Fundamental Theorem of Calculus: Definite Intervals - AP Calculus BC

If $Q'(x)=q(x)$ and $Q(x)=\sqrt{x}+\cos x$ , what is $\int_{1}^{4} q(x)\,dx$ ?
Answer: $Q(4)-Q(1)$ This problem uses the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral using the given antiderivative relationship. Since Q'(x) = q(x) and Q(x) = √x + cos(x), we find ∫[1 to 4] q(x)dx = Q(4) - Q(1). We calculate Q(4) = √4 + cos(4) = 2 + cos(4) and Q(1) = √1 + cos(1) = 1 + cos(1). Therefore, ∫[1 to 4] q(x)dx = Q(4) - Q(1) = (2 + cos(4)) - (1 + cos(1)) = 1 + cos(4) - cos(1). The incorrect choice Q(1) - Q(4) would reverse the sign of the answer, a common error when students forget the proper order. Always apply the pattern: integral from a to b equals F(b) - F(a).
If $H'(x)=6x-5$ and $H(1)=2$ , what is $\int_{1}^{4}(6x-5)\,dx$ ?
Answer: $H(4)-H(1)$ This problem demonstrates using the Fundamental Theorem of Calculus Part 2 when given a derivative and initial condition. Since H'(x) = 6x - 5, we know H(x) is an antiderivative of 6x - 5. By FTC Part 2, ∫[1 to 4] (6x - 5)dx = H(4) - H(1). While we don't need the explicit form of H(x) to apply the theorem, we can verify: H(x) = 3x² - 5x + C, and using H(1) = 2 gives C = 4. Thus H(4) = 3(16) - 5(4) + 4 = 32 and H(1) = 2, so the integral equals 30. The distractor H(1) - H(4) would give -30, incorrectly reversing the evaluation order. For definite integrals, always evaluate at the upper limit first, then subtract the lower limit value.
Given $Q(x)=\frac{x^5}{5}$ is an antiderivative of $x^4$ , what is $\int_{-2}^{1} x^4\,dx$ ?
Answer: $Q(1)-Q(-2)$ 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, Q(x) = $x^5$ / 5 is given as an antiderivative of $x^4$ , so the integral from -2 to 1 is Q(1) - Q(-2). Evaluating Q(1) gives 1/5, and Q(-2) = (-32)/5, so 1/5 - (-32/5) = 33/5. This demonstrates correct handling of even powers over negative intervals. A tempting distractor like Q(-2) - Q(1) fails because it swaps the order, yielding -33/5, which is incorrect for the given limits. Always remember to subtract the antiderivative at the lower limit from its value at the upper limit for any definite integral evaluation.
Let $H'(x)=h(x)$ and $H(x)=\sin x+3x$ . What is $\int_{0}^{\pi} h(x)\,dx$ ?
Answer: $H(\pi)-H(0)$ This problem uses the Fundamental Theorem of Calculus Part 2, which connects derivatives and definite integrals through antiderivatives. Given that H'(x) = h(x) and H(x) = sin(x) + 3x, we can evaluate ∫[0 to π] h(x)dx = H(π) - H(0). We calculate H(π) = sin(π) + 3π = 0 + 3π = 3π and H(0) = sin(0) + 3(0) = 0 + 0 = 0. Therefore, ∫[0 to π] h(x)dx = H(π) - H(0) = 3π - 0 = 3π. A common mistake is computing H(0) - H(π) = -3π, which reverses the correct order and produces a negative result when the answer should be positive. Always remember: definite integral equals F(upper) - F(lower), maintaining this order consistently.
If $P'(x)=p(x)$ and $P(x)=\frac{1}{x}+x$ , evaluate $\int_{2}^{5} p(x)\,dx$ .
Answer: $P(5)-P(2)$ 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.
Let $R(x)=\arctan x$ be an antiderivative of $\frac{1}{1+x^2}$ . Evaluate $\int_{1}^{4} \frac{1}{1+x^2}\,dx$ .
Answer: $R(4)-R(1)$ 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, R(x) = arctan x is given as an antiderivative of 1/(1 + $x^2$ ), so the integral from 1 to 4 is R(4) - R(1). Evaluating R(4) gives arctan(4), and R(1) = arctan(1) = π/4, so arctan(4) - π/4. This applies the theorem to inverse trigonometric functions. A tempting distractor like R(1) - R(4) fails because it reverses the subtraction, giving a negative value instead of the positive integral. Always remember to subtract the antiderivative at the lower limit from its value at the upper limit for any definite integral evaluation.
Given $F'(x)=\frac{1}{1+x^2}$ and $F(0)=0$ , what is the value of $\int_{0}^{2}\frac{1}{1+x^2}\,dx$ ?
