📝 Integral curves and families of antiderivatives (25 MCQs)

📖 Explanation: Integral curves of the same function are vertical translations of each other. The constant difference $C_2 - C_1 = 3$ is preserved for all $x$. Since the curves differ by 3 units vertically at $x=2$, they will differ by exactly 3 units at $x=5$. The value $f(5)$ is irrelevant because both curves share the same derivative and hence the same slope at every point.

📖 Explanation: An antiderivative of $f(x)$ is any function whose derivative is $f(x)$. The derivative of $x^3 + 2$ is $3x^2$, and the derivative of $x^3 - 5$ is also $3x^2$. Therefore, both are valid antiderivatives. The constant of integration $C$ can be any real number, and different choices of $C$ produce different integral curves, all belonging to the same family $y = x^3 + C$.

📖 Explanation: The student confuses the concept of vertical translation with parallelism. While integral curves $y = \sin x + C$ are vertical translations of each other, they are not lines, and their slopes are not constant. The slope at any point $x$ is $\cos x$, which varies. Therefore, the curves are not parallel lines; they are sine waves shifted vertically. This is a common misconception: vertical translation does not imply parallelism in non-linear functions.

📖 Explanation: The slope of the tangent line to any integral curve of $f$ at a point is given by the derivative of the curve at that point. Since $y = F(x) + 5$ is an antiderivative of $f$, its derivative is F'(x) = f(x). The constant 5 does not affect the derivative. Therefore, the slope at $x = a$ is $f(a)$. This is a direct Easy of the definition of antiderivatives.

📖 Explanation: An integral curve of $f(x) = 2x$ must have derivative $2x$. The derivative of $y = x^2 + 3$ is $2x$. The derivative of $y = x^2 - 1$ is $2x$. The derivative of $y = x^2 + \pi$ is $2x$. However, the derivative of $y = (x+1)^2$ is $2(x+1) = 2x + 2$, which is not equal to $2x$. Thus, $y = (x+1)^2$ is not an integral curve of $f(x) = 2x$, even though it is a parabola. This tests the ability to verify antiderivatives by differentiation.

📖 Explanation: In a slope field for a first-order differential equation $dy/dx = f(x)$, the slope at any point $(x,y)$ depends only on $x$. This means that for each $x$, all curves have the same slope, regardless of their vertical position. If two integral curves intersected, they would share a point $(x,y)$, and thus have the same derivative at that point. However, two different solutions to the differential equation cannot intersect because that would violate the uniqueness of solutions. Therefore, integral curves for functions of $x$ only are non-intersecting vertical translates.

📖 Explanation: The general integral curve of $f(x) = e^x$ is $y = e^x + C$. To find the specific curve passing through $(0,0)$, we substitute: $0 = e^0 + C$ which gives $0 = 1 + C$, so $C = -1$. Thus the equation is $y = e^x - 1$. This is a classic initial condition problem where the constant of integration is determined by a point on the curve, demonstrating the selection of a specific member from the family of antiderivatives.

📖 Explanation: The constant $C$ represents the vertical translation of the antiderivative $F(x)$. It does not represent the $y$-intercept unless $F(0)=0$. It is not the slope, as the slope is F'(x) = f(x). It is also not the area, though area functions are related to antiderivatives. The key idea is that all antiderivatives differ by a constant, which shifts the entire graph vertically. This is a fundamental concept in integral calculus, linking differentiation and integration.

📖 Explanation: If $f(x) = mx + b$, then its antiderivative is $F(x) = \frac{m}{2}x^2 + bx + C$. This is a quadratic function, whose graph is a parabola. This connects the derivative of a quadratic to a linear function. Students often forget that integrating a linear function (degree 1) results in a quadratic function (degree 2), not a line. The constant of integration $C$ determines the vertical position of the parabola, while the coefficients $m/2$ and $b$ determine its shape and orientation.

📖 Explanation: At a local maximum of an integral curve $y = F(x)$, the derivative F'(x) must be zero and change sign from positive to negative. Since F'(x) = f(x), it follows that $f(2) = 0$. This is a direct Easy of the first derivative test, but in the context of integral curves. The graph of the integral curves provides visual information about where the derivative of the curves, and hence the original function $f$, is zero. It reinforces the connection between the geometry of antiderivatives and the algebra of derivatives.

📖 Explanation: The antiderivative of $\sin x$ is $-\cos x + C$. Therefore, any integral curve of $f(x) = \sin x$ must be of the form $y = -\cos x + C$. Option A, $y = -\cos x + 2$, is of this form with $C=2$. Option B is the antiderivative of $-\sin x$. Option C is the derivative of $-\cos x$ and not an antiderivative of $\sin x$. Option D is not a correct antiderivative. This question tests the memorization of basic trigonometric antiderivatives and the ability to identify them.

📖 Explanation: The fact that two different integral curves pass through $(1,2)$ and $(1,5)$ is impossible for a function of $x$ only. Integral curves for a function $f(x)$ are vertical translations of each other. If they pass through the same $x$-value, they must have the same $y$-value because they are translated vertically. The difference in $y$-values at $x=1$ would be a constant, but they cannot both exist for the same $f$ unless $f$ is not single-valued. The problem contains an inherent contradiction; therefore, the value of $f(1)$ cannot be determined from this invalid scenario. This tests the student's ability to recognize logical inconsistencies.

📖 Explanation: The integral curves are $y = \frac{x^4}{4} + C$. The second derivative is y&#039;&#039; = 3x^2. Since $3x^2$ is positive for $x \neq 0$ and zero at $x=0$, the concavity is up for $x>0$ and down for $x<0$. Thus, the concavity changes at $x=0$, which is a point of inflection. This requires computing the second derivative and analyzing its sign, linking the original function $f(x)$ to the concavity of its integral curves. It tests a deeper understanding of the relationship between a function and its antiderivatives.

