AP Calculus AB

AP Calculus AB Practice Test: Practice Test 1

Practice Test 1 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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A population satisfies $\dfrac{dP}{dt}=4P$ . What is the general solution for $P$ as a function of $t$ ?

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Question 1

A population satisfies $\dfrac{dP}{dt}=4P$ . What is the general solution for $P$ as a function of $t$ ?

$P=4e^{t}+C$
$P=Ce^{4t}$ (correct answer)
$P=e^{4t}$
$\ln|P|=4t$
$P=C+e^{4t}$

Explanation: This differential equation requires the separation of variables technique to find the general solution. We can rewrite $\frac{dP}{dt} = 4P$ as $\frac{dP}{P} = 4dt$ , separating the variables P and t on opposite sides. Integrating both sides gives us $\ln|P| = 4t + C_1$ , where $C_1$ is an arbitrary constant. Exponentiating both sides yields $P = e^{4t + C_1} = e^{C_1} \cdot e^{4t}$ . Since $e^{C_1}$ is just another arbitrary positive constant, we can write this as $P = Ce^{4t}$ where $C$ can be any real number. Choice A incorrectly treats this as a simple integration problem without recognizing the multiplicative relationship. The key to recognizing when to use separation of variables is seeing the derivative equal to a product/quotient where one factor contains only the dependent variable and the other contains only the independent variable.

Question 2

A rate model uses $w(x)=(2x+1)\,3^{x}$ . What is $w'(x)$ ?

$2\cdot 3^{x}$
$(2x+1)\,3^{x}\ln 3$
$2+3^{x}\ln 3$
$2\cdot 3^{x}+(2x+1)\,3^{x}\ln 3$ (correct answer)
$(2x+1)\,3^{x}$

Explanation: The function $w(x) = (2x+1) \cdot 3^x$ requires the product rule for a product involving an exponential function. Let $u = 2x+1$ with $u' = 2$ , and $v = 3^x$ with $v' = 3^x \ln 3$ . Applying the product rule: $w'(x) = 2 \cdot 3^x + (2x+1) \cdot 3^x \ln 3$ . This gives us $w'(x) = 2 \cdot 3^x + (2x+1)3^x \ln 3$ . A common mistake with exponential functions like $3^x$ is forgetting the chain rule factor of $\ln 3$ . Remember that $(a^x)' = a^x \ln a$ for any positive constant $a \neq 1$ . When products involve exponential functions, use the product rule just as with any other functions, but be careful with the exponential derivatives.

Question 3

A savings account balance is $B(t)$ . Which expression represents the instantaneous earning rate at $t=12$ months?

$\dfrac{B(12)-B(0)}{12}$
$\dfrac{B(18)-B(6)}{18-6}$
$\lim_{h\to 0}\dfrac{B(12+h)-B(12)}{h}$ (correct answer)
$B(12)-B(11)$
$\dfrac{B(13)-B(12)}{13-12}$

Explanation: The instantaneous earning rate at $t=12$ months represents how fast the account balance is growing at that exact time, which is the derivative of the balance function. Choice C correctly expresses this as $\lim_{h\to 0}\frac{B(12+h)-B(12)}{h}$ , capturing the precise rate of change through the limit definition of the derivative. Choices A and B calculate average earning rates over different time periods, while choices D and E approximate using one-month differences but without the limiting process. Students often confuse the earnings over one month (like $B(12)-B(11)$ ) with the instantaneous rate, but this gives an average rate for that month, not the rate at the instant $t=12$ . The distinction: instantaneous rates use limits to find the rate at a single moment, while finite differences give average rates over intervals.

Question 4

A continuous function $g$ on $[0,6]$ has critical points $x=2,5$ and values $g(0)=1,g(2)=4,g(5)=0,g(6)=3$ ; where is the absolute minimum?

