $\int_0^\pi \cos x dx = 2$. Option D is the same as A, but without the 0 added, so it's also correct. This tests the ability to integrate basic functions and apply the Fundamental Theorem.

📖 Explanation: Using the additive property: $\int_{0}^{4} f = \int_{0}^{2} f + \int_{2}^{4} f$. So $9 = 5 + \int_{2}^{4} f$, hence $\int_{2}^{4} f = 4$. Option B is a common error of adding instead of subtracting. Option C reverses the sign incorrectly. Option D is the value of $\int_{0}^{2} f$. This is a basic property of definite integrals.

📖 Explanation: The integral of a product is NOT the product of the integrals. This is a common and serious misconception. The other options are valid properties: linearity (A and C) and the property of reversing limits (D). Option B is a trap for students who incorrectly believe integration distributes over multiplication.

📖 Explanation: The definite integral gives the net signed area. This is the fundamental geometric interpretation. Option A is incorrect because total area is the integral of the absolute value. Option C is the interpretation of the average value, not the integral itself. Option D is the arc length. This question tests the student's understanding of the core meaning of the definite integral.

📖 Explanation: For an even function, $f(-x) = f(x)$, so the integral from -2 to 2 is twice the integral from 0 to 2. Therefore, $10 = 2 \int_{0}^{2} f(x) dx$, so $\int_{0}^{2} f(x) dx = 5$. Option B is the value of the symmetric integral, not the half. Option C is the result if the function were odd. Option D is incorrect because evenness provides a definitive relationship. This tests symmetry properties of integrals.

📖 Explanation: Let $u = 1+x^2$, $du = 2x dx$. Then $x dx = du/2$. When $x=0$, $u=1$; when $x=1$, $u=2$. The integral becomes $\frac{1}{2} \int_{1}^{2} u^{-1/2} du = \frac{1}{2} [2\sqrt{u}]_1^2 = \sqrt{2} - 1$. Option B is a common error in evaluating the limits. Option C is the value of $\int_{0}^{1} 1/\sqrt{1+x^2} dx$. Option D is missing the subtraction. This tests substitution in definite integrals.

📖 Explanation: $\int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} \frac{1-\cos 2x}{2} dx = \frac{\pi}{2} - \frac{\sin 2x}{4} \big|_0^\pi = \pi/2$. The student likely used the incorrect identity $\sin^2 x = 1$ or thought the average value was 1. Option A is a trap for students who think the area under $\sin^2 x$ is $\pi$. Option B is incorrect because the antiderivative of $\sin^2 x$ is not $-\cos x$. This highlights a common trigonometric integration error.

📖 Explanation: By the Fundamental Theorem of Calculus Part 2, if $F(x) = \int_{a}^{x} f(t) dt$, then F'(x) = f(x). Here $f(t) = \cos(t^2)$, so F'(x) = \cos(x^2). Option B incorrectly applies the chain rule to the variable of integration. Option C is the derivative of $\sin(x^2)$. Option D is a common error of treating the integrand as $\cos t$. This is a direct Easy of the FTC, but the $t^2$ inside makes it look more complex.

📖 Explanation: The average value of $f$ on [0,2] is $\frac{1}{2} \int_0^2 f(x) dx = 1.5$. Since the average is 1.5, the function must be at least 1.5 at some point, and at most 1.5 at some point (or be constantly 1.5). Option A is not necessarily true because the minimum could be above 1.5 (e.g., $f(x)=2$). Option B is false because the average value is not necessarily attained at a specific point (though it is for continuous functions by the MVT). Option D is true, but the question asks for what must be true, and C is a logical consequence of the average being 1.5.

📖 Explanation: With $u = 1-x$, $du = -dx$, and $x = 1-u$. When $x=0$, $u=1$; when $x=1$, $u=0$. The integral becomes $\int_{1}^{0} (1-u) u^5 (-du)$. Option B is the simplified version after flipping limits: $\int_{0}^{1} (1-u) u^5 du$. But Option A is the correct initial substitution before simplification. Option C is missing the $-du$. Option D is missing the $(1-u)$ factor. This tests the mechanics of substitution in definite integrals, including changing the limits.