Definite Integrals

A definite integral is not just area — it is signed area. Regions above the x-axis count as positive and regions below count as negative, so the two can cancel each other out. The real key is the Fundamental Theorem of Calculus: find one antiderivative F, then subtract the value at the lower limit F(a) from the value at the upper limit F(b), and that is your integral. Drag the two endpoints a and b below to watch how the positive and negative areas of sin x add up and cancel.

Signed Area & the Definite Integral

Lower a0

Upper b3.141592653589793

👀 Signed Area

①Area above x-axis is positive, below is negative

②a=0, b=π → area 2 (all positive)

③a=0, b=2π → area 0 (cancellation!)

④This is the heart of the definite integral

Fundamental Theorem of Calculus

FTC

∫_a^b f(x) dx = F(b) − F(a) (F' = f)

Definite integral = antiderivative at upper minus at lower

🌉 Bridge Between Derivative and Integral

①Differentiation and integration are inverse operations

②Find F with F'(x) = f(x); the integral is F(b) − F(a)

③One of the most beautiful theorems in mathematics

FTC II

ddx ∫_a^x f(t) dt = f(x)

Differentiating the integral with variable upper limit gives back the integrand

Properties of Definite Integrals

Interval Splitting

∫_a^b f dx = ∫_a^c f dx + ∫_c^b f dx

Split the interval, integrate parts, then add

Even / Odd Function

f(−x) = f(x) → ∫_−a^a f dx = 2∫₀^a f dx

Even functions are symmetric: integrate half and double

Odd Function Property

f(−x) = −f(x) → ∫_−a^a f dx = 0

Symmetric integral of an odd function is always zero

⚖️ Power of Symmetry

①Even: y-axis symmetric → halve work × 2

②Odd: origin symmetric → cancellation → 0

③Cuts exam computation in half

Improper Integrals

Upper t5

∞ Infinite but Finite?

①1/x² decays so fast its area is finite even over an infinite tail

②∫₁^t 1/x² dx = 1 − 1/t → 1 as t→∞

③But ∫₁^∞ 1/x dx = ln t → ∞ — divergent!

Convergence Test

∫₁^∞ 1x^p dx : p > 1 converges, p ≤ 1 diverges

p-test: p=2 converges, p=1 (harmonic) diverges

Worked Examples

Example 1

Evaluate the definite integral ∫₁² (3x² + 2) dx.

Find an antiderivative F(x).

F(x) = x³ + 2x

Apply the Fundamental Theorem: F(2) − F(1).

(2³ + 2·2) − (1³ + 2·1) = 12 − 3 = 9

▸ 9

A definite integral is computed by finding an antiderivative F and taking F(upper) − F(lower) (FTC).

Example 2

Evaluate ∫₋₂² (x³ + x²) dx.

Split: x³ is odd (integral 0), x² is even (half the interval, times 2).

∫₋₂² x³ dx = 0, ∫₋₂² x² dx = 2∫₀² x² dx

Evaluate only the even part.

2 · 2³3 = 163

▸ 163

On symmetric intervals, odd parts vanish and even parts are computed on half and doubled — a big shortcut.

Wrap-up

FTC

∫_a^b f(x) dx = F(b) − F(a), F'(x) = f(x)

Computing a definite integral reduces to finding an antiderivative

CSAT-style

For g(x) = ∫₁^x (2t + 1) dt, what is g'(2) + g(1)?

①3

②4

③5

④6

⑤7

▸ ③ 5

By the second form of the FTC, g'(x) = 2x + 1.

g'(x) = 2x + 1 ⇒ g'(2) = 5

g(1) = 0 because the interval has zero width.

g(1) = 0 ⇒ g'(2) + g(1) = 5

🎯 Exam Points

①FTC: compute F(b) − F(a)

②Signed area: area below the axis counts negative

③Even/odd: cuts work in half on symmetric intervals

④Improper: p > 1 converges, p ≤ 1 diverges

⑤FTC II: derivative of an integral with variable upper limit = f(x)

← Previous

Integration Methods

Functions Defined by Integrals

Was this helpful? Support seegongsik