Answer: $F(2)-F(0)$ This problem utilizes the Fundamental Theorem of Calculus Part 2, which states that if F is an antiderivative of f, then the definite integral $\int_a^b f(x) \, dx$ equals $F(b) - F(a)$ . Here, $F'(x) = \frac{1}{1 + x^2}$ , so F is an antiderivative of the integrand. To evaluate the integral $\int_0^2 \frac{1}{1 + x^2} \, dx$ , 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.
A position function satisfies $s'(t)=6t-1$ and $s(t)=3t^2-t$ . Find $\int_{-1}^{2} (6t-1)\,dt$ .
Answer: $s(2)-s(-1)$ 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, $s(t) = 3t^2 - t$ is given as an antiderivative of 6t - 1, so the integral from -1 to 2 is $s(2) - s(-1)$ . Evaluating $s(2)$ gives $3*(4) - 2 = 12 - 2 = 10$ , and $s(-1) = 3*(1) - (-1) = 3 + 1 = 4$ , so $10 - 4 = 6$ . This shows the theorem works across negative and positive bounds. A tempting distractor like $s(-1) - s(2)$ fails because it reverses the limits, giving -6 instead of the correct positive area. Always remember to subtract the antiderivative at the lower limit from its value at the upper limit for any definite integral evaluation.
If an antiderivative of $g$ is $G(x)=\ln(x)+x^2$ , find $\int_{1}^{e} g(x)\,dx$ .
Answer: $G(e)-G(1)$ This problem applies the Fundamental Theorem of Calculus Part 2, where we use the given antiderivative G(x) = ln(x) + x² to evaluate the definite integral of g(x). Since G is an antiderivative of g, we know G'(x) = g(x), so ∫[1 to e] g(x)dx = G(e) - G(1). We calculate G(e) = ln(e) + e² = 1 + e² and G(1) = ln(1) + 1² = 0 + 1 = 1. Therefore, ∫[1 to e] g(x)dx = G(e) - G(1) = (1 + e²) - 1 = e². Students might incorrectly choose G(1) - G(e), reversing the subtraction order and getting a negative result. The key strategy is to always subtract the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit.
Given $N(x)=\sqrt{x}$ is an antiderivative of $\frac{1}{2\sqrt{x}}$ , evaluate $\int_{1}^{9} \frac{1}{2\sqrt{x}}\,dx$ .
Answer: $N(9)-N(1)$ 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, N(x) = √x is given as an antiderivative of 1/(2√x), so the integral from 1 to 9 is N(9) - N(1). Evaluating N(9) gives √9 = 3, and N(1) = √1 = 1, so 3 - 1 = 2. This confirms the theorem for root functions. A tempting distractor like N(1) - N(9) fails because it reverses the limits, giving -2 instead of the correct positive value. Always remember to subtract the antiderivative at the lower limit from its value at the upper limit for any definite integral evaluation.
A cost rate is $c(x)=6x-5$ . Using $C(x)=3x^2-5x$ , evaluate $\int_{2}^{7} c(x)\,dx$ .
Answer: $C(7)-C(2)$ 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.
If $R'(x)=\frac{1}{1-x}$ and $R(x)=-\ln(1-x)$ , find $\int_{0}^{1/2} \frac{1}{1-x}\,dx$ .
Answer: $R(1/2)-R(0)$ This problem demonstrates the Fundamental Theorem of Calculus Part 2 with a logarithmic antiderivative. Given R'(x) = 1/(1-x) and R(x) = -ln(1-x), we evaluate ∫[0 to 1/2] 1/(1-x)dx = R(1/2) - R(0). Computing: R(1/2) = -ln(1-1/2) = -ln(1/2) = -ln(1/2) = ln(2), and R(0) = -ln(1-0) = -ln(1) = 0. Therefore, the integral equals R(1/2) - R(0) = ln(2) - 0 = ln(2). The distractor R(0) - R(1/2) would give 0 - ln(2) = -ln(2), incorrectly suggesting a negative area. When applying FTC Part 2, always use F(upper) - F(lower) to ensure the correct signed result.
If $H'(x)=\cos x$ and $H(x)=\sin x$ , what is the value of $\int_{\pi/6}^{\pi/2} \cos x\,dx$ ?
Answer: $H(\pi/2)-H(\pi/6)$ This problem demonstrates the Fundamental Theorem of Calculus Part 2 with trigonometric functions. Given that H'(x) = cos x and H(x) = sin x, we can evaluate ∫[π/6 to π/2] cos x dx = H(π/2) - H(π/6). Computing the values: H(π/2) = sin(π/2) = 1, and H(π/6) = sin(π/6) = 1/2. Therefore, the integral equals H(π/2) - H(π/6) = 1 - 1/2 = 1/2. The distractor H(π/6) - H(π/2) incorrectly reverses the order, giving 1/2 - 1 = -1/2. When using FTC Part 2, maintain the correct order: evaluate the antiderivative at the upper bound first, then subtract the value at the lower bound.
