This is your first proper encounter with integration, the reverse operation of differentiation. Everything in this set of notes covers indefinite integrals: given a derivative, how do you recover the original function? This sits at the foundation of integral calculus and directly feeds into the Fundamental Theorem of Calculus, which links antiderivatives to areas under curves.
You should already be comfortable with differentiation rules (power rule, trig derivatives, chain rule) before tackling these problems. If you can differentiate a polynomial or a trig function confidently, you are ready.
An antiderivative of f(x) is any function F(x) whose derivative equals f(x). Finding one means reversing the differentiation rules you already know. The indefinite integral collects every possible antiderivative by adding an arbitrary constant C.
Antiderivative
A function F(x) such that F'(x) = f(x). In simple terms, it is the function you get when you "undo" a derivative.
Indefinite integral
The family of all antiderivatives of a function, written as ∫ f(x) dx = F(x) + C. Think of it as the general answer to "what was differentiated to give me this?"
Constant of integration (C)
The arbitrary constant added to every indefinite integral, reflecting the fact that infinitely many functions share the same derivative. If you forget C on an indefinite integral, you lose marks.
Power rule for integration
The rule ∫ xʳ dx = xʳ⁺¹ / (r + 1) + C, valid for all r ≠ −1. Think of it as the reverse of the power rule for differentiation: bump the exponent up by one, then divide by the new exponent.
Linearity of integration
The property that ∫ [a·f(x) + b·g(x)] dx = a·∫ f(x) dx + b·∫ g(x) dx. In simple terms, you can split an integral across sums and pull constants out front, then integrate each piece separately.
The power rule is the workhorse of indefinite integration. If you can differentiate xⁿ, you can integrate xʳ by running the rule backwards.
Standard form: ∫ xʳ dx = xʳ⁺¹ / (r + 1) + C, for any r ≠ −1.
The rule works for negative and fractional exponents just as well as for positive integers. Rewrite roots and reciprocal powers as fractional exponents before applying it.
Example (fractional/negative exponent):
Integrate ∫ (2 − x⁻⁵ᐟ⁴) dx.
Split into two terms: 2∫ dx − ∫ x⁻⁵ᐟ⁴ dx.
Apply the power rule to x⁻⁵ᐟ⁴: the new exponent is −5/4 + 1 = −1/4, and dividing by −1/4 gives a factor of −4, so ∫ x⁻⁵ᐟ⁴ dx = −4x⁻¹ᐟ⁴ + C.
Putting it together: I = 2x + 4x⁻¹ᐟ⁴ + C.
Example (expanding before integrating):
Integrate ∫ x(2 + 9x³) dx.
Expand first: x(2 + 9x³) = 2x + 9x⁴.
Integrate term by term: ∫ (2x + 9x⁴) dx = x² + (9/5)x⁵ + C.
Linearity means you can break an integral of a sum into a sum of integrals, and pull constants through the integral sign.
∫ [af(x) + bg(x)] dx = a∫ f(x) dx + b∫ g(x) dx
This lets you handle any polynomial (or polynomial-like expression) by integrating one term at a time.
Example (polynomial):
Integrate ∫ (5 + 8x − 7x²) dx.
Split and apply the power rule to each term:
∫ 5 dx = 5x
∫ 8x dx = 4x²
∫ 7x² dx = (7/3)x³
Result: I = 5x + 4x² − (7/3)x³ + C.
Example (expand first, then use linearity):
Integrate ∫ (3 − √x)(5 + √x) dx.
Expand the product: (3 − √x)(5 + √x) = 15 − 2√x − x.
Now integrate term by term:
∫ 15 dx = 15x
∫ 2√x dx = 2 · (2/3)x³ᐟ² = (4/3)x√x
∫ x dx = (1/2)x²
Result: I = 15x − (4/3)x√x − (1/2)x² + C.
The basic trig antiderivatives follow directly from reversing the derivative rules you already know:
∫ cos x dx = sin x + C (because d/dx [sin x] = cos x)
∫ sin x dx = −cos x + C (because d/dx [−cos x] = sin x)
Example (integrating a second derivative twice):
Given f''(x) = 16x − 3 cos x, find f(x).
First integration gives f'(x): ∫ (16x − 3 cos x) dx = 8x² − 3 sin x + C.