📖 Explanation: A horizontal tangent line to the curve $y = F(x)$ at $x=3$ means that the slope of the tangent, which is F&#039;(3), is zero. Since $F$ is an antiderivative of $f$, F&#039;(x) = f(x). Therefore, F&#039;(3) = f(3) = 0. This question connects the geometric property of a horizontal tangent to the algebraic condition that the derivative is zero, and then maps this back to the original function $f$. It emphasizes the inverse relationship between the derivative of $F$ and the function $f$.

📖 Explanation: The derivative of $\tan^{-1} x$ is $\frac{1}{1+x^2}$. Therefore, if $y = \tan^{-1} x + C$ is the family of integral curves, then $f(x)$ must be the derivative of $\tan^{-1} x$, which is $\frac{1}{1+x^2}$. This is a direct Easy of the definition of antiderivatives, where the function $f$ is the derivative of the antiderivative. It tests the ability to recall basic derivative formulas of inverse trigonometric functions.

📖 Explanation: The student is correct. The derivative of $y = e^x + C$ is $e^x$, regardless of the value of $C$. Therefore, the family $y = e^x + C$ represents all possible integral curves of $f(x) = e^x$. This reinforces the idea that the constant of integration generates a family of curves, all of which are valid antiderivatives. It also highlights that $C$ can be any real number, not just positive, which is a common point of confusion.

📖 Explanation: If $y = x^3 + 2x + 1$ is an integral curve, then $f(x)$ is its derivative. The derivative of $x^3 + 2x + 1$ is $3x^2 + 2$. This is a straightforward Easy of the power rule and the definition of integral curves. The constant 1 disappears during differentiation, as expected. This tests the basic skill of moving from an antiderivative back to its derivative.

📖 Explanation: The function $f(x) = |x|$ is not differentiable at $x=0$. However, it is integrable. The antiderivative is piecewise: $F(x) = \frac{x^2}{2} + C$ for $x \ge 0$ and $F(x) = -\frac{x^2}{2} + C$ for $x < 0$. This function is continuous and has a derivative equal to $|x|$ everywhere except possibly at $x=0$. At $x=0$, the derivative from the left is 0 and from the right is 0, so the derivative exists and is 0. Therefore, the integral curves are actually smooth and have no corner at $x=0$. This is a sophisticated point that challenges the misconception that an antiderivative inherits the discontinuities of its derivative.

📖 Explanation: The position $s(t)$ is an antiderivative of the velocity $v(t)$. Since $v(t) = 2t$, the general antiderivative is $s(t) = t^2 + C$. Using the initial condition $s(0)=3$, we find $0^2 + C = 3$, so $C = 3$. Thus, $s(t) = t^2 + 3$. This is a classic Easy of integration to rectilinear motion, demonstrating how initial conditions select a specific integral curve from the family of possible position functions.

📖 Explanation: The slope field for $dy/dx = x^2$ indicates that the slopes are positive for all $x \neq 0$ and zero at $x=0$. The integral curves are parabolas of the form $y = \frac{x^3}{3} + C$. To pass through the origin, we need $0 = 0 + C$, so $C = 0$. Thus, the curve is $y = \frac{x^3}{3}$. This question requires the student to interpret the slope field and select the correct antiderivative based on a given point. It combines visual and algebraic reasoning.

📖 Explanation: If $F(x)$ is even, then $F(-x) = F(x)$. Differentiating both sides with respect to $x$ using the chain rule gives -F&#039;(-x) = F&#039;(x). Since F&#039;(x) = f(x), we have $-f(-x) = f(x)$, which means $f(-x) = -f(x)$. Therefore, $f(x)$ is an odd function. This is an Easy of the chain rule and the definitions of even and odd functions, linking the symmetry of an antiderivative to the symmetry of its derivative. It tests a higher-level concept of function symmetry in the context of integration.

📖 Explanation: The antiderivative of $x^2$ is $\frac{x^3}{3} + C$, not $\frac{x^3}{3} + Cx$. The student incorrectly included the variable $x$ in the constant term. The derivative of $Cx$ is $C$, which would add an extra term $C$ to the derivative, making it $x^2 + C$, not $x^2$. This is a common algebraic mistake where students treat the constant of integration as if it can be a term involving $x$. The constant of integration must be a true constant with respect to the variable of integration.

📖 Explanation: If $f(x) = a_n x^n + \dots$, then an antiderivative $F(x) = \frac{a_n}{n+1} x^{n+1} + \dots$. The degree of the antiderivative is one more than the degree of the original function. This is a general principle of polynomial integration: the power rule increases the exponent by one. It tests the understanding of how the degree of a polynomial changes under integration and how the leading coefficient is transformed.

📖 Explanation: If $F(x)$ is an antiderivative, the family of integral curves is $y = F(x) + C$. Therefore, the family is $y = x^2 + \sin x + C$. The other options are either derivatives of $F$ or completely different antiderivatives. This question is a direct test of the definition of the family of integral curves and the role of the constant of integration.

📖 Explanation: The function $\ln |x|$ is defined for all $x \neq 0$. Therefore, the integral curves $y = \ln |x| + C$ have a domain of all real numbers except zero. This is because the derivative $f(x) = \frac{1}{x}$ is also undefined at $x=0$. This question tests the understanding of the domain of logarithmic functions and how discontinuities in the derivative translate to restrictions on the domain of the antiderivative. It is a common point of confusion where students might incorrectly assume the domain is only positive numbers.