For $q(x)=\frac{1}{\sqrt{x}}$ with antiderivative $Q(x)=2\sqrt{x}$ , compute $\int_{4}^{9} q(x)\,dx$ .
Answer: $Q(9)-Q(4)$ This problem tests the Fundamental Theorem of Calculus Part 2 with a radical function. Given Q(x) = 2√x as an antiderivative of q(x) = 1/√x, we compute Q(9) - Q(4). Evaluating: Q(9) = 2√9 = 2(3) = 6 and Q(4) = 2√4 = 2(2) = 4, so the integral equals 6 - 4 = 2. The FTC pattern requires subtracting the antiderivative at the lower bound from the antiderivative at the upper bound. Choice D incorrectly evaluates the original function q(x) at the bounds, giving 1/√9 - 1/√4 = 1/3 - 1/2 = -1/6, which confuses the integrand with its antiderivative. To evaluate a definite integral, always use the antiderivative function at both bounds, then subtract.
A tank fills at rate $r(t)=\frac{1}{1+t^2}$ . Using $R(t)=\arctan t$ , evaluate $\int_{0}^{3} r(t)\,dt$ .
Answer: $\arctan(3)-\arctan(0)$ This problem applies the Fundamental Theorem of Calculus Part 2 to find the total amount of water added to a tank. Since R(t) = arctan t is an antiderivative of r(t) = 1/(1+t²), we evaluate the integral as R(3) - R(0). We have R(3) = arctan(3) and R(0) = arctan(0) = 0, giving us arctan(3) - 0 = arctan(3). The subtraction follows the standard FTC pattern: antiderivative at upper bound minus antiderivative at lower bound. Choice E, arctan(3-0), incorrectly attempts to evaluate the arctangent of the difference of bounds rather than the difference of arctangent values. Remember to evaluate the antiderivative function at each bound separately, then subtract.
If an antiderivative of $f(x)=4x^3-9$ is $A(x)=x^4-9x$ , what is $\int_{-2}^{1} f(x)\,dx$ ?
Answer: $A(1)-A(-2)$ This problem applies the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral over a negative-to-positive interval. Since A(x) = x⁴ - 9x is an antiderivative of f(x) = 4x³ - 9, we compute ∫[-2 to 1] f(x)dx = A(1) - A(-2). Evaluating: A(1) = 1⁴ - 9(1) = 1 - 9 = -8, and A(-2) = (-2)⁴ - 9(-2) = 16 + 18 = 34. Therefore, the integral equals A(1) - A(-2) = -8 - 34 = -42. The distractor A(-2) - A(1) would give 34 - (-8) = 42, missing the negative sign. When applying FTC Part 2, always maintain the order F(upper) - F(lower), regardless of whether bounds are positive or negative.
If $H'(x)=e^{2x}$ , what is the value of $\int_{1}^{3} e^{2x}\,dx$ in terms of $H$ ?
Answer: $H(3)-H(1)$ This problem employs the Fundamental Theorem of Calculus Part 2, which evaluates definite integrals using an antiderivative's values at the bounds. With H'(x) = $e^{2x}$ , H is an antiderivative of the integrand. Apply the FTC by evaluating H at 3 and subtracting H at 1. Thus, the integral from 1 to 3 equals H(3) - H(1). A tempting distractor is choice A, H(1) - H(3), which inverts the subtraction and gives the negative result, possibly from confusing upper and lower limits. A transferable evaluation strategy is to identify an antiderivative and subtract its lower limit value from the upper limit value for reliable integral computation.
An antiderivative of $p(x)$ is $P(x)=\ln(x)+\dfrac{1}{x}$ . Evaluate $\int_{1}^{e} p(x)\,dx$ .
Answer: $P(e)-P(1)$ 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.
If $M(x)=\frac{1}{3}(x+2)^3$ is an antiderivative of $(x+2)^2$ , find $\int_{-1}^{2} (x+2)^2\,dx$ .
Answer: $M(2)-M(-1)$ 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, M(x) = (1/3) $(x + 2)^3$ is given as an antiderivative of $(x + 2)^2$ , so the integral from -1 to 2 is M(2) - M(-1). Evaluating M(2) gives (1/3) $(4)^3$ = (1/3)(64) = 64/3, and M(-1) = (1/3) $(1)^3$ = 1/3, so 64/3 - 1/3 = 63/3 = 21. This shows the theorem for polynomial expansions. A tempting distractor like M(-1) - M(2) fails because it inverts the order, resulting in -21. Always remember to subtract the antiderivative at the lower limit from its value at the upper limit for any definite integral evaluation.
Given $T'(x)=\sec^2 x$ on $[0,\pi/4]$ , evaluate $\int_{0}^{\pi/4}\sec^2 x\,dx$ using $T$ .
Answer: $T(\pi/4)-T(0)$ 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.

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