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.

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.

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.

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.

This problem draws on the Fundamental Theorem of Calculus Part 2 to express definite integrals in terms of antiderivatives. Since $A'(t) = \sin t$, $A$ is an antiderivative of the velocity function $v(t)$. The FTC requires evaluating $A$ at $\pi$ and subtracting $A$ at 0. Therefore, the integral from 0 to $\pi$ equals $A(\pi) - A(0)$. A tempting distractor is choice A, $A(0) - A(\pi)$, 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.

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.

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.

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.

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.

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.