The Disappearing Constant in Definite Integral Antiderivatives
Including the '+C' constant in both the upper and lower substitutions when evaluating a definite integral means it cancels out and never appears in your final answer. If you write ∫₁² (3x² + 2) dx = [x³ + 2x + C]₂₁ and keep the '+C', you still get the right number because the constants subtract away.
Suppose you hit this late at night: your homework asks you to compute ∫₁² (3x² + 2) dx. You remember to add '+C' when finding the antiderivative, but now you see examples omitting it for definite integrals. Do you keep it, or drop it?
The x³ + 2x + C That Seems Unsettled
Integral calculus teaches that every indefinite integral needs '+C', the constant of integration, since any function differing by a constant has the same derivative. But for definite integrals, you usually see the answer as just a number, no '+C'. Here is where students get tangled: the method for definite integrals looks like this:
∫₁² (3x² + 2) dx = [x³ + 2x + C]₂₁
If you feel compelled to include the '+C', here's how the algebra plays out:
Wrong path:
- Find the indefinite integral: x³ + 2x + C
- Plug in bounds: (2³ + 2×2 + C) - (1³ + 2×1 + C) = (8 + 4 + C) - (1 + 2 + C)
- Simplify: (12 + C) - (3 + C) = 12 + C - 3 - C = 9
The '+C' cancels. That's the check.
But some students get spooked and write:
Final answer: 9 + C
This is not correct. For a definite integral, the answer must be a single number, not a family of functions. The constant cancels automatically if you plug it in twice—there is no leftover '+C' in the final boxed answer.
Why the Constant Vanishes: The Algebra Says So
The core rule here is the Fundamental Theorem of Calculus, which states:
If F(x) is any antiderivative of f(x), then ∫ₐᵇ f(x) dx = F(b) - F(a).
Any antiderivative, including one with '+C'. If you write F(x) + C, then:
F(b) + C - [F(a) + C] = F(b) - F(a).
The two '+C's subtract. There is no ambiguity: it cannot stick around. The only time you see '+C' is after an indefinite integral, not a definite one.
Leaving a '+C' in your definite integral answer makes it incomplete. It tells the grader you missed how the theorem works. The sign that you understood: your boxed answer is a number, not an expression with '+C'.
Exception: When the Constant Actually Stays
There's one situation where a constant survives in an answer, and it's not in definite integrals. Suppose you’re asked to solve:
∫ (3x² + 2) dx = 10
Now you are solving an indefinite integral and setting it equal to a specific value. You must include '+C' when solving for x or setting up an equation. If you write:
x³ + 2x + C = 10 → x³ + 2x = 10 - C
The solution is not a single x—it depends on C. Here, '+C' does not cancel, and dropping it would lose possible solutions. That makes this problem a true exception: for antiderivatives set equal to a value (not definite integrals), the constant is necessary and real.
A Harder Variant: Nonzero Lower Bound and a Shifting Constant
Suppose you are told:
∫₃⁵ (4x) dx = [2x² + K]₅₃
You still get:
(2×25 + K) - (2×9 + K) = (50 + K) - (18 + K) = 50 - 18 = 32
Even if the “constant” is written as K, it cancels. The name doesn't matter. Dropping it is valid. The only exception: if the limits themselves are symbolic (e.g., ∫ₐᵇ f(x) dx), the same rule applies—the constant cancels, and the final answer is F(b) - F(a), no '+C'.
One More to Try
Evaluate ∫₀¹ (6x + 3) dx.
- Antiderivative: 3x² + 3x + C
- Plug in bounds: [3(1)² + 3×1 + C] - [3(0)² + 3×0 + C] = (3 + 3 + C) - (0 + 0 + C) = 6 + C - C = 6
The '+C' disappears. That's the only answer that shows you understood how definite integrals work.
Summary
Including the '+C' constant in both the upper and lower substitutions when evaluating a definite integral means it cancels out and never appears in your final answer.
