Questions: Fundamental Theorem of Calculus Part 2

Question 1

Evaluate ∫₁⁴ 3x² dx using FTC Part 2. What is the correct answer?

• A. 192 — computed as F(1) - F(4) where F(x) = x³
• B. 63 — computed as F(4) - F(1) where F(x) = x³
• C. 63 + C — the constant of integration must be retained
• D. −63 — computed as F(1) - F(4)

Answer: (B)

The antiderivative of 3x² is x³. FTC Part 2 gives [x³]₁⁴ = 4³ - 1³ = 64 - 1 = 63. Option C incorrectly includes +C — in definite integrals, the constant cancels: (F(b) + C) - (F(a) + C) = F(b) - F(a). Options A and D reverse the order to F(a) - F(b), giving a negative result. Order matters: it is always F(b) - F(a), upper limit minus lower limit.

Question 2

A student evaluates ∫₀² x³ dx and gets −4, reasoning that F(0) - F(2) = 0 - 4 = -4 where F(x) = x⁴/4. What went wrong?

• A. The antiderivative is wrong — the correct antiderivative of x³ is 3x²
• B. The order of subtraction is reversed — FTC Part 2 requires F(b) - F(a), not F(a) - F(b)
• C. The student should have used a Riemann sum to verify the result
• D. The student forgot to include +C before evaluating at the endpoints

Answer: (B)

The antiderivative F(x) = x⁴/4 is correct, but FTC Part 2 requires F(b) - F(a): F(2) - F(0) = 16/4 - 0 = 4. The student computed F(a) - F(b) = F(0) - F(2) = -4. Reversing the limits changes the sign of the definite integral. This is one of the most common computational errors in applying FTC Part 2 — the direction of integration (lower limit to upper limit) is encoded in the subtraction order.

Question 3

When applying FTC Part 2, you is expected to use the specific antiderivative that satisfies F(a) = 0; otherwise the formula gives an incorrect answer.

• T. True
• F. False

Answer: False

Any antiderivative works. If F and G are both antiderivatives of f, they differ by a constant: G(x) = F(x) + C. Then G(b) - G(a) = [F(b) + C] - [F(a) + C] = F(b) - F(a). The constant C cancels regardless of its value. This is why you don't write '+C' in definite integral evaluations and why choosing the simplest antiderivative (C = 0) is always valid.

Question 4

FTC Part 2 is a computational shortcut that approximates the definite integral; using Riemann sums would give a more exact result.

• T. True
• F. False

Answer: False

FTC Part 2 gives the exact value of the definite integral — not an approximation. Riemann sums are approximations that approach the exact integral only in the limit as the number of rectangles approaches infinity. FTC Part 2 is the theorem that establishes the exact equality between the definite integral and the antiderivative evaluation — it doesn't shortcut around precision, it provides it.

Question 5

Why does the constant of integration (+C) disappear when applying FTC Part 2 to evaluate a definite integral?

Model answer: Because it cancels algebraically. Any antiderivative F of f differs from any other by a constant C. When you compute F(b) - F(a), you get [G(b) + C] - [G(a) + C] = G(b) - G(a), and C vanishes. Every valid antiderivative therefore produces the same value for the definite integral, which is why you can freely choose the most convenient one and omit +C from definite integral calculations.

This cancellation reflects a deep property of the definite integral: it measures a net quantity (area, displacement, accumulated change) that is independent of where you set the 'zero point' of the antiderivative. Shifting the antiderivative up or down by C shifts both F(b) and F(a) by the same amount, so the difference is unchanged. Recognizing this makes the formula intuitive — the constant represents where you started counting, and starting from a different point doesn't change how much accumulates between a and b.