AP Calculus BC Quiz: Fundamental Theorem Of Calculus Accumulation Functions

What this quiz covers

This quiz focuses on Fundamental Theorem Of Calculus Accumulation Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 1
2. $\sqrt{3}$ (correct answer)
3. 3
4. There is no such value of $x$.

Explanation: To find a local minimum of $F(x)$, we analyze the first derivative, $F'(x)$. By the Fundamental Theorem of Calculus, $F'(x) = \frac{x^2 - 3}{x^2 + 1}$. A local minimum can occur at a critical point where $F'(x)$ changes from negative to positive. Set $F'(x)=0$: $\frac{x^2 - 3}{x^2 + 1} = 0 \implies x^2 - 3 = 0 \implies x = \pm\sqrt{3}$. Since we are looking for a positive value of $x$, we consider $x=\sqrt{3}$. The denominator $x^2+1$ is always positive. For $0 < x < \sqrt{3}$, $x^2-3$ is negative, so $F'(x) < 0$. For $x > \sqrt{3}$, $x^2-3$ is positive, so $F'(x) > 0$. Since $F'(x)$ changes from negative to positive at $x=\sqrt{3}$, $F(x)$ has a local minimum there.

Question 2

1. $x=1$
2. $x=3$
3. $x=5$ (correct answer)
4. $x=0$

Explanation: To find local extrema of $F$, we examine its derivative, $F'(x)$. By the Fundamental Theorem of Calculus, $F'(x) = (x-1)^2(x-5)$. Critical points occur when $F'(x)=0$, which are at $x=1$ and $x=5$. For a local minimum, $F'(x)$ must change from negative to positive. The term $(x-1)^2$ is always non-negative. The sign of $F'(x)$ is determined by the term $(x-5)$. For $x<5$, $F'(x) < 0$. For $x>5$, $F'(x) > 0$. Therefore, a local minimum occurs at $x=5$. At $x=1$, the derivative does not change sign.

Question 3

1. $(0, \infty)$
2. $(-\infty, 0)$ (correct answer)
3. $(-\infty, \infty)$
4. $(-1, 1)$

Explanation: The concavity of $h(x)$ is determined by the sign of its second derivative, $h''(x)$. First, we find $h'(x) = e^{-x^2/2}$ using the Fundamental Theorem of Calculus. Then, we differentiate again to find $h''(x)$: $h''(x) = \frac{d}{dx}(e^{-x^2/2}) = e^{-x^2/2} \cdot (-x)$. The graph of $h(x)$ is concave up when $h''(x) > 0$. Since $e^{-x^2/2}$ is always positive, we need $-x > 0$, which implies $x < 0$. So, the interval is $(-\infty, 0)$.

Question 4

1. $g(x) = \int_0^x \sin(t^2) \, dt$
2. $g(x) = 4 + \sin(x^2)$
3. $g(x) = 4 - \int_x^0 \sin(t^2) \, dt$
4. $g(x) = 4 + \int_0^x \sin(t^2) \, dt$ (correct answer)

Explanation: By the Fundamental Theorem of Calculus, if $g'(x) = f(x)$, then $g(x) = g(a) + \int_a^x f(t) \, dt$. Here, $f(x) = \sin(x^2)$, $a=0$, and $g(0)=4$. So, $g(x) = 4 + \int_0^x \sin(t^2) \, dt$.

Question 5

1. $\int_{-1}^{x} e^{2t}\,dt$
2. $e^{2t}$
3. $\int_{-1}^{x} e^{2x}\,dt$
4. $e^{2x}$ (correct answer)
5. $\int_{-1}^{x} e^{2t}\,dx$

Explanation: 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.

Question 6

1. $\displaystyle (3x-4)e^{x}$ (correct answer)
2. $\displaystyle \int_5^x (3t-4)e^{t}\,dt$
3. $\displaystyle (3t-4)e^{t}$
4. $\displaystyle 3e^{x}$
5. $\displaystyle (3x-4)e^{5}$

Explanation: 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.

