If with antiderivative , evaluate .
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AP Calculus BC Quiz
Practice Fundamental Theorem Of Calculus Definite Intervals in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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If s(t)=sec2t with antiderivative S(t)=tant, evaluate ∫0π/4s(t)dt.
This quiz focuses on Fundamental Theorem Of Calculus Definite Intervals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.
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If s(t)=sec2t with antiderivative S(t)=tant, evaluate ∫0π/4s(t)dt.
Explanation: This problem requires using the Fundamental Theorem of Calculus Part 2 with a trigonometric antiderivative. Given S(t) = tan t as an antiderivative of s(t) = sec² t, we compute S(π/4) - S(0). Evaluating: S(π/4) = tan(π/4) = 1 and S(0) = tan(0) = 0, so the integral equals 1 - 0 = 1. The FTC tells us to subtract the antiderivative at the lower bound from the antiderivative at the upper bound. Choice D incorrectly evaluates the original function s(t) = sec² t at the bounds instead of using the antiderivative, which would give sec²(π/4) - sec²(0) = 2 - 1 = 1 (coincidentally the same answer but wrong method). Always apply FTC by evaluating the antiderivative function, not the integrand, at the bounds.
Let p(x)=cosx with antiderivative P(x)=sinx. Find ∫π/3πp(x)dx.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to a trigonometric integral. Since P(x) = sin x is an antiderivative of p(x) = cos x, we evaluate P(π) - P(π/3). Computing: P(π) = sin(π) = 0 and P(π/3) = sin(π/3) = √3/2, so the integral equals 0 - √3/2 = -√3/2. The evaluation follows the pattern of antiderivative at upper bound minus antiderivative at lower bound. Choice D incorrectly evaluates the original function p(x) = cos x at the bounds instead of using the antiderivative, which would give cos(π) - cos(π/3) = -1 - 1/2 = -3/2. Always use the antiderivative function when applying FTC Part 2 to evaluate definite integrals.
A cost rate is c(x)=6x−5. Using C(x)=3x2−5x, evaluate ∫27c(x)dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2 to find total cost over an interval. Given C(x) = 3x² - 5x as an antiderivative of c(x) = 6x - 5, we evaluate C(7) - C(2). Computing: C(7) = 3(49) - 5(7) = 147 - 35 = 112 and C(2) = 3(4) - 5(2) = 12 - 10 = 2, so the integral equals 112 - 2 = 110. The FTC requires subtracting the antiderivative at the lower bound from the antiderivative at the upper bound. Choice D incorrectly uses the original function c(x) instead of its antiderivative, evaluating c(7) - c(2), which demonstrates confusion between a function and its antiderivative. Remember: definite integrals are evaluated using antiderivatives, not the original function.
An antiderivative of m(x)=x3 is M(x)=6x. Evaluate ∫49m(x)dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2 with a radical function. Given that M(x) = 6√x is an antiderivative of m(x) = 3/√x, we evaluate ∫[4 to 9] m(x)dx = M(9) - M(4). Computing the values: M(9) = 6√9 = 6(3) = 18, and M(4) = 6√4 = 6(2) = 12. Therefore, the integral equals M(9) - M(4) = 18 - 12 = 6. The distractor M(4) - M(9) reverses the subtraction, giving 12 - 18 = -6, which would be incorrect since m(x) is positive on [4,9]. Remember that FTC Part 2 always evaluates as F(upper) - F(lower) to maintain proper orientation.
A particle’s velocity is v(t)=sint. If A′(t)=v(t), find ∫0πsintdt in terms of A.
Explanation: This problem draws on the Fundamental Theorem of Calculus Part 2 to express definite integrals in terms of antiderivatives. Since A′(t)=sint, A is an antiderivative of the velocity function v(t). The FTC requires evaluating A at π and subtracting A at 0. Therefore, the integral from 0 to π equals A(π)−A(0). A tempting distractor is choice A, A(0)−A(π), which reverses the order and yields a negative value, perhaps from misapplying the net change interpretation. A transferable evaluation strategy is to use F(b)−F(a) for any integral from a to b, ensuring accurate accumulation of the function's rate.
Given T′(x)=sec2x on [0,π/4], evaluate ∫0π/4sec2xdx using T.
Explanation: This problem relies on the Fundamental Theorem of Calculus Part 2 to evaluate definite integrals using antiderivatives. Given T'(x) = sec²x on [0, π/4], T is an antiderivative of the integrand. The FTC requires T at π/4 minus T at 0. Thus, the integral from 0 to π/4 equals T(π/4) - T(0). A tempting distractor is choice A, T(0) - T(π/4), which inverts the evaluation and produces the negative, possibly from confusing the theorem's formula. A transferable evaluation strategy is to always use the antiderivative's value at the upper limit minus the lower limit for precise integral calculation.
Let S′(x)=x1 for x>0. What is the value of ∫19x1dx?
