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Calculus I (Differential & intro Integral) Unit 6: Antiderivatives & the definite integral
Integration adds up small pieces of change. The definite integral gives net signed area, and the Fundamental Theorem of Calculus connects it to antiderivatives.
Covers: Riemann sums, definite integral properties, Fundamental Theorem of Calculus, u-substitution
Study guide
Integration starts with accumulation. A Riemann sum approximates the area under a curve by adding rectangles: Σ f(xᵢ)Δx. Left, right and midpoint sums use different heights, and trapezoidal sums average left and right. For increasing functions, left sums underestimate and right sums overestimate. As the rectangles get thinner, the sum approaches the definite integral ∫ₐᵇ f(x) dx, the net signed area: area above the x-axis counts positive and area below counts negative.
Definite integrals have useful properties: reversing the limits changes the sign, ∫ₐᵃ f = 0, integrals over adjacent intervals add, and constants factor out.
An antiderivative F of f satisfies F′ = f. The indefinite integral ∫ f(x) dx = F(x) + C includes a constant because antiderivatives differ by constants. Basic rules reverse derivative rules: ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C for n ≠ −1, ∫ 1/x dx = ln|x| + C, ∫ eˣ dx = eˣ + C, ∫ cos x dx = sin x + C.
The Fundamental Theorem of Calculus links derivatives and integrals. Part 1: d/dx ∫ₐˣ f(t) dt = f(x), so accumulation functions are antiderivatives. Part 2: ∫ₐᵇ f(x) dx = F(b) − F(a) for any antiderivative F.
U-substitution reverses the chain rule. Choose u as an inner function whose derivative also appears, rewrite the integral in u, integrate, and substitute back, or change the limits for a definite integral.
If the upper limit is a function of x, d/dx ∫ₐ^g(x) f(t) dt = f(g(x))·g′(x). If both limits vary, the derivative is f(g(x))·g′(x) − f(h(x))·h′(x), where h(x) is the lower limit. Tabulated data often has subintervals of different widths. Multiply each height by its own width, and for a trapezoidal sum add (width)·(left value + right value)/2 for each piece.
More basic antiderivatives: ∫ sin x dx = −cos x + C, ∫ sec²x dx = tan x + C, ∫ 1/(1 + x²) dx = arctan x + C, and ∫ 1/√(1 − x²) dx = arcsin x + C. For g(x) = ∫ₐˣ f(t) dt, find each value from geometric areas (triangles, trapezoids, semicircles), counting area below the axis as negative. When x < a, g(x) = −∫ₓᵃ f(t) dt, so area above the axis to the left of a makes g negative.
Since g′ = f for g(x) = ∫ₐˣ f(t) dt, it follows that g″ = f′. So g is concave up where f is increasing, and g has inflection points where f changes from increasing to decreasing or the reverse.
Key ideas
- Riemann sums (left, right, midpoint, trapezoid) approximate area with rectangles or trapezoids.
- The definite integral is the limit of Riemann sums as the rectangles get infinitely thin.
- Area below the x-axis counts as negative in a definite integral.
- FTC part 1: the derivative of an accumulation function ∫ₐˣ f(t) dt is f(x).
- FTC part 2: evaluate ∫ₐᵇ f(x) dx with any antiderivative F as F(b) − F(a).
- u-substitution reverses the chain rule: pick u as the inside function whose derivative appears.
Formulas and rules
Common mistakes
- Dropping + C on indefinite integrals.
- Forgetting to change the bounds (or switch back to x) after u-substitution.
Words to know
- FTC part 2
- ∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f.
- FTC part 1
- d/dx ∫ₐˣ f(t) dt = f(x)
- ∫ xⁿ dx (n ≠ −1)
- xⁿ⁺¹/(n + 1) + C
- ∫ 1/x dx
- ln|x| + C
- Definite integral
- Net signed area: area above the axis minus area below.
- Reversing limits of integration
- ∫ₐᵇ f = −∫ᵦᵃ f.
- U-substitution
- Use when an inner function's derivative also appears in the integrand.
- Riemann sum
- Σ f(xᵢ)Δx: rectangles approximating area.
- Left / right Riemann sum
- Heights from left or right endpoints; for increasing f, left under- and right overestimates.
- Trapezoidal rule
- Averages left and right sums using trapezoids.
- Antiderivative
- F with F′ = f.
- Indefinite integral
- ∫ f(x) dx = F(x) + C
- ∫ cos x dx
- sin x + C
- ∫ eˣ dx
- eˣ + C
- Accumulation function
- g(x) = ∫ₐˣ f(t) dt
Sample practice questions
45 more questions like these, from level 1 to 5, are in the practice set.
∫ 4x³ dx
- 4x⁴ + C
- x⁴ + C
- x⁴
- 12x² + C
Show the answer and steps
Answer: B) x⁴ + C
- Reverse power rule: ∫ xⁿ dx = xⁿ⁺¹/(n + 1).
- ∫ 4x³ dx = 4x⁴/4 + C = x⁴ + C.
Find ∫ −7x⁴ dx.
- −(7/5)x⁵
- −7x⁵ + C
- −28x³ + C
- −(7/5)x⁵ + C
Show the answer and steps
Answer: D) −(7/5)x⁵ + C
- ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C.
- −7 × x⁵/5 = −(7/5)x⁵ + C.
d/dx ∫₁ˣ √(t² + 1) dt
- 0
- √(x² + 1)
- √(x² + 1) − √2
- x/√(x² + 1)
Show the answer and steps
Answer: B) √(x² + 1)
- FTC part 1: d/dx ∫ₐˣ f(t) dt = f(x).
- f(x) = √(x² + 1).