Definite integrals turn antiderivatives into numbers. Where Part 1 asked "what function has this derivative?", this part asks "what is the value of the integral from a to b?" The Fundamental Theorem of Calculus (FTC) bridges the two: it says you can evaluate a definite integral by finding any antiderivative and subtracting its values at the endpoints.
This is arguably the most important single result in the course. Nearly every later topic (area between curves, volumes, arc length, work) depends on it. You should be comfortable with the indefinite integration techniques from Part 1 before working through these problems.
The Fundamental Theorem of Calculus lets you evaluate ∫ₐᵇ f(x) dx by finding an antiderivative F, then computing F(b) − F(a). This works for polynomials, trig functions, piecewise functions, and absolute-value integrals, provided you split the domain correctly at any points where the function's definition changes.
Definite integral
The integral ∫ₐᵇ f(x) dx, which evaluates to a single number representing the net signed area between f(x) and the x-axis on [a, b]. Think of it as "the total accumulation of f from a to b."
Fundamental Theorem of Calculus (FTC), Part 1
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ₐᵇ f(x) dx = F(b) − F(a). In simple terms, to evaluate a definite integral, find the antiderivative and plug in the endpoints.
Fundamental Theorem of Calculus (FTC), Part 2
If f is continuous, then d/dx [∫₀ˣ f(t) dt] = f(x). In simple terms, differentiating an integral with a variable upper limit gives you back the integrand.
Additivity of integrals
The property that ∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx = ∫ₐᶜ f(x) dx. Think of it as being able to break a journey into legs and add them up.
Piecewise function
A function defined by different formulas on different intervals. To integrate one, split the integral at the boundary points and apply the correct formula on each sub-interval.
Sometimes you are given partial information (the value of one integral and a graph) and must combine pieces to find the answer.
Example: Given that ∫₃⁸ f(x) dx = 161/8 and the graph shows f(x) = (7/4)x on [2, 3], find F(8) − F(2).
By additivity: ∫₂⁸ f(x) dx = ∫₂³ f(x) dx + ∫₃⁸ f(x) dx.
Compute the first piece: ∫₂³ (7/4)x dx = (7/8)[x²]₂³ = (7/8)(9 − 4) = 35/8.
Add the pieces: F(8) − F(2) = 35/8 + 161/8 = 196/8 = 49/2.
The key technique here is reading the function from the graph on the interval where it is not given algebraically, then using additivity to stitch the pieces together.
The standard procedure: find an antiderivative F(x), then compute F(b) − F(a). No + C needed, because it cancels.
Example (polynomial):
Evaluate ∫₀¹ (3 − 6x + 6x²) dx.
Antiderivative: F(x) = 3x − 3x² + 2x³.
F(1) − F(0) = (3 − 3 + 2) − 0 = 2.
Example (simplify first, then integrate):
Evaluate ∫₁² (2x³ − x² + 2) / x² dx.
Divide each term by x²: (2x³ − x² + 2)/x² = 2x − 1 + 2x⁻².
Antiderivative: F(x) = x² − x − 2/x.
F(2) − F(1) = (4 − 2 − 1) − (1 − 1 − 2) = 1 − (−2) = 3.
The lesson here: when the integrand is a fraction with a monomial denominator, divide each term in the numerator separately before integrating.
When the integrand is explicitly written as d/dx of something, the FTC gives you the answer immediately: the integral of a derivative is just the function itself, evaluated at the endpoints.
Example:
Evaluate ∫₀⁴ (d/dx)(2 + x²)¹ᐟ² dx.
Since the integrand is the derivative of (2 + x²)¹ᐟ², the antiderivative is simply (2 + x²)¹ᐟ².
I = [(2 + x²)¹ᐟ²]₀⁴ = (2 + 16)¹ᐟ² − (2 + 0)¹ᐟ² = √18 − √2 = 3√2 − √2 = 2√2.
No computation of the derivative is needed. Recognising that the integrand is already a derivative saves time.
These require you to know the standard trig antiderivatives and the exact trig values at standard angles (especially π/4).
Example:
Evaluate ∫₀^(π/4) (6 sin x + 7 cos x) dx.
Antiderivative: F(x) = −6 cos x + 7 sin x.
F(π/4) − F(0) = (−6/√2 + 7/√2) − (−6 + 0) = 1/√2 + 6 = (√2 + 12)/2.
Note: 1/√2 = √2/2, which is where the factor of 1/2 in the answer comes from.
Example:
Evaluate ∫₀^(π/4) (5 sec² θ − 9 sin θ) dθ.
Recall that ∫ sec² θ dθ = tan θ and ∫ sin θ dθ = −cos θ.
Antiderivative: F(θ) = 5 tan θ + 9 cos θ.
F(π/4) − F(0) = (5·1 + 9/√2) − (0 + 9) = 5 + 9/√2 − 9 = −4 + 9/√2.
You need these exact values memorised: sin(π/4) = cos(π/4) = 1/√2 = √2/2, tan(π/4) = 1.
Whenever the integrand contains an absolute value or is defined piecewise, you must split the integral at the boundary points and handle each piece separately.