Explanation: This problem invokes the Fundamental Theorem of Calculus Part 2 for expressing integrals in terms of antiderivatives. S'(x) = 1/√x for x > 0, making S an antiderivative over the positive domain. Evaluate S at 9 and subtract S at 1 per the FTC. Therefore, the integral from 1 to 9 equals S(9) - S(1). A tempting distractor is choice A, S(1) - S(9), which swaps the terms and yields a negative result, perhaps from misremembering the subtraction order. A transferable evaluation strategy is to systematically compute the difference F(b) - F(a) for integrals from a to b, ensuring domain conditions are met.
If R′(x)=cos(5x), what is ∫−π/5π/5cos(5x)dx in terms of R?
Explanation: This problem uses the Fundamental Theorem of Calculus Part 2 to find definite integrals through antiderivative evaluation. With R'(x) = cos(5x), R is an antiderivative of the integrand. Apply the FTC by subtracting R at -π/5 from R at π/5. Thus, the integral from -π/5 to π/5 equals R(π/5) - R(-π/5). A tempting distractor is choice B, R(-π/5) - R(π/5), which reverses the order and gives the negative, often due to symmetric limit confusion. A transferable evaluation strategy is to adhere to F(upper) - F(lower) when using antiderivatives, even for symmetric intervals.
An antiderivative of f(x)=2xcos(x2) is U(x)=sin(x2). Find ∫0π2xcos(x2)dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2, using antiderivatives to compute definite integrals. U(x) = sin(x²) is an antiderivative of 2x cos(x²). Evaluate U at √π and subtract U at 0 according to the FTC. Therefore, the integral from 0 to √π equals U(√π) - U(0). A tempting distractor is choice A, U(0) - U(√π), which reverses the order and gives a negative value, likely from mishandling the non-standard upper limit. A transferable evaluation strategy is to compute F(b) - F(a) for any definite integral, verifying the antiderivative matches the integrand via differentiation.
If P(x)=e2x is an antiderivative of 2e2x, evaluate ∫252e2xdx.
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2, which states that the definite integral of a function over an interval equals the antiderivative evaluated at the upper limit minus the antiderivative evaluated at the lower limit. Here, P(x) = e^{2x} is given as an antiderivative of 2e^{2x}, so the integral from 2 to 5 is P(5) - P(2). Evaluating P(5) gives e^{10}, and P(2) = e^4, so e^{10} - e^4. This directly applies the theorem for exponential functions. A tempting distractor like P(2) - P(5) fails because it inverts the order, resulting in the negative integral value. Always remember to subtract the antiderivative at the lower limit from its value at the upper limit for any definite integral evaluation.
Given R′(x)=e2x and R(0)=1, find ∫−11e2xdx.
Explanation: This problem demonstrates using the Fundamental Theorem of Calculus Part 2 when given R'(x) = e^(2x) and an initial condition. Since R'(x) = e^(2x), R(x) is an antiderivative of e^(2x). By FTC Part 2, ∫[-1 to 1] e^(2x)dx = R(1) - R(-1). We can find R(x) = (1/2)e^(2x) + C, and using R(0) = 1 gives (1/2)e^0 + C = 1, so C = 1/2. Thus R(1) = (1/2)e^2 + 1/2 and R(-1) = (1/2)e^(-2) + 1/2, making the integral equal to (1/2)(e^2 - e^(-2)). The incorrect choice R(-1) - R(1) would give the negative of this value. Remember: FTC Part 2 always uses the pattern upper bound evaluation minus lower bound evaluation.
An antiderivative of p(x) is P(x)=ln(x)+x1. Evaluate ∫1ep(x)dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2 using a given antiderivative function. Since P(x) = ln(x) + 1/x is an antiderivative of p(x), we have P'(x) = p(x). By FTC Part 2, ∫[1 to e] p(x)dx = P(e) - P(1). Evaluating at the bounds: P(e) = ln(e) + 1/e = 1 + 1/e and P(1) = ln(1) + 1/1 = 0 + 1 = 1. Therefore, the integral equals (1 + 1/e) - 1 = 1/e. The incorrect choice P(1) - P(e) would give -1/e, which has the wrong sign due to reversed bounds. When using FTC Part 2, maintain the correct order: antiderivative at upper bound minus antiderivative at lower bound.
Given T′(x)=x3 for x>0 and T(1)=0, evaluate ∫14x3dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2 to a logarithmic derivative. Given T'(x) = 3/x and T(1) = 0, we know T(x) is an antiderivative of 3/x, specifically T(x) = 3ln(x) + C. Using T(1) = 0 gives 3ln(1) + C = 0, so C = 0 and T(x) = 3ln(x). By FTC Part 2, ∫[1 to 4] (3/x)dx = T(4) - T(1) = 3ln(4) - 3ln(1) = 3ln(4) - 0 = 3ln(4). The incorrect choice T(1) - T(4) would give -3ln(4), reversing the sign. For any definite integral, maintain the correct evaluation order: antiderivative at upper limit minus antiderivative at lower limit.
An antiderivative of m(x) is M(x)=5x5−2x2. Compute ∫20m(x)dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2 to an integral with reversed bounds. Given that M(x) = x^5/5 - 2x² is an antiderivative of m(x), we evaluate ∫[2 to 0] m(x)dx = M(0) - M(2). Note the integral goes from 2 to 0, so the upper bound is 0 and the lower bound is 2. Evaluating: M(0) = 0 and M(2) = 32/5 - 8 = -8/5, so the integral equals 0 - (-8/5) = 8/5. The distractor M(2) - M(0) would give -8/5, which incorrectly treats 2 as the upper bound. When bounds are reversed, carefully identify which is truly the upper bound in the direction of integration.
