This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if $P(x)$ is defined as the integral from a constant to x of $g(t) , dt$, then $P'(x)$ equals $g(x)$. Here, $P(x)$ accumulates the growth rate $g(t) = e^{2t}$ from t=-1 to t=x, so its derivative $P'(x)$ should simply be the integrand evaluated at x, which is $e^{2x}$. FTC Part 1 works because the derivative of the accumulated population up to x is precisely the growth rate at that point x. This is like how the rate of change of total bacteria at time x is exactly the growth rate at that instant. A tempting distractor is choice B, $e^{2t}$, but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the instantaneous rate of growth. Given G(x) = ∫₅ˣ (3t-4)eᵗ dt, the FTC Part 1 tells us that G'(x) equals the integrand evaluated at x, which gives us (3x-4)eˣ. The derivative of an accumulation function is simply the integrand with the dummy variable t replaced by the upper limit x. Choice C incorrectly retains the dummy variable t, failing to evaluate at the upper limit. Choice D attempts to differentiate the integrand rather than evaluate it. When you see a derivative of an integral where x appears only in the limits, apply FTC Part 1 by substituting the upper limit into the integrand.

This problem requires applying the Fundamental Theorem of Calculus Part 1, which states that the derivative of an accumulation function equals the integrand evaluated at the upper limit. Since A(x) = ∫₀ˣ v(t) dt where v(t) = t²sin t, we have A'(x) = v(x) = x²sin x. The FTC Part 1 tells us that when we differentiate an integral with respect to its upper limit, we simply substitute that upper limit into the integrand. Choice A incorrectly suggests the derivative is the integral itself, failing to apply the FTC. To recognize FTC Part 1 problems, look for derivatives of integrals where the variable appears only in the limits of integration.

This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1) to find the derivative of an accumulation function. The FTC Part 1 states that if $V(x) = \int_{\text{constant}}^x f(t), dt$, then $V'(x) = f(x)$, provided f is continuous at x. Here, the integrand $f(t) = \sqrt{1 + t^4}$, so $V'(x) = \sqrt{1 + x^4}$. This is because the instantaneous rate of inflow at x equals the integrand evaluated at x. A tempting distractor is choice E, which looks like the integral of the derivative of the integrand, but that misapplies differentiation under the integral sign and ignores FTC's direct result. To recognize FTC Part 1 problems, look for questions asking for the derivative of a function defined as a definite integral with a variable upper limit.

This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if $B(x)$ is defined as the integral from a constant to x of $d(t) , dt$, then $B'(x)$ equals $d(x)$. Here, $B(x)$ accumulates the drain rate $d(t) = \frac{t}{t+1}$ from t=1 to t=x, so its derivative $B'(x)$ should simply be the integrand evaluated at x, which is $\frac{x}{x+1}$. FTC Part 1 works because the derivative of the accumulated battery drain up to x is precisely the drain rate at that point x. This is like how the rate of change of total drain at time x is exactly the drainage rate at that instant. A tempting distractor is choice B, $\frac{t}{t+1}$, but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.

This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if R(x) is defined as the integral from a constant to x of f(t) dt, then R'(x) equals f(x). Here, R(x) accumulates the flow rate f(t) = √(1-t) from t=-1 to t=x, so its derivative R'(x) should simply be the integrand evaluated at x, which is √(1-x). FTC Part 1 works because the derivative of the accumulated volume up to x is precisely the flow rate at that point x. This is like how the rate of change of total water flowed at time x is exactly the flow rate at that instant. A tempting distractor is choice B, √(1-t), but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.

This problem demonstrates the Fundamental Theorem of Calculus Part 1 for finding the derivative of an accumulation function. Given $D(t) = \int_{5}^{t}\ln(x),dx$, we want $D'(t)$. By FTC Part 1, when we differentiate an integral with respect to its upper limit, we get the integrand evaluated at that limit. Thus, $D'(t) = \ln(t)$, which is the integrand with $x$ replaced by $t$. Choice B ($\frac{1}{t}$) is the derivative of $\ln(t)$, which students might select if they incorrectly differentiate the result, but FTC Part 1 only requires substitution. When differentiating $\int_{a}^{t} f(x),dx$, the result is $f(t)$—no additional differentiation needed.

This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if V(x) is defined as the integral from a constant to x of a(t) dt, then V'(x) equals a(x). Here, V(x) accumulates the acceleration a(t) = √(t+4) from t=2 to t=x, so its derivative V'(x) should simply be the integrand evaluated at x, which is √(x+4). FTC Part 1 works because the derivative of the accumulated velocity up to x is precisely the acceleration at that point x. This is like how the rate of change of speed at time x is exactly the acceleration at that instant. A tempting distractor is choice D, √(t+4), but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.

This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if O(x) is defined as the integral from a constant to x of q(t) dt, then O'(x) equals q(x). Here, O(x) accumulates the output rate q(t) = ln(t+5) from t=0 to t=x, so its derivative O'(x) should simply be the integrand evaluated at x, which is ln(x+5). FTC Part 1 works because the derivative of the accumulated output up to x is precisely the output rate at that point x. This is like how the rate of change of total output at time x is exactly the production rate at that instant. A tempting distractor is choice C, ln(t+5), but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.

This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if S(x) is defined as the integral from a constant to x of v(t) dt, then S'(x) equals v(x). Here, S(x) accumulates the speed v(t) = sin t + t from t=0 to t=x, so its derivative S'(x) should simply be the integrand evaluated at x, which is sin x + x. FTC Part 1 works because the derivative of the accumulated distance up to x is precisely the speed at that point x. This is like how the rate of change of position at time x is exactly the velocity at that instant. A tempting distractor is choice D, sin t + t, but it fails because it doesn't replace the dummy variable t with the upper limit x, leaving it in terms of t instead of expressing the derivative as a function of x. To recognize FTC Part 1 problems, look for a function defined as an integral with a variable upper limit and a request for its derivative, which will always be the integrand with t swapped for x.