📝 Fundamental theorem of calculus part 1 (24 MCQs)

📖 Explanation: The Fundamental Theorem of Calculus requires the integrand $f(x)$ to be continuous on the entire closed interval $[a, b]$. Here, $f(x) = \frac{1}{x^2}$ has an infinite discontinuity at $x = 0$, which lies within $[-1, 2]$. Thus, the theorem's conditions are violated, and the computation is invalid. The negative result is a consequence of this misEasy, not a valid area.

📖 Explanation: The displacement of a particle is the net change in position, which is the integral of its velocity function $v(t)$ over the time interval. By the Fundamental Theorem of Calculus, this integral is equal to the difference of the antiderivative $s(t) = t^3 - 6t^2 + 9t$ evaluated at the limits. Option A calculates the total distance traveled, not displacement, as it integrates the absolute value of velocity.

📖 Explanation: This is a direct restatement of the Fundamental Theorem of Calculus Part 2 (which is often presented as Part 1 in some texts). It states that the derivative of an integral with a variable upper limit is the integrand evaluated at that limit, provided the integrand is continuous. This theorem establishes the inverse relationship between differentiation and integration.

📖 Explanation: The Fundamental Theorem of Calculus Part 1 guarantees that if $f$ is continuous on the closed interval $[a, b]$, then the definite integral can be evaluated using any antiderivative $F$ such that F' = f. While differentiability of the antiderivative is implied, the core condition for the integrand is continuity. Options C and D are not required for the theorem's Easy.

📖 Explanation: By the Fundamental Theorem of Calculus, F'(x) = \sqrt{x^3 + 1}. Evaluating this at $x=0$ gives F'(0) = \sqrt{0^3 + 1} = 1. The value of $F(0)$ is $\int_2^0 \sqrt{t^3+1} \, dt$, which is negative because the integral is from 2 down to 0, but the question asks about the derivative. The function is well-defined for all real $x$ as the integrand is continuous everywhere.

📖 Explanation: The statement is false. The definite integral being zero implies that the net signed area is zero, which can happen if the positive and negative areas cancel out. For example, $\int_{-1}^1 x \, dx = 0$ even though $f(x)=x$ is not identically zero. A function can be positive over some parts of the interval and negative over others, leading to a net area of zero without the function being zero everywhere.

📖 Explanation: The absolute value function must be split at the point where the expression inside changes sign, which is $x=1$. On $[0,1]$, $x-1 \le 0$, so $|x-1| = -(x-1)$. On $[1,2]$, $x-1 \ge 0$, so $|x-1| = x-1$. The integral then becomes $\int_0^1 -(x-1) \, dx + \int_1^2 (x-1) \, dx = [x - x^2/2]_0^1 + [x^2/2 - x]_1^2 = (1/2) + (1/2) = 1$. Option B correctly splits the integral. Option A is incorrect because $\frac{(x-1)^2}{2}$ is only an antiderivative of $x-1$, not $|x-1|$.

📖 Explanation: The Fundamental Theorem of Calculus Part 2 states that $\frac{d}{dx} \int_a^x f(t) \, dt = f(x)$. Therefore, the derivative of the integral function $F(x) = \int_0^x f(t) \, dt$ at any point $x$ is simply the value of the integrand $f$ at that point. This is independent of the areas or slopes; it directly evaluates the function defining the area's rate of change.

📖 Explanation: By the Fundamental Theorem of Calculus, both $F(x)$ and $G(x)$ are antiderivatives of $f(x) = x^2$. The derivative of any antiderivative of a function is that function itself. Adding a constant to a function does not change its derivative. Therefore, F'(x) = x^2 and G'(x) = x^2, so the derivatives are equal.

📖 Explanation: $\int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2$. This is correct. Option B is incorrect because $\frac{1}{x}$ is not continuous on $[-1, 1]$. Option C is incorrect: $\int_1^2 x^{-2} dx = [-x^{-1}]_1^2 = -1/2 - (-1) = 1/2$. Option D is incorrect: $\int_0^1 e^x dx = e^1 - e^0 = e - 1$.

📖 Explanation: To find g'(x), we use the property that $\int_{x^2}^{5} \cos t \, dt = - \int_{5}^{x^2} \cos t \, dt$. Then, by the Fundamental Theorem of Calculus and the Chain Rule, g'(x) = -\frac{d}{dx} \int_5^{x^2} \cos t \, dt = -\cos(x^2) \cdot (2x) = -2x \cos(x^2). The lower limit is constant, and the upper limit is a function of $x$, requiring the chain rule. Option A ignores the chain rule and the reversed limits.

📖 Explanation: F'(x) = f(x). A local maximum of $F$ occurs where its derivative changes from positive to negative, which is a sign change of $f(x)$ from positive to negative. The points where $f(x)$ is greatest are just the largest values of the derivative, not necessarily extrema of $F$. A sign change from negative to positive indicates a local minimum of $F$.

📖 Explanation: The integrand $f(x) = \frac{1}{x-2}$ has a vertical asymptote at $x=2$, which is within the interval of integration $[1, 3]$. The function is not continuous on this interval, so the Fundamental Theorem of Calculus cannot be applied directly. The antiderivative $\ln|x-2|$ is not defined (and its derivative is not $f(x)$) at $x=2$. The result 0 is misleading and incorrect, as the improper integral would diverge.