$x=2$
$x=0$
$x=6$
$x=5$ (correct answer)
There is no absolute minimum on $[0,6]$

Explanation: This problem applies the Candidates Test with given function values to find the absolute minimum. The Candidates Test requires checking all critical points (x = 2, 5) and endpoints (x = 0, 6) for a continuous function on [0, 6]. We have g(0) = 1, g(2) = 4, g(5) = 0, and g(6) = 3, so we compare these four values to find the smallest. The minimum value is 0, which occurs at x = 5, making this the location of the absolute minimum. Students might incorrectly choose x = 0 thinking the endpoint with the second-smallest value is the answer, but the Candidates Test requires finding the actual minimum value among all candidates. Remember the Candidates Test checklist: identify all critical points, include both endpoints, evaluate the function at all candidates, and compare to find extrema.

Question 5

A measurement model is $r(t)=\arcsin\!\left(\frac{t-4}{3}\right)$ . What is $r'(t)$ ?

$\dfrac{1}{\sqrt{1-\left(\frac{t-4}{3}\right)^2}}$
$\dfrac{1}{3\sqrt{1-\left(\frac{t-4}{3}\right)^2}}$ (correct answer)
$\dfrac{1}{3\sqrt{1+\left(\frac{t-4}{3}\right)^2}}$
$-\dfrac{1}{3\sqrt{1-\left(\frac{t-4}{3}\right)^2}}$
$\cos\!\left(\frac{t-4}{3}\right)$

Explanation: This problem involves differentiating arcsin with a fractional linear argument. The derivative of arcsin(u) is 1/√(1-u²), and for u = (t-4)/3, we need the chain rule. Computing: r'(t) = 1/√(1-((t-4)/3)²) · d/dt((t-4)/3) = 1/√(1-((t-4)/3)²) · (1/3) = 1/(3√(1-((t-4)/3)²)). Option A omits the factor of 1/3 from differentiating the inner function. When differentiating expressions like (t-a)/b, the derivative is 1/b, which must be included.

Question 6

A tank contains $V(t)$ liters; it leaks at a rate proportional to the amount present. Which differential equation models $V$ ?

$\dfrac{dV}{dt}=k$
$\dfrac{dV}{dt}=-kV$ (correct answer)
$\dfrac{dV}{dt}=kV+t$
$\dfrac{dV}{dt}=-\dfrac{k}{V}$
$\dfrac{dt}{dV}=-kV$

Explanation: This problem requires setting up a differential equation from a rate description involving proportional decay. The tank leaks at a rate proportional to the current amount V(t), which means the rate of change of volume is negative and directly related to the volume itself. Since "proportional to the amount present" means the leak rate equals k times V(t), we get dV/dt = -kV where k > 0. The negative sign is crucial because the volume is decreasing due to leaking. Choice A represents constant rate change, not proportional to volume, while choice D has the wrong form with V in the denominator. Always ensure the sign reflects whether the quantity is increasing or decreasing, and check that proportional relationships translate to multiplication by a constant.

Question 7

The table suggests $c(x)$ approaches $-1$ as $x$ approaches $4$ . Which limit expression correctly represents this behavior?

$\displaystyle c(4)=-1$
$\displaystyle \lim_{x\to 4} c(x)=-1$ (correct answer)
$\displaystyle \lim_{x\to -1} c(x)=4$
$\displaystyle \lim_{x\to 4^-} c(x)=1$
$\displaystyle \lim_{x\to 4} c(4)=-1$

Explanation: Table suggests c(x) to -1 at x=4. \lim_{x\to 4} c(x)=-1 represents. Valid both sides. Error: \lim_{x\to 4} c(4)=-1. c(4)=-1 value. Separate. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 8

For $q(x)=\begin{cases}\frac{x^2-16}{x-4},&x\ne4\\9,&x=4\end{cases}$ , what value should replace $q(4)$ to remove the discontinuity?

$9$
$4$
$0$
$-8$
$8$ (correct answer)

Explanation: The function q(x) has a removable discontinuity at x = 4 because the numerator x² - 16 = (x - 4)(x + 4) contains the factor (x - 4) that cancels with the denominator. After canceling, we get q(x) = x + 4 for x ≠ 4. To make q continuous, we need q(4) = lim(x→4) (x + 4) = 4 + 4 = 8. A common mistake is keeping the current value of 9 or thinking the function should equal 0 at the discontinuity. The strategy is to recognize difference of squares factorizations, cancel common factors, then evaluate at the point of discontinuity.

Question 9

A model for revenue is $R(t)=5t^4-3t^2+7$ . What is $R'(t)$ ?