Second integration gives f(x): ∫ (8x² − 3 sin x) dx = (8/3)x³ + 3 cos x + Cx + D.
The constants C and D are arbitrary. One valid answer: f(x) = (8/3)x³ + 3 cos x − 5x + 7.
Notice how the sign of the trig term flips at each stage. Differentiating sin gives cos; differentiating cos gives −sin. Integrating reverses this chain, so integrating −3 cos x gives −3 sin x, and integrating −3 sin x gives +3 cos x.
Rule | Formula |
|---|---|
Power rule | ∫ xʳ dx = xʳ⁺¹ / (r + 1) + C, r ≠ −1 |
Constant | ∫ a dx = ax + C |
Cosine | ∫ cos x dx = sin x + C |
Sine | ∫ sin x dx = −cos x + C |
Linearity | ∫ [af(x) + bg(x)] dx = a∫f(x) dx + b∫g(x) dx |
Students often forget the constant of integration on indefinite integrals. Every indefinite integral needs a + C. Definite integrals do not, because the constant cancels when you subtract.
When integrating x⁻⁵ᐟ⁴, students sometimes add 1 to the exponent incorrectly. The new exponent is −5/4 + 1 = −1/4, not +1/4. Write the arithmetic out if you are unsure.
A common sign error with trig: the antiderivative of sin x is −cos x, not +cos x. Check by differentiating your answer.
When integrating a product like (3 − √x)(5 + √x), students try to integrate the factors separately. You cannot. Expand the product first, then integrate term by term.
⚠️ The power rule for integration is tested constantly, often with negative and fractional exponents designed to trip you up.
⚠️ Forgetting + C on an indefinite integral is one of the most common point losses in exams. Markers look for it.
⚠️ Problems that give you f''(x) and ask for f(x) require you to integrate twice. Each integration introduces its own constant.
⚠️ Expand products and simplify quotients before integrating. If you see a product of two expressions in an integral, your first instinct should be to multiply it out.
True or false: ∫ x⁻¹ dx = x⁰ / 0 + C. (False. The power rule does not apply when r = −1.)
Fill in the blank: ∫ cos x dx = ______ + C. (sin x)
True or false: ∫ [f(x) · g(x)] dx = [∫ f(x) dx] · [∫ g(x) dx]. (False. There is no product rule for integrals. Expand first.)
Fill in the blank: The antiderivative of 8x² is ______. ((8/3)x³ + C)
True or false: If F'(x) = f(x), then F(x) + 7 is also an antiderivative of f(x). (True. Adding a constant does not change the derivative.)
Q: If f''(x) = 16x − 3 cos x, which of the following could be f(x)?
A: f(x) = (8/3)x³ + 3 cos x − 5x + 7. Integrate twice: f'(x) = 8x² − 3 sin x + C, then f(x) = (8/3)x³ + 3 cos x + Cx + D. Any values of C and D are valid.
Q: Determine ∫ (2 − x⁻⁵ᐟ⁴) dx.
A: I = 2x + 4x⁻¹ᐟ⁴ + C. Apply the power rule to x⁻⁵ᐟ⁴: new exponent is −1/4, and 1/(−1/4) = −4.
Q: Determine ∫ (5 + 8x − 7x²) dx.
A: I = 5x + 4x² − (7/3)x³ + C. Integrate each term separately using the power rule.
Q: Calculate ∫ (3 − √x)(5 + √x) dx.
A: Expand first to get 15 − 2√x − x, then integrate: I = 15x − (4/3)x√x − (1/2)x² + C.
This connects directly to the Fundamental Theorem of Calculus (covered in Part 2), which says that the definite integral ∫ₐᵇ f(x) dx = F(b) − F(a), where F is any antiderivative of f. Everything you practise here with indefinite integrals becomes the tool you use to evaluate definite integrals.
The power rule for integration is also the foundation for computing areas, volumes of revolution, and work problems later in the course. The trig antiderivatives reappear heavily in integration by substitution and integration by parts.
antiderivative, indefinite integral, power rule for integration, constant of integration, integration of polynomials, trig integrals, integrating cos x, integrating sin x, fractional exponents, negative exponents, double integration, second derivative to original function, linearity of integrals, expand before integrating, integral calculus, UT Austin, HW 02