Derivatives measure how fast something changes. Integrals do the reverse: they add up small pieces to find a total, like the area under a curve, the distance traveled from a velocity graph or the total water that flowed into a tank. This guide covers the core ideas you need for a first calculus course.
Two kinds of integrals
- An indefinite integral,
∫ f(x) dx, is a family of functions whose derivative is f(x). It always ends with "+ C." - A definite integral,
∫ from a to b of f(x) dx, is a number: the signed area between the curve and the x-axis from x = a to x = b. Area above the axis counts as positive and area below as negative.
Antiderivatives and the power rule
An antiderivative of f is any function F with F′ = f. Since the derivative of a constant is zero, every antiderivative has a "+ C." The most-used rule is the reverse power rule:
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, for n ≠ −1.
Other basics: ∫ k dx = kx + C, ∫ eˣ dx = eˣ + C, ∫ (1/x) dx = ln|x| + C, ∫ cos x dx = sin x + C and ∫ sin x dx = −cos x + C. Integrals split over sums, and constants can be pulled out front.
Worked example 1: an indefinite integral
Find ∫ (6x² − 4x + 5) dx.
- Integrate term by term: 6 · x³/3 − 4 · x²/2 + 5x + C.
- Simplify: 2x³ − 2x² + 5x + C.
- Check by differentiating: 6x² − 4x + 5. Correct.
The Fundamental Theorem of Calculus
This theorem connects the two kinds of integrals. If F is any antiderivative of f, then:
∫ from a to b of f(x) dx = F(b) − F(a).
So to find an area, you find an antiderivative and subtract its values at the endpoints. The + C cancels, so you can leave it out.
Worked example 2: area under a curve
Find the area under y = x² from x = 1 to x = 3.
- Antiderivative: F(x) = x³/3.
- F(3) − F(1) = 27/3 − 1/3 = 26/3 ≈ 8.67 square units.
Worked example 3: distance from velocity
A car's velocity is v(t) = 4t + 2 m/s. How far does it travel from t = 0 to t = 5 seconds?
∫ from 0 to 5 of (4t + 2) dt = [2t² + 2t] from 0 to 5 = (50 + 10) − 0 = 60 m.
u-substitution
u-substitution undoes the chain rule. Look for a function and its derivative inside the integral.
Example: ∫ 2x(x² + 1)³ dx.
- Let u = x² + 1, so du = 2x dx.
- The integral becomes ∫ u³ du = u⁴/4 + C.
- Substitute back: (x² + 1)⁴ / 4 + C.
Estimating with rectangles
Before the theorem, areas were estimated with Riemann sums: split the interval into rectangles, add their areas and make the rectangles thinner. For y = x² on [0, 2] with four right-endpoint rectangles of width 0.5, the heights are 0.25, 1, 2.25 and 4, so the estimate is 0.5 × 7.5 = 3.75. The exact area is 8/3 ≈ 2.67; right endpoints overestimate here because the function is increasing.
Common mistakes
- Forgetting + C on indefinite integrals.
- Applying the power rule to 1/x. Its integral is ln|x|.
- Subtracting in the wrong order: it is F(b) − F(a).
- In u-substitution, forgetting to replace dx or to switch back to x.
Practice questions
- Find ∫ (3x² + 2) dx.
- Evaluate ∫ from 0 to 2 of x³ dx.
- Find ∫ cos(3x) dx.
- Evaluate ∫ from 1 to e of (1/x) dx.
Answers: (1) x³ + 2x + C. (2) 16/4 − 0 = 4. (3) sin(3x)/3 + C. (4) ln e − ln 1 = 1.
Last reviewed 2026-10-23 by the BrainBoost team.