$20t^3-6t$ (correct answer)
$5t^3-3t+7$
$20t^4-6t^2$
$20t^3-3t^2$
$20t^3+6t$

Explanation: This problem requires applying the power rule to find the derivative of a revenue function. The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1). For R(t) = 5t^4 - 3t^2 + 7, we apply the power rule to each term: the derivative of 5t^4 is 5·4t^3 = 20t^3, the derivative of -3t^2 is -3·2t^1 = -6t, and the derivative of the constant 7 is 0. Therefore, R'(t) = 20t^3 - 6t. A common error (choice C) is to forget to reduce the exponent by 1, keeping t^4 and t^2 instead of t^3 and t. When differentiating polynomials, multiply the coefficient by the exponent and reduce the exponent by 1 for each term.

Question 10

A rectangle has one side on the $x$ -axis and top corners on $y=9-x^2$ ; if the right top corner is at $x$ , which $x$ maximizes area $A=2x(9-x^2)$ ?

$x=0$
$x=3$
$x=\sqrt{\frac{9}{2}}$
$x=\sqrt{6}$
$x=\sqrt{3}$ (correct answer)

Explanation: This problem involves solving optimization problems by maximizing the area A = 2x(9 - x²) of a rectangle under y=9-x². To find the maximum, compute the derivative A'(x) = 2(9 - x²) - 4x² = 18 - 6x² and set it to zero, yielding x = √3 as the critical point. Evaluating A at x = √3 gives the maximum. The interval is implicit for x > 0, and no endpoints are needed. A tempting distractor is x = 3, where area is zero, but this is outside the domain or a minimum. A transferable optimization-solving strategy is to identify critical points by setting the derivative to zero, evaluate the function at those points and any endpoints if applicable, and compare values to determine the extreme.

Question 11

Let $h(x)=x^2+2x$ (restricted to $x\ge -1$ ) and $h(3)=15$ . What is $(h^{-1})'(15)$ ?

$8$
$\dfrac{1}{8}$ (correct answer)
$\dfrac{1}{h'(15)}$
$h'(3)$
$\dfrac{1}{15}$

Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where h(b) = a. Here, h(3) = 15, so b = 3 and a = 15. Compute h'(x) = 2x + 2, so h'(3) = 8, thus (h⁻¹)'(15) = 1/8. A tempting distractor is A, 8, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to h, not to h⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

Question 12

For $f(x)=\sin\!\left(\ln(3x^2-5x+4)\right)$ , what is $f'(x)$ ?

$\cos\!\left(\ln(3x^2-5x+4)\right)$
$\cos\!\left(\ln(3x^2-5x+4)\right)\cdot\dfrac{1}{3x^2-5x+4}$
$\cos\!\left(\ln(3x^2-5x+4)\right)\cdot\dfrac{6x-5}{3x^2-5x+4}$ (correct answer)
$\sin\!\left(\ln(3x^2-5x+4)\right)\cdot\dfrac{6x-5}{3x^2-5x+4}$
$\dfrac{6x-5}{3x^2-5x+4}$

Explanation: This function has a three-layer composition: sine (outer), natural log (middle), and a polynomial (inner). Using the chain rule, we differentiate from outside to inside: the derivative of sin(u) is cos(u), times the derivative of ln(v) which is 1/v, times the derivative of 3x²-5x+4 which is 6x-5. Combining these gives cos(ln(3x²-5x+4)) · (1/(3x²-5x+4)) · (6x-5). A common error is forgetting to multiply by the derivative of the innermost function (6x-5). The pattern to remember: for sin(ln(polynomial)), you'll always have cos(ln(...)) multiplied by a fraction with the polynomial's derivative in the numerator and the polynomial itself in the denominator.

Question 13

The graph of $f$ is increasing and concave down on $(0,5)$ . Which graph could represent $f'$ on $(0,5)$ ?

A curve above the $x$ -axis that is increasing on $(0,5)$ .
A curve below the $x$ -axis that is decreasing on $(0,5)$ .
A curve above the $x$ -axis that is decreasing on $(0,5)$ . (correct answer)
A curve that crosses the $x$ -axis within $(0,5)$ from negative to positive.
A horizontal line at $0$ on $(0,5)$ .