📖 Explanation: The definite integral is additive over adjacent intervals. This property can be stated as $\int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx$. Applying this property, $\int_0^5 f(x) \, dx = \int_0^2 f(x) \, dx + \int_2^5 f(x) \, dx = 5 + (-2) = 3$. This is a fundamental property that follows from the Riemann sum definition or the Fundamental Theorem.

📖 Explanation: The total distance traveled is the integral of speed, which is the absolute value of velocity: $\int_0^3 |2t-4| \, dt$. The velocity changes sign at $t=2$. So, the integral is $\int_0^2 (4-2t) \, dt + \int_2^3 (2t-4) \, dt = [4t - t^2]_0^2 + [t^2 - 4t]_2^3 = (8-4) + (9-12) - (4-8) = 4 + (-3) - (-4) = 5$. Therefore, both B and C are correct expressions for the total distance. Option A calculates the displacement, which is -3.

📖 Explanation: The Fundamental Theorem of Calculus states that F'(x) = \frac{1}{1+x^2}. Therefore, F'(1) = \frac{1}{1+1^2} = \frac{1}{2}. The value $\pi/4$ is the value of $F(1)$, which is $\int_1^1 \frac{1}{1+t^2} \, dt = 0$. It is important to distinguish between the function's value and its derivative. This question tests the direct Easy of the theorem.

📖 Explanation: This limit is the definition of the derivative of the function $F(x) = \int_2^x e^{t^2} \, dt$ at the point $x=2$. By the Fundamental Theorem of Calculus, F'(x) = e^{x^2}. Therefore, the limit is F'(2) = e^{2^2} = e^4. This question connects the limit definition of the derivative with the Fundamental Theorem, requiring recognition of the derivative form.

📖 Explanation: The Fundamental Theorem of Calculus gives \int_1^4 f'(x) \, dx = f(4) - f(1). Therefore, f(4) = f(1) + \int_1^4 f'(x) \, dx = 3 + \int_1^4 2x \, dx. Evaluating the integral: $\int_1^4 2x \, dx = [x^2]_1^4 = 16 - 1 = 15$. Thus, $f(4) = 3 + 15 = 18$. This is a classic Easy of the theorem to reconstruct a function from its derivative and an initial condition.

📖 Explanation: The Fundamental Theorem of Calculus does not require the integrand $f(x)$ to be non-negative. If $f(x)$ is negative on some parts of the interval, the definite integral computes the net signed area. The theorem applies to continuous functions (or more generally, integrable functions with an antiderivative). The other statements are true: continuity is a key condition, any antiderivative works, and the result is independent of the constant of integration.

📖 Explanation: Both the left endpoint sum (Option B) and the right endpoint sum (Option A) for a continuous function will converge to the same definite integral $\int_0^1 x^2 \, dx$ as $n \to \infty$. The Fundamental Theorem then provides a way to evaluate this limit exactly: $\int_0^1 x^2 \, dx = [x^3/3]_0^1 = 1/3$. This shows the connection between the Riemann sum definition and the Fundamental Theorem as a tool for exact evaluation.

📖 Explanation: The Fundamental Theorem of Calculus states that $F(x) = \int_0^x t^3 \, dt$ is an antiderivative of $f(x) = x^3$. Adding a constant to any antiderivative yields another antiderivative. Therefore, both $F(x)$ and $G(x) = F(x) + 2$ are antiderivatives of $x^3$, differing only by a constant. This illustrates the general solution to an antiderivative problem.

📖 Explanation: The antiderivative of $(x-1)^2$ is $\frac{(x-1)^3}{3}$. The definite integral is $\left[ \frac{(x-1)^3}{3} \right]_0^2 = \frac{(1)^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left( -\frac{1}{3} \right) = \frac{2}{3}$. The function is always non-negative, so the area cannot be zero. The symmetry argument fails because the curve is not symmetric about the x-axis; it's a parabola shifted and the area under it is positive. The correct Easy of the theorem gives $2/3$.

📖 Explanation: Since the integrand is defined piecewise, the integral over $[0,2]$ must be split at the point where the definition changes, which is $x=1$. $\int_0^2 f(x) \, dx = \int_0^1 x \, dx + \int_1^2 (2-x) \, dx$. Using the Fundamental Theorem: $[x^2/2]_0^1 + [2x - x^2/2]_1^2 = (1/2 - 0) + (4-2) - (2-1/2) = 1/2 + 2 - 3/2 = 1$. This is the net area under the piecewise linear function, which forms a triangle of base 2 and height 1.

📖 Explanation: By definition, $F(x) = \int_0^x f(t) \, dt$ represents the net signed area between the curve $y = f(t)$ and the t-axis from 0 to $x$. Therefore, $F(2)$ is the net area from 0 to 2. The derivative F'(x) = f(x), so $f(2)$ is the rate of change of the area (the slope of $F$), not the area itself. Option B and D refer to $f(2)$, not $F(2)$.