Question 7

1. $\displaystyle \int_0^x t^2\sin t\,dt$
2. $\displaystyle x^2\sin x$ (correct answer)
3. $\displaystyle \sin x$
4. $\displaystyle x\sin x$
5. $\displaystyle x^2\cos x$

Explanation: 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.

Question 8

1. $\int_{0}^{x}\sqrt{1+t^4}\,dt$
2. $\sqrt{1+x^4}$ (correct answer)
3. $\int_{0}^{x}\sqrt{1+x^4}\,dt$
4. $\sqrt{1+0^4}$
5. $\int_{0}^{x}\frac{4t^3}{2\sqrt{1+t^4}}\,dt$

Explanation: 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.

Question 9

1. $\int_{1}^{x} \frac{t}{t+1}\,dt$
2. $\frac{t}{t+1}$
3. $\int_{1}^{x} \frac{x}{x+1}\,dt$
4. $\int_{1}^{x} \frac{t}{t+1}\,dx$
5. $\frac{x}{x+1}$ (correct answer)

Explanation: 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.

Question 10

1. $\sqrt{1-x}$ (correct answer)
2. $\sqrt{1-t}$
3. $\int_{-1}^{x} \sqrt{1-t}\,dt$
4. $\int_{-1}^{x} \sqrt{1-x}\,dt$
5. $\int_{-1}^{x} \sqrt{1-t}\,dx$

Explanation: 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.

Question 11

1. $\ln(t)$ (correct answer)
2. $\dfrac{1}{t}$
3. $\int_{5}^{t}\ln(x)\,dx$
4. $t\ln(t)-t$
5. $\ln(x)$

Explanation: 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.

Question 12

1. $\sqrt{x+4}$ (correct answer)
2. $\int_{2}^{x} \sqrt{t+4}\,dt$
3. $\int_{2}^{x} \sqrt{x+4}\,dt$
4. $\sqrt{t+4}$
5. $\int_{2}^{x} \sqrt{t+4}\,dx$

Explanation: 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.

Question 13

1. $\ln(x+5)$ (correct answer)
2. $\int_{0}^{x} \ln(t+5)\,dt$
3. $\ln(t+5)$
4. $\int_{0}^{x} \ln(x+5)\,dt$
5. $\int_{0}^{x} \ln(t+5)\,dx$

Explanation: 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.

Question 14

1. $\sin x+x$ (correct answer)
2. $\int_{0}^{x} (\sin t+t)\,dt$
3. $\int_{0}^{x} (\sin x+x)\,dt$
4. $\sin t+t$
5. $\int_{0}^{x} (\sin t+t)\,dx$

Explanation: 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.

Question 15

1. $\int_{-2}^{x} (t^3-2)\,dt$
2. $t^3-2$
3. $x^3-2$ (correct answer)
4. $\int_{-2}^{x} (x^3-2)\,dt$
5. $\int_{-2}^{x} (t^3-2)\,dx$

Explanation: This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if W(x) is defined as the integral from a constant to x of F(t) dt, then W'(x) equals F(x). Here, W(x) accumulates the force F(t) = t³ - 2 from t=-2 to t=x, so its derivative W'(x) should simply be the integrand evaluated at x, which is x³ - 2. FTC Part 1 works because the derivative of the accumulated work up to x is precisely the force at that point x. This is like how the rate of change of total work at position x is exactly the force at that instant. A tempting distractor is choice B, t³ - 2, 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.

Question 16

1. $\int_{1}^{x} (t^2+3t)\,dt$
2. $x^2+3x$ (correct answer)
3. $\int_{1}^{x} (x^2+3x)\,dt$
4. $\int_{1}^{x} (t^2+3t)\,dx$
5. $t^2+3t$

Explanation: This problem requires applying the first part of the Fundamental Theorem of Calculus (FTC Part 1), which states that if A(x) is defined as the integral from a constant to x of r(t) dt, then A'(x) equals r(x). Here, A(x) accumulates the net inflow rate r(t) = t² + 3t from t=1 to t=x, so its derivative A'(x) should simply be the integrand evaluated at x, which is x² + 3x. FTC Part 1 works because the derivative of the accumulated area under the curve r(t) up to x is precisely the height of the function at that point x. This is like how the rate of change of total water in the tank at time x is exactly the inflow rate at that instant. A tempting distractor is choice E, t² + 3t, 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.