Explanation: This question tests graph-derivative reasoning by connecting the function's monotonicity and concavity to its derivative's behavior. On (0,5), f increasing means f' > 0, so above the x-axis, and concave down means f'' < 0, so f' is decreasing. Thus, f' is positive and decreasing on (0,5), matching choice C. This ensures the slope starts higher and reduces while staying positive, consistent with slowing increase and downward curvature. A tempting distractor is A, positive and increasing, but that would imply f'' > 0, making f concave up instead of down. For checking sketches, always link concavity to whether f' is increasing or decreasing, and confirm signs match monotonicity.

Question 14

Let $Q(x)=\int_{5}^{x}\ln(1+t)\,dt$ for $x>-1$ . What is $Q'(x)$ ?

$\ln(1+x)$ (correct answer)
$\dfrac{1}{1+x}$
$\int_{5}^{x}\ln(1+t)\,dt$
$\ln(1+t)$
$\left[(1+t)\ln(1+t)-t\right]_{5}^{x}$

Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function Q(x) is defined as the integral from 5 to x of ln(1 + t) dt, which accumulates the area under the curve of f(t) = ln(1 + t) from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative Q'(x) equals the integrand evaluated at x, which is ln(1 + x). This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from 5 to x, but that actually computes Q(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.

Question 15

An object’s position is $s(x)=9x^4+2x^3-5$ . Find $s'(x)$ .

$36x^3+6x^2$ (correct answer)
$36x^4+6x^3$
$9x^3+2x^2$
$36x^3+6x^2-5$
$12x^3+6x^2$

Explanation: This problem requires applying the power rule to find the derivative of a polynomial representing an object's position. The power rule dictates that for c x^n, the derivative is c n x^{n-1}, handling each term separately. Differentiating 9x^4 gives 36x^3, while 2x^3 becomes 6x^2, and the constant -5 becomes 0. Combining these yields s'(x) = 36x^3 + 6x^2. A tempting distractor is choice D, which wrongly includes the constant -5, but constants disappear in derivatives. When differentiating polynomials, apply the power rule independently to each term and combine the results, remembering constants become zero.

Question 16

For $h(x)=\ln\!\left(\dfrac{x^2+1}{\sqrt{5x-2}}\right)$ , which differentiation rules should be applied to find $h'(x)$ ?

Power rule and product rule
Quotient rule only
Chain rule only
Chain rule and quotient rule (correct answer)
Implicit differentiation only

Explanation: This question assesses your ability to identify the appropriate differentiation procedures for a logarithmic function. The function $h(x)=\ln\left(\frac{x^2+1}{\sqrt{5x-2}}\right)$ involves the natural logarithm of a quotient. The chain rule is needed first because we have a composite function (logarithm of something). Inside the logarithm, we have a quotient $\frac{x^2+1}{\sqrt{5x-2}}$ , which requires the quotient rule when differentiating. Note that we could alternatively use logarithm properties to rewrite this as $\ln(x^2+1) - \frac{1}{2}\ln(5x-2)$ before differentiating, but the question asks which rules are needed for the function as given. The key recognition strategy is: when you see $\ln$ of a fraction, you'll need the chain rule for the outer function and the quotient rule for the inner fraction.

Question 17

A bacterium culture grows at rate $g(t)$ cells/hour, $t$ hours after start. What does $\int_{2}^{5} g(t)\,dt$ represent?

The number of cells present at $t=5$ , in cells
The net increase in cells from $t=2$ to $t=5$ , in cells (correct answer)
The growth rate at $t=5$ , in cells per hour
The average growth rate from $t=2$ to $t=5$ , in cells
The change in growth rate from $t=2$ to $t=5$ , in cells per hour

Explanation: This problem tests the interpretation of definite integrals when the integrand is a growth rate. Since g(t) represents the bacterium culture's growth rate in cells per hour, integrating this rate from t=2 to t=5 hours gives the total increase in cell count during this time period. By the definition of definite integrals as accumulation, ∫₂⁵ g(t)dt represents the net number of cells added to the culture between hour 2 and hour 5. Choice D is incorrect because it suggests the integral gives an average growth rate, but that would require dividing by the time interval (3 hours). The units confirm our interpretation: (cells/hour) × hours = cells, verifying the integral measures net cell increase.