Example (absolute value):
Evaluate ∫₃⁷ |x − 4| dx.
The expression inside the absolute value changes sign at x = 4. Split:
On [3, 4]: |x − 4| = 4 − x, so I₁ = ∫₃⁴ (4 − x) dx = [4x − x²/2]₃⁴ = (16 − 8) − (12 − 9/2) = 8 − 15/2 = 1/2.
On [4, 7]: |x − 4| = x − 4, so I₂ = ∫₄⁷ (x − 4) dx = [x²/2 − 4x]₄⁷ = (49/2 − 28) − (8 − 16) = −7/2 + 8 = 9/2.
Total: I = 1/2 + 9/2 = 5.
Example (piecewise function):
Evaluate ∫₀² f(x) dx where f(x) = 5x − x² on [0, 1] and f(x) = 5x − 1 on (1, 2].
Split at the boundary: I = ∫₀¹ (5x − x²) dx + ∫₁² (5x − 1) dx.
I₁ = [5x²/2 − x³/3]₀¹ = 5/2 − 1/3 = 13/6.
I₂ = [5x²/2 − x]₁² = (10 − 2) − (5/2 − 1) = 8 − 3/2 = 13/2.
Total: I = 13/6 + 13/2 = 13/6 + 39/6 = 52/6 = 26/3.
Result | Formula |
|---|---|
FTC Part 1 | ∫ₐᵇ f(x) dx = F(b) − F(a) |
FTC Part 2 | d/dx [∫₀ˣ f(t) dt] = f(x) |
Additivity | ∫ₐᵇ f dx + ∫ᵇᶜ f dx = ∫ₐᶜ f dx |
∫ sec² x dx | tan x + C |
∫ sin x dx | −cos x + C |
∫ cos x dx | sin x + C |
Students often add + C to definite integrals. The constant of integration is only for indefinite integrals; it cancels out in F(b) − F(a).
With absolute value integrals, students frequently forget to split the domain. If you integrate |x − 4| as if it were just (x − 4) across the whole interval, you will get the wrong sign on part of the region.
When a function is piecewise, students sometimes use the wrong formula on the wrong interval. Always check which piece applies to which sub-interval before integrating.
On trig integrals, a common error is writing F(π/4) using decimal approximations instead of exact values. Keep √2/2 in exact form throughout.
⚠️ The FTC is the backbone of nearly every definite-integral problem on the exam. Know it cold.
⚠️ Absolute value and piecewise integrals appear frequently. Splitting the domain correctly is the entire difficulty; the integration itself is usually straightforward.
⚠️ Trig integrals at π/4 test whether you know exact values. Memorise sin(π/4), cos(π/4), tan(π/4) and their reciprocals.
⚠️ Problems that write the integrand as d/dx [something] are testing whether you recognise the FTC shortcut. Do not expand the derivative; just evaluate the original expression at the endpoints.
True or false: ∫₀¹ (3 − 6x + 6x²) dx requires a + C in the final answer. (False. Definite integrals evaluate to a number.)
Fill in the blank: ∫₀^(π/4) cos x dx = ______. (√2/2, which is sin(π/4).)
True or false: ∫₃⁷ |x − 4| dx can be evaluated without splitting the integral. (False. The absolute value changes the formula at x = 4.)
Fill in the blank: If ∫₁⁵ f(x) dx = 10 and ∫₅⁸ f(x) dx = 7, then ∫₁⁸ f(x) dx = ______. (17, by additivity.)
Q: Evaluate ∫₀¹ (3 − 6x + 6x²) dx.
A: Antiderivative is 3x − 3x² + 2x³. Evaluate at 1: 3 − 3 + 2 = 2. Evaluate at 0: 0. Answer: 2.
Q: Evaluate ∫₃⁷ |x − 4| dx.
A: Split at x = 4. I₁ = ∫₃⁴ (4 − x) dx = 1/2. I₂ = ∫₄⁷ (x − 4) dx = 9/2. Total: 5.
Q: Evaluate ∫₀^(π/4) (6 sin x + 7 cos x) dx.
A: Antiderivative is −6 cos x + 7 sin x. At π/4: −6/√2 + 7/√2 = 1/√2. At 0: −6. Result: 1/√2 + 6 = (√2 + 12)/2.
Q: Evaluate ∫₀⁴ (d/dx)(2 + x²)^(1/2) dx.
A: By the FTC, this equals (2 + x²)^(1/2) evaluated from 0 to 4: √18 − √2 = 3√2 − √2 = 2√2.
Q: Evaluate ∫₀² f(x) dx where f(x) = 5x − x² on [0, 1] and f(x) = 5x − 1 on (1, 2].
A: Split at x = 1. I₁ = 13/6, I₂ = 13/2. Total: 26/3.
This connects to area between curves, which uses definite integrals to find the area enclosed by two functions. Absolute value integrals are a special case of this: |f(x)| is the area between f(x) and the x-axis regardless of sign.
The FTC Part 2 (differentiating an integral) reappears in Part 3 of these notes and is essential for problems involving accumulation functions and rate-of-change applications.
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