Question 17

1. $\displaystyle \int_{-2}^x \frac{1}{1+t^4}\,dt$
2. $\displaystyle \frac{1}{1+t^4}$
3. $\displaystyle \frac{1}{1+x^4}$ (correct answer)
4. $\displaystyle \frac{4x^3}{(1+x^4)^2}$
5. $\displaystyle \frac{1}{1+(-2)^4}$

Explanation: This problem demonstrates the Fundamental Theorem of Calculus Part 1, which connects accumulation functions to their rate functions. Since F(x) = $\int_{-2}^x \frac{1}{1+t^4} \, dt$, applying FTC Part 1 gives us F'(x) = $\frac{1}{1+x^4}$. The theorem states that differentiating an integral with respect to its upper limit yields the integrand evaluated at that upper limit. We replace the dummy variable t with x to get our answer. Choice B incorrectly keeps the dummy variable t, while choice D mistakenly tries to differentiate the integrand using the chain rule. To apply FTC Part 1 correctly, identify that x appears only as an integration limit, then substitute x for the dummy variable in the integrand.

Question 18

1. $\int_{3}^{x}\frac{t}{\sqrt{t^2+9}}\,dt$
2. $\frac{3}{\sqrt{3^2+9}}$
3. $\frac{x}{\sqrt{x^2+9}}$ (correct answer)
4. $\int_{3}^{x}\frac{x}{\sqrt{x^2+9}}\,dt$
5. $\int_{3}^{x}\frac{9}{(t^2+9)^{3/2}}\,dt$

Explanation: 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 P(x) = ∫ from a constant lower limit to x of f(t) dt, then P'(x) = f(x), provided f is continuous at x. For this, f(t) = t / √(t² + 9), so P'(x) = x / √(x² + 9). This principle shows the derivative is the integrand substituted with x. A tempting distractor is choice E, which integrates an expression resembling the derivative of the integrand, but that confuses FTC with unnecessary differentiation. 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.

Question 19

1. $\int_{5}^{x}\sin(t^2)\,dt$
2. $\sin(5^2)$
3. $\sin(x^2)$ (correct answer)
4. $\int_{5}^{x}\sin(x^2)\,dt$
5. $\int_{5}^{x}2t\cos(t^2)\,dt$

Explanation: 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 G(x) = ∫ from a constant lower limit to x of f(t) dt, then G'(x) = f(x), provided f is continuous at x. The integrand is f(t) = sin(t²), so G'(x) = sin(x²). This means the derivative captures the integrand's value at the upper limit x. A tempting distractor is choice E, which is the integral of the chain rule derivative of the integrand, but FTC Part 1 simply evaluates the original integrand at x without further differentiation. 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.

Question 20

1. $\int_{2}^{x}(t^3-4t+1)\,dt$
2. $x^3-4x+1$ (correct answer)
3. $2^3-4\cdot2+1$
4. $\int_{2}^{x}(x^3-4x+1)\,dt$
5. $\int_{2}^{x}(3t^2-4)\,dt$

Explanation: 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 A(x) = ∫ from a constant lower limit to x of f(t) dt, then A'(x) = f(x), provided f is continuous at x. In this case, the integrand f(t) = t³ - 4t + 1, so replacing t with x gives A'(x) = x³ - 4x + 1. This works because the derivative of the accumulated area up to x is the height of the function at x. A tempting distractor is choice E, which is the integral of the derivative of the integrand, but that incorrectly applies the chain rule and does not follow FTC Part 1. 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.