Question 18

Let $X(x)=\int_{-2}^{x}\dfrac{1}{(t^2+1)^{1/2}}\,dt$ . What is $X'(x)$ ?

$\dfrac{1}{\sqrt{x^2+1}}$ (correct answer)
$\dfrac{-x}{(x^2+1)^{3/2}}$
$\int_{-2}^{x}\dfrac{1}{(t^2+1)^{1/2}}\,dt$
$\dfrac{1}{\sqrt{t^2+1}}$
$\left[\ln\left|t+\sqrt{t^2+1}\right|\right]_{-2}^{x}$

Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function X(x) is defined as the integral from -2 to x of $\frac{1}{\sqrt{t^2+1}}$ dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative X'(x) is simply the integrand evaluated at t = x, so X'(x) = $\frac{1}{\sqrt{x^2+1}}$ . This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 19

If $M(x)=\int_{4}^{x}\left(2^{t}+\sqrt{t}\right)dt$ for $x\ge 0$ , what is $M'(x)$ ?

$2^{x}+\sqrt{x}$ (correct answer)
$\ln(2)\,2^{x}+\dfrac{1}{2\sqrt{x}}$
$\displaystyle \int_{4}^{x}\left(\ln(2)\,2^{t}+\dfrac{1}{2\sqrt{t}}\right)dt$
$\displaystyle \int_{4}^{x}\left(2^{t}+\sqrt{t}\right)dt$
$\displaystyle (2^{t}+\sqrt{t})\big|_{t=4}$

Explanation: This problem demonstrates the Fundamental Theorem of Calculus Part 1, where differentiating an accumulation function yields the integrand evaluated at the variable upper limit. Since $M(x) = \int_{4}^{x}(2^t + \sqrt{t})\,dt$ , we have $M'(x) = 2^x + \sqrt{x}$ by replacing all instances of t with x in the integrand. The solution doesn't require finding antiderivatives or differentiating the integrand terms. Choice B ( $\ln(2) \cdot 2^x + \frac{1}{2\sqrt{x}}$ ) shows the derivatives of $2^x$ and $\sqrt{x}$ , which would be incorrect as FTC Part 1 calls for evaluation, not differentiation. To recognize FTC Part 1 situations, look for derivatives of definite integrals where x appears as a limit of integration.

Question 20

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ ; which statement correctly describes where absolute extrema can occur?

Only at points where $f'(x)=0$ .
Only at endpoints $a$ and $b$ .
Only at points where $f'(x)$ does not exist.
At endpoints or at critical points in $(a,b)$ . (correct answer)
Nowhere, because differentiability prevents extrema.

Explanation: This problem examines where absolute extrema can occur for continuous functions that are differentiable on the interior of closed intervals. For a continuous function $f$ on $[a,b]$ that is differentiable on $(a,b)$ , absolute extrema can occur at two types of locations: the endpoints $a$ and $b$ , or at critical points in the interior $(a,b)$ where $f'(x)=0$ or $f'(x)$ doesn't exist. Since $f$ is differentiable throughout $(a,b)$ , interior critical points must satisfy $f'(x)=0$ . The EVT guarantees absolute extrema exist, and they must occur at these candidate points. Choice A incorrectly restricts to only points where $f'(x)=0$ , ignoring endpoints. Choice B incorrectly restricts to only endpoints. The complete principle: absolute extrema occur at endpoints or interior critical points (where derivative equals zero or doesn't exist).

Question 21

A measured angle satisfies $h(x)=\arccos(x^2)$ . What is $h'(x)$ ?

$\dfrac{2x}{\sqrt{1-x^4}}$
$\dfrac{-2x}{\sqrt{1-x^4}}$ (correct answer)
$\dfrac{-x}{\sqrt{1-x^4}}$
$\dfrac{-2x}{1-x^4}$
$-2x\sin(x^2)$

Explanation: This problem involves differentiating arccos of a power function. The derivative of arccos(u) is -1/√(1-u²) times the derivative of u. With u = x², we multiply by 2x (the derivative of x²). This yields -2x/√(1-x⁴), where (x²)² = x⁴ appears under the radical. A student might write -2x/(1-x⁴) without the square root, mixing up the formulas for arccos and arctan derivatives. Remember: arcsin and arccos derivatives always have square roots in their denominators, distinguishing them from arctan.

