Difficulty: Intermediate | Prerequisites: Basic differentiation, power rule for antiderivatives
An antiderivative of f is any function F with F' = f. The general antiderivative includes +C. An initial value problem (IVP) uses conditions like f(a) = b to pin down C (and D, for second-order problems). These problems test whether you can integrate, then use algebra to find constants from given information.
General antiderivative
The family of all functions F(x) + C whose derivative is f(x). The +C is essential because any constant vanishes under differentiation.
In simple terms, the general antiderivative is the "reverse derivative" plus an unknown constant.
Particular antiderivative
A specific member of the antiderivative family, found by using an initial condition to determine C.
Think of it as pinning down the one curve (out of infinitely many vertical shifts) that passes through a given point.
Initial value problem (IVP)
A problem that gives you a derivative (or second derivative) and one or more conditions on the function's value at specific points. You integrate, then solve for the constants.
In simple terms, you are given the rate of change and a starting point, and you reconstruct the original function.
Arbitrary constant
The +C that appears in every indefinite integral. It represents the fact that antiderivatives are unique only up to a vertical shift.
Students who forget +C on the general antiderivative lose marks. Students who include +C when evaluating a definite integral are doing unnecessary work (it cancels).
Finding general antiderivatives
Use the reverse power rule: the antiderivative of x^n is x^(n+1)/(n+1) + C, for n not equal to -1.
Example: the general antiderivative of f(x) = 6x^2 - 6x + 5 is F(x) = 2x^3 - 3x^2 + 5x + C.
Always include +C for a general antiderivative. Omitting it is a common mark-losing error.
First-order IVPs: one derivative, one condition
Given f'(t) and a value like f(pi/2) = 2, integrate f' to get f(t) + C, then solve for C.
Example: f'(t) = (4/3)cos(t/3) - 4sin(2t/3). Integrating gives f(t) = 4sin(t/3) + 6cos(2t/3) + C.
Using f(pi/2) = 2: plug in to get 4sin(pi/6) + 6cos(pi/3) + C = 2, so 2 + 3 + C = 2, giving C = -3.
Then f(pi) = 4sin(pi/3) + 6cos(2pi/3) - 3 = 2sqrt(3) - 3 - 3 = -6 + 2sqrt(3).
Unique antiderivatives with F(a) = 0
Example: find the unique antiderivative F of f(x) = 20e^(5x-2) with F(0) = 0.
The general antiderivative is F(x) = 4e^(5x-2) + C.
F(0) = 0 gives 4e^(-2) + C = 0, so C = -4e^(-2).
Result: F(x) = 4e^(-2)(e^(5x) - 1).
Second-order IVPs: two integrations, two conditions
Given f''(t), you integrate twice. The first integration gives f'(t) + C, and you use a condition on f' to find C. The second integration gives f(t) + D, and you use a condition on f to find D.
Example: f''(t) = 15t - 6, with f'(1) = 6 and f(1) = 5.
First integration: f'(t) = (15/2)t^2 - 6t + C. Using f'(1) = 6: 15/2 - 6 + C = 6, so C = 9/2.
Second integration: f(t) = (5/2)t^3 - 3t^2 + (9/2)t + D. Using f(1) = 5: 5/2 - 3 + 9/2 + D = 5, so D = 1.
Result: f(t) = (5/2)t^3 - 3t^2 + (9/2)t + 1. Evaluating at t = -1 gives f(-1) = -9.
Example: f''(x) = 9x, with f(-1) = 5 and f'(-1) = 2.
First integration: f'(x) = (9/2)x^2 + C. Using f'(-1) = 2: 9/2 + C = 2, so C = -5/2.
Second integration: f(x) = (3/2)x^3 - (5/2)x + D. Using f(-1) = 5: -3/2 + 5/2 + D = 5, so D = 4.
Result: f(x) = (3/2)x^3 - (5/2)x + 4. Then f(1) = 3/2 - 5/2 + 4 = 3.
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\int e^{ax} \, dx = \frac{e^{ax}}{a} + C\int \cos(kx) \, dx = \frac{\sin(kx)}{k} + C\int \sin(kx) \, dx = -\frac{\cos(kx)}{k} + CFor second-order IVPs, integrate twice and use two conditions to find both constants.
Students often forget the +C when finding the general antiderivative. Without +C, the answer is incomplete and loses marks.
When solving for C in an IVP, students sometimes plug into the wrong function. If you are given f'(1) = 6, you plug into f', not into f.
In second-order IVPs, students sometimes try to find both constants simultaneously. The correct approach is sequential: find C first using the condition on f', then find D using the condition on f.
Students sometimes write the antiderivative of sin(kx) as cos(kx)/k, forgetting the negative sign. The correct form is -cos(kx)/k.
General antiderivative questions ("find the most general F") always require +C. This is an easy-marks question if you remember it.
IVPs with trigonometric or exponential functions are common. Be comfortable antidifferentiating sin(kx), cos(kx), and e^(ax) with the correct coefficients.
Second-order IVPs (given f'', find f at a point) test your ability to integrate twice and manage two constants sequentially. These are worth the most points and have the most room for algebraic error.
Watch for problems that give f(x) = 20e^(5x-2) and ask for the unique antiderivative with F(0) = 0. Factoring out the constant term cleanly is the key step.
Fill in the blank: The general antiderivative of 6x^2 - 6x + 5 is ______. Answer: 2x^3 - 3x^2 + 5x + C.
True or false: If F'(x) = f(x) and G'(x) = f(x), then F(x) = G(x). Answer: False. F(x) = G(x) + C for some constant C.
Fill in the blank: If f''(x) = 9x and f'(-1) = 2, then f'(x) = ______. Answer: (9/2)x^2 - 5/2.
True or false: The antiderivative of e^(5x-2) is e^(5x-2)/5 + C. Answer: True.
Q: Find the most general antiderivative of f(x) = 6x^2 - 6x + 5.
A: F(x) = 2x^3 - 3x^2 + 5x + C.
Q: Find f(pi) when f'(t) = (4/3)cos(t/3) - 4sin(2t/3) and f(pi/2) = 2.
A: f(t) = 4sin(t/3) + 6cos(2t/3) - 3. At t = pi, f(pi) = -6 + 2sqrt(3).
Q: Find the unique antiderivative F of f(x) = 20e^(5x-2) such that F(0) = 0.
A: F(x) = 4e^(-2)(e^(5x) - 1).
Q: Given f''(t) = 15t - 6, f'(1) = 6, and f(1) = 5, find f(-1).
A: f(-1) = -9. Integrate twice, using f'(1) = 6 to find C = 9/2 and f(1) = 5 to find D = 1.
Q: Given f''(x) = 9x, f(-1) = 5, and f'(-1) = 2, find f(1).
A: f(1) = 3. The function is f(x) = (3/2)x^3 - (5/2)x + 4.
Antiderivatives are the bridge between differentiation and the Fundamental Theorem of Calculus. Every definite integral evaluation relies on finding an antiderivative first.
Initial value problems are the simplest form of differential equations. The techniques here (integrate, solve for constants) extend directly to separable differential equations and linear first-order ODEs later in the course.
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