Every question so far has run in one direction: given a function, find its derivative. This section runs it backward for the first time. Given a function f, an antiderivative is any function F with F'(x)=f(x). The moment that question gets asked, something new happens that never came up computing a derivative directly: the answer isn't unique.
Someone hands you a car's current speed and asks where it started. You genuinely can't know. The same speed reading is consistent with starting a mile up the road, a mile down it, or right here. Knowing the speed at every instant pins down how POSITION CHANGES perfectly, but it says nothing about where position started. An antiderivative has exactly that gap built in. The +C is the honest admission that the starting point was never in the data to begin with.
An antiderivative of a function f is any function F whose own derivative gives f back: F'(x)=f(x). If F'(x) = f(x), then so is (F(x)+5)', and (F(x)-100)', and (F(x)+C)' for literally any constant C, since a constant's own derivative is always 0. That's not a technicality to patch over. It's a genuine fact about how much information a derivative throws away. Two functions with the same derivative on an interval differ by a constant, never anything more exotic. That's why "+C" isn't an approximation or a safety net. It's the complete description of every antiderivative f has. The indefinite integral, written \int f(x)\,dx, is the notation for this entire family at once: not a single function, but every antiderivative f has, bundled into one expression.
Every antiderivative rides the same slopes
Slide C — the curve rides through the same field of slopes no matter where it starts.
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A steeper set of slopes, same idea
Slide C again — a different f' still produces a whole parallel family.
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Worked example
A first antiderivative, with +C
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1
Find \displaystyle\int (4x^3 - 6x) \, dx.
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2
Reverse the power rule on each term: raise the exponent by 1, then divide by the new exponent. For 4x^3: raise to x^4, divide by 4. The 4's cancel, leaving x^4.
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3
For -6x: raise to x^2, divide by 2, giving -3x^2.
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4
Add the constant of integration, since nothing in the problem pins down a starting value: \boxed{\int (4x^3-6x)\,dx = x^4 - 3x^2 + C}.
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Where this goes next
Every remaining section of this chapter
+C isn't optional bookkeeping. The moment a problem supplies extra information, such as an initial value or a point the curve passes through, that information pins C down to one specific number. A whole family of curves becomes a single one.
Check your understanding (counts toward your progress)
If F(x) = x^3 - 4x + C is an antiderivative of f(x)=3x^2-4 and F(0)=7, what is C?
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Why
F(0) = 0-0+C = C, so C=7 directly from the given initial value.
Correct answer
Also accepted:
Why
Plug x=0 into F(x) and set the result equal to the given value F(0)=7.
F(x) = 8x^{4}. Which f(x) makes F an antiderivative of f?
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Why
F is an antiderivative of f exactly when F'=f — differentiating F(x)=8x^{4} gives F'(x)=32x^{3}, so that's f.
Find the antiderivative of this quadratic trinomial: \displaystyle\int (x^2 - 3x + 5)\,dx.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Term by term power rule on each of the three terms, plus C.
Correct answer
Also accepted:
Why
Apply the power rule \int x^n\,dx = \frac{x^{n+1}}{n+1} to each of the three terms separately.
F(x) = 14x^{5}. Which f(x) makes F an antiderivative of f?
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Why
F is an antiderivative of f exactly when F'=f — differentiating F(x)=14x^{5} gives F'(x)=70x^{4}, so that's f.
Practice: Reversing the power rule, one term at a time (4 problems · doesn’t affect your progress)
Which of these is a valid antiderivative of f(x) = 2x?
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x^2 works, but so does x^2+5, x^2-100, and every other shift — restricting to one misses the whole family.
x^2+5 works, but it's only ONE member of the family, not the complete answer.
\dfrac{d}{dx}[2x^2] = 4x \neq 2x — this isn't an antiderivative of 2x at all.
Why
Every antiderivative of 2x has the form x^2+C — any two antiderivatives of the same function differ only by a constant.
x^2+C for any constant C
Find the antiderivative of six times x-squared: \displaystyle\int 6x^2\,dx.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Raise the exponent to 3, divide by 3 — \frac{6}{3}x^3 = 2x^3, plus C.
Correct answer
Also accepted:
Why
Find \displaystyle\int \sqrt{x}\,dx.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int x^{1/2}\,dx = \dfrac{x^{3/2}}{3/2} = \dfrac{2}{3}x^{3/2}, plus C.
Correct answer
Also accepted:
Why
Rewrite as x^{1/2} first.
Find \displaystyle\int \dfrac{x^3+2}{x^2}\,dx.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Split into x + 2x^{-2}, then integrate term by term — \dfrac{x^2}{2} - 2x^{-1}, plus C.
Correct answer
Also accepted:
Why
Split into two fractions first: \dfrac{x^3}{x^2} + \dfrac{2}{x^2}.