Question 22

A curve has slope $y'(x)=25x^4$ . Which is an antiderivative of $y'(x)$ ?

$5x^5+C$ (correct answer)
$25x^5+C$
$5x^5$
$100x^3+C$
$$ \frac{25}{4}x^4+C$

Explanation: Finding the antiderivative involves basic reasoning about reversing the power rule for differentiation. To find the antiderivative of 25x^4, increase the exponent to 5 and divide by 5, yielding 5x^5. There are no other terms, so add the constant C. This reverses the differentiation process directly. A tempting distractor is choice B, which multiplies by 5 again incorrectly. To master antiderivatives, always increase the exponent by one, divide by the new exponent, and include the constant C for indefinite integrals.

Question 23

A reservoir has water depth $h(t)$ ; the outflow rate is proportional to $\sqrt{h(t)}$ . Which differential equation models $h$ decreasing?

$\dfrac{dh}{dt}=-k\sqrt{h}$ (correct answer)
$\dfrac{dh}{dt}=k\sqrt{h}$
$\dfrac{dh}{dt}=-kh$
$\dfrac{dh}{dt}=-\dfrac{k}{\sqrt{h}}$
$\dfrac{dt}{dh}=-k\sqrt{h}$

Explanation: This problem models reservoir drainage where outflow rate follows Torricelli's law, making flow rate proportional to the square root of height. Since "outflow rate is proportional to √h(t)" and the reservoir is decreasing, we have dh/dt = -k√h where k > 0. The negative sign indicates decreasing water depth as water flows out. Choice B has the wrong sign representing filling rather than draining. Choice C would be simple exponential decay. Torricelli's law for fluid flow under gravity creates square root dependencies between height and flow rate in drainage problems.

Question 24

A profit function is $P(x)=x^9-3x^4+2x^2$ . What is $P'(x)$ ?

$9x^8-12x^3+4x$ (correct answer)
$9x^9-12x^4+4x^2$
$9x^8-3x^3+4x$
$9x^8+12x^3+4x$
$8x^8-12x^3+4x$

Explanation: This problem requires applying the power rule to find P'(x) for the profit function P(x) = x^9 - 3x^4 + 2x^2. Using the power rule: x^9 becomes 9x^8, -3x^4 becomes -3·4x^3 = -12x^3, and 2x^2 becomes 2·2x^1 = 4x. Thus, P'(x) = 9x^8 - 12x^3 + 4x. Choice C incorrectly shows -3x^3 instead of -12x^3, forgetting to multiply -3 by 4. When applying the power rule to polynomials, always multiply the coefficient by the exponent before reducing the exponent by 1.

Question 25

Which definite integral matches $\sum_{i=1}^{n}\left(\frac{4i}{n}\right)^3\frac{4}{n}$ ?

$\displaystyle \int_{0}^{4}x^3\,dx$ (correct answer)
$\displaystyle \int_{0}^{1}(4x)^3\,dx$
$\displaystyle \int_{0}^{4}(4x)^3\,dx$
$\displaystyle \int_{0}^{1}x^3\,dx$
$\displaystyle \int_{1}^{4}x^3\,dx$

Explanation: This question tests your ability to translate a Riemann sum into its corresponding definite integral notation. The Riemann sum uses $Δx = 4/n$ , indicating an interval width of 4. The argument $4i/n$ increases from approximately 0 to 4 as i goes from 1 to n, mapping to the integral limits from 0 to 4. The integrand $x^3$ corresponds directly to the cubed expression in the sum. A tempting distractor is choice B, which fails because using $(4x)^3$ over 0 to 1 scales by $4^3 = 64$ but integrates over width 1, resulting in 64 times the integral $\int_0^1 x^3 \, dx$ , which is too large. A transferable translation strategy is to identify $Δx$ as $(b - a)/n$ , determine a and b from the range of the argument as the index varies from 1 to n, and set the integrand to match the function of that argument.

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