Drill 1 · Math · Fundamental Theorem of Calculus
AP Calculus AB: Fundamental Theorem of Calculus Part 1 (Drill 1) is a Math practice drill covering Fundamental Theorem of Calculus. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice applying the Fundamental Theorem of Calculus Part 1 to differentiate accumulation functions, including cases where the chain rule is required. These AP Calculus AB skills are central to both the multiple-choice and free-response sections.
Question 1. If \( F(x) = \displaystyle\int_3^x (t^2+1)\,dt \), then \( F'(x) = \)
Explanation: Choice B is correct. By FTC Part 1, \( \frac{d}{dx}\int_3^x f(t)\,dt = f(x) \), so \( F'(x) = x^2+1 \). The lower bound of 3 is a constant and does not affect the derivative. Choice A is incorrect because \( \frac{x^3}{3}+x-12 \) is \( F(x) \) itself, the antiderivative evaluated from 3 to \( x \), not its derivative. The student computed \( F(x) \) using FTC Part 2 instead of differentiating. Choice C is incorrect because \( 2x \) is the derivative of \( x^2+1 \), the student differentiated the integrand rather than evaluating it at the upper limit; FTC Part 1 does not call for an additional differentiation step. Choice D is incorrect because \( x^2-9 \) results from subtracting \( f(3) \) from \( f(x) \): \( (x^2+1)-(3^2+1) = x^2-9 \). The student mistakenly believed the lower bound must be substituted and subtracted, confusing FTC Part 1 with FTC Part 2.
Question 2. If \( G(x) = \displaystyle\int_0^{x^3} \sin(t)\,dt \), then \( G'(x) = \)
Explanation: Choice B is correct. When the upper limit is a function of \( x \), FTC Part 1 requires the chain rule. Let \( u = x^3 \); then \( G'(x) = \sin(u) \cdot \frac{du}{dx} = \sin(x^3) \cdot 3x^2 \). Choice A is incorrect because it applies FTC Part 1 without the chain rule, treating the upper limit as simply \( x \) rather than \( x^3 \), the most common error on this question type. Choice C is incorrect because the student differentiated the integrand \( \sin(t) \) to obtain \( \cos(t) \), which is not part of FTC Part 1; the integrand is evaluated at the upper limit, not differentiated. Choice D is incorrect because the student both differentiated the integrand (\( \sin \to \cos \)) and applied the chain rule factor \( 3x^2 \), committing two errors simultaneously.
Question 3. If \( H(x) = \displaystyle\int_x^{5} e^{t^2}\,dt \), then \( H'(x) = \)
Explanation: Choice B is correct. Since \( x \) appears in the lower limit, rewrite by reversing the limits: \( H(x) = -\int_5^x e^{t^2}\,dt \). FTC Part 1 then gives \( H'(x) = -e^{x^2} \). The upper limit is the constant 5, so no chain rule is needed beyond the negative sign. Choice A is incorrect because it applies FTC Part 1 without reversing the limits, omitting the negative sign, the most common error when the variable is in the lower bound. Choice C is incorrect because \( 2x \) would be the chain rule factor if the lower limit were \( x^2 \) rather than \( x \); since the lower limit here is simply \( x \), no additional chain rule multiplication is needed. Choice D is incorrect because the student evaluates the integral at both bounds as if using FTC Part 2, writing \( e^{25}-e^{x^2} \) as a fixed value rather than differentiating the accumulation function, confusing FTC Part 1 (differentiation) with FTC Part 2 (evaluation).
Question 4. Let \( A(x) = \displaystyle\int_0^x f(t)\,dt \), where \( f \) is continuous, \( f(t) > 0 \) on \( (0,2) \), \( f(2) = 0 \), and \( f(t) < 0 \) on \( (2,5) \). Which of the following must be true?
Explanation: Choice C is correct. By FTC Part 1, \( A'(x) = f(x) \). Since \( f(x) > 0 \) on \( (0,2) \), \( A \) is increasing there; since \( f(x) < 0 \) on \( (2,5) \), \( A \) is decreasing there. Because \( A' \) changes from positive to negative at \( x=2 \), the First Derivative Test guarantees a local maximum at \( x=2 \). Choice A is incorrect because \( A'(x) = f(x) < 0 \) on \( (2,5) \), so \( A \) is decreasing, not increasing, a student who confuses the sign of \( f \) with the behavior of \( A \) will choose this. Choice B is incorrect because a local minimum requires \( A' \) to change from negative to positive; here the sign change is from positive to negative, which produces a local maximum. Confusing the direction of the sign change is a classic First Derivative Test error. Choice D is incorrect because \( A(x) \) begins the interval \( (2,5) \) at a positive value (since \( A \) was accumulating positive area on \( (0,2) \)) and then decreases, whether \( A \) ever becomes negative depends on the relative areas above and below the \( x \)-axis, which cannot be determined from the given information alone.
Question 5. Let \( P(x) = \displaystyle\int_{x^2}^{7} \sqrt{1+t^3}\,dt \). Which of the following equals \( P'(x) \)?
Explanation: Choice C is correct. This problem requires three steps in order. First, the variable is in the lower limit, so reverse the limits: \( P(x) = -\int_7^{x^2}\sqrt{1+t^3}\,dt \). Second, apply FTC Part 1 with the chain rule: let \( u = x^2 \), so \( \frac{d}{dx}\int_7^u\sqrt{1+t^3}\,dt = \sqrt{1+u^3} \cdot \frac{du}{dx} \). Third, substitute \( u = x^2 \) and \( \frac{du}{dx} = 2x \): \( \sqrt{1+(x^2)^3} \cdot 2x = 2x\sqrt{1+x^6} \). Including the negative from the limit reversal: \( P'(x) = -2x\sqrt{1+x^6} \). Choice A is incorrect because it omits both the limit reversal (missing the negative) and the chain rule factor \( 2x \), the student applies FTC Part 1 as if the upper limit were simply \( x \). Choice B is incorrect because the student correctly reverses the limits and applies the chain rule factor \( 2x \), but substitutes \( x \) into the integrand instead of \( x^2 \), writing \( \sqrt{1+x^3} \) instead of \( \sqrt{1+(x^2)^3} = \sqrt{1+x^6} \). Choice D is incorrect because the student correctly substitutes \( x^2 \) into the integrand and applies the chain rule, arriving at the correct magnitude, but forgets to reverse the limits, omitting the negative sign.