Given F′(x)=f(x) and F(x)=x3−4x, what is ∫−12f(x)dx?
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2, which states that if F'(x) = f(x), then ∫[a to b] f(x)dx = F(b) - F(a). Since we're given that F'(x) = f(x) and F(x) = x³ - 4x, we can directly evaluate the definite integral from -1 to 2. We calculate F(2) = 2³ - 4(2) = 8 - 8 = 0 and F(-1) = (-1)³ - 4(-1) = -1 + 4 = 3. Therefore, ∫[-1 to 2] f(x)dx = F(2) - F(-1) = 0 - 3 = -3. A common error would be reversing the order to get F(-1) - F(2) = 3, which gives the wrong sign. Remember: for definite integrals, always evaluate the antiderivative at the upper limit minus the antiderivative at the lower limit.
If P′(x)=p(x) and P(x)=x1+x, evaluate ∫25p(x)dx.
Explanation: This problem demonstrates the Fundamental Theorem of Calculus Part 2, using the relationship between P'(x) = p(x) and the given antiderivative P(x) = 1/x + x. To find ∫[2 to 5] p(x)dx, we evaluate P(5) - P(2). We calculate P(5) = 1/5 + 5 = 0.2 + 5 = 5.2 and P(2) = 1/2 + 2 = 0.5 + 2 = 2.5. Therefore, ∫[2 to 5] p(x)dx = P(5) - P(2) = 5.2 - 2.5 = 2.7. Students often mistakenly compute P(2) - P(5) = -2.7, which gives the negative of the correct answer by reversing the subtraction. The consistent pattern for evaluating definite integrals is always antiderivative at upper bound minus antiderivative at lower bound.
A particle’s velocity is v(t)=3t2−4t+1. What is ∫14v(t)dt using an antiderivative V(t)=t3−2t2+t?
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral. Since V(t) is given as an antiderivative of v(t), we can evaluate the integral as V(4) - V(1). Substituting into V(t) = t³ - 2t² + t, we get V(4) = 64 - 32 + 4 = 36 and V(1) = 1 - 2 + 1 = 0, so the integral equals 36 - 0 = 36. The order matters: we always subtract the antiderivative evaluated at the lower bound from the antiderivative evaluated at the upper bound. Choice A incorrectly adds the values instead of subtracting, which is a common error when first learning FTC. Remember: for any definite integral from a to b, evaluate the antiderivative at b minus the antiderivative at a.
Given an antiderivative G(x)=ln(x)+x1 for g(x)=x1−x21, find ∫1eg(x)dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral using a given antiderivative. Since G(x)=ln(x)+x1 is an antiderivative of g(x)=x1−x21, we know that ∫1eg(x)dx=G(e)−G(1). Evaluating at the bounds: G(e)=ln(e)+e1=1+e1, and G(1)=ln(1)+11=0+1=1. Therefore, the integral equals G(e)−G(1)=(1+e1)−1=e1. The distractor G(1)−G(e) reverses the subtraction order, yielding −e1 instead of e1. Remember that FTC Part 2 always uses the pattern: antiderivative at upper limit minus antiderivative at lower limit.
If F'(x)=(3x^2-4x+1) and F(x)=x3−2x2+x, what is ∫−12(3x2−4x+1)dx?
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 2, which states that if F'(x) = f(x), then ∫[a to b] f(x)dx = F(b) - F(a). Since we're given that F'(x) = 3x² - 4x + 1 and F(x) = x³ - 2x² + x, we can directly apply the theorem. To evaluate ∫[-1 to 2] (3x² - 4x + 1)dx, we compute F(2) - F(-1). First, F(2) = 2³ - 2(2²) + 2 = 8 - 8 + 2 = 2, and F(-1) = (-1)³ - 2(-1)² + (-1) = -1 - 2 - 1 = -4. Therefore, the integral equals F(2) - F(-1) = 2 - (-4) = 6. The distractor F(-1) - F(2) reverses the order, giving -6 instead of 6. When applying FTC Part 2, always evaluate the antiderivative at the upper bound minus the lower bound: F(upper) - F(lower).
Given F′(x)=1+x21 and F(0)=0, what is the value of ∫021+x21dx?
Explanation: This problem utilizes the Fundamental Theorem of Calculus Part 2, which states that if F is an antiderivative of f, then the definite integral ∫abf(x)dx equals F(b)−F(a). Here, F′(x)=1+x21, so F is an antiderivative of the integrand. To evaluate the integral ∫021+x21dx, apply the FTC by computing F at the upper limit and subtracting F at the lower limit. Thus, the integral equals F(2)−F(0). A tempting distractor is choice B, F(0)−F(2), which reverses the order and would yield the negative of the correct value, potentially confusing the signs of the limits. A transferable evaluation strategy is to always compute the antiderivative at the upper limit minus its value at the lower limit for any definite integral.