The rule, reversed: raise by one, divide by the new power.
∫ xⁿ dx = xⁿ⁺¹n + 1 + C
the power rule run backwards
∫ x² dx = x³3 + C, ∫ x dx = x²2 + C
up one, divide by the new power
∫ 5 dx = 5x + C
a constant is 5x⁰ — it gains an x
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- Antiderivatives
Antiderivatives: The Reverse Power Rule
Integration asks the backwards question — which function has THIS as its derivative? Raise the power by one, divide by it, and always add C.
Integrate
∫ 2x dx
Which function differentiates to 2x? x² does — and so does x² + 5, x² − 100, and every other vertical shift, which is why the answer must carry + C: x² + C. B forgets the C, C differentiates instead of integrating, and A never divided by the new power.
Why the + C is not optional.
x², x² + 5 and x² − 100 all have the same derivative, 2x — vertical shifts never change slope. So "the" antiderivative of 2x is a whole family of curves, one for every constant. The + C is the family's name tag, and exams deduct for losing it.
Sums: integrate term by term.
∫ (3x² − 4x + 1) dx
three terms, three reversals
3 · x³3 − 4 · x²2 + x
coefficients ride along, exactly as before
= x³ − 2x² + x + C
one C covers the whole answer
Rewrite as powers first — negative ones too.
Find ∫ 5x³ dx and ∫ (x − 1)(x + 3) dx.
5x³ = 5x−3
bottom power → negative power
5 · x−2/(−2) = −52x² + C
−3 + 1 = −2; divide by −2
(x − 1)(x + 3) = x² + 2x − 3
expand — there is no product rule backwards here
= x³3 + x² − 3x + C
term by term
Raising a negative power moves it toward zero
x−3 goes to x−2, not x−4 (that is differentiating). Then mind the double sign: a minus in front times the new −2 below. Check by differentiating: −52x² = −2.5x−2 → 5x−3 ✓. The rule fails only for x−1, which would divide by zero.
A fraction over a single power splits term by term first. x⁴ + 6x² = x² + 6x−2, so its integral is x³3 + 6x−1/(−1) = x³3 − 6x + C.
One known point pins down C.
(a) f′(x) = 6x − 2 and f(2) = 9: find f(x). (b) A particle has velocity v(t) = 3t² + 2t and starts at s = 5. Where is it at t = 3?
(a) f(x) = 3x² − 2x + C
integrate, keep the C
12 − 4 + C = 9 → C = 1
put in the known point
f(x) = 3x² − 2x + 1
one curve from the family
(b) s(t) = ∫ v dt = t³ + t² + C, s(0) = 5 → C = 5
position is the antiderivative of velocity
s(3) = 27 + 9 + 5 = 41
not v(3) = 33
Same for “a curve through (p, q) with slope …”: the slope is f′, the point fixes C. Leaving C out of s(3) gives 36 — the trap answer that forgets where it started.
Variations: fractional powers, a point with negatives, two constants.
Fractional powers. Find ∫ 3√x dx and ∫ 3√x dx.
3√x = 3x−1/2 → 3 · x1/2 ÷ ½
−½ + 1 = ½
= 6√x + C
dividing by ½ doubles
3√x = x1/3 → x4/3 ÷ 43 = ¾x4/3 + C
dividing by 43 multiplies by ¾
What is different: the new power is a fraction, so “divide by the new power” means multiply by its flip.
3√x does not give 1.5√x
Multiplying by ½ instead of dividing by it gives 1.5√x. Differentiate to catch it: 1.5√x → 0.75√x, not 3√x. And 6√x → 6 · ½x−1/2 = 3√x ✓.
A point with negative coordinates. f′(x) = 2x + 3 and the curve passes through (−1, −6). Find f(x).
f(x) = x² + 3x + C
integrate term by term
(−1)² + 3(−1) + C = −6 → 1 − 3 + C = −6
x = −1 in brackets
C = −4 → f(x) = x² + 3x − 4
C may be negative
What is different: C comes out negative, and every sign in the substitution counts.
Two constants. f″(x) = 6x, f′(0) = −2 and f(0) = 1. Find f(x).
f′(x) = 3x² + C₁, and f′(0) = −2 → C₁ = −2
first integration, first condition
f(x) = x³ − 2x + C₂, and f(0) = 1 → C₂ = 1
second integration, second condition
f(x) = x³ − 2x + 1
check: f″ = 6x ✓
What is different: integrating twice brings in two constants. Find each one before integrating again.
Every integral carries its own free checker.
Differentiate your answer. If the original does not come back, the answer is wrong — no exceptions.
| Claimed integral | Differentiates to | Verdict |
|---|---|---|
| x³ − 2x² + x + C | 3x² − 4x + 1 | matches ✓ |
| x³ − 4x² + x + C | 3x² − 8x + 1 | −8x ≠ −4x ✗ |
On the exam
Expect an integral with a root or 1x² (rewrite as powers), an f(x) from f′ and one point, or position from velocity. Since every choice can be differentiated in a few seconds, the fastest route is often to differentiate the options and see which one gives back the integrand — then check it passes through the given point.
Check the worked answer by differentiating it.
x³
→ 3x²
first term back ✓
−2x²
→ −4x
second term back ✓
x + C
→ 1 + 0
the C vanishes, as it should ✓
Now you try
Practice problem
Integrate
∫ 3x² dx
Raise the power: x³, divide by the new power: 3x³3 = x³ + C — and differentiating x³ indeed returns 3x². A went the wrong direction (that is the derivative). C forgot to divide by 3; D divided but dropped the coefficient 3 that should have cancelled it.