Functions Defined by Integrals

As you slowly push the upper limit of a definite integral outward, the signed area gathered so far becomes a brand-new function F(x) of x. Where f is positive the area adds on and F grows, and where f is negative the area is shaved off and F shrinks. Remarkably, differentiating this area-built F hands back the original integrand f, and that is the Fundamental Theorem of Calculus. Drag the upper-limit x slider to watch the shaded area pile up and to see exactly when the value of F rises and falls.

Area Piles Up Into a Function

Upper limit x1.5

👀 See It

①F(x) is the signed area of f from 0 to x

②Where f is positive, area is added and F increases

③Where f is negative, F decreases

④At an x where f changes sign, F has an extremum

Differentiating Returns the Integrand

Fundamental Theorem (Part 2)

ddx ∫_a^x f(t) dt = f(x)

With a constant lower limit a, differentiating in the upper limit x gives f(x)

Value at the lower limit

F(a) = ∫_a^a f(t) dt = 0

If the interval is a single point, the area is 0

Monotonicity and Extrema of F

Monotonicity test for F

F'(x) = f(x) ⇒ f>0 means F increases, f<0 means F decreases

F has an extremum where f(x)=0 and changes sign

Compute It Directly

Example 1

Find the extremum of F(x)=∫₀^x (t−2) dt.

Differentiating by the theorem returns the integrand.

F'(x) = x − 2

F'(x)=0 ⇒ x=2; the sign goes −→+, so a local min.

F(2) = ∫₀² (t−2) dt = [t²2 − 2t]₀² = −2

▸ local min −2

Without integrating F, you can judge monotonicity first via F′=f.

Example 2

For F(x)=∫₀^x (t²−3t+2) dt, find F'(1).

By the theorem F'(x)=f(x), so substitute x into t.

Substitute x=1.

F'(1) = 1 − 3 + 2 = 0

▸ 0

No need to integrate F — just evaluate f(1) directly.

Wrap-up

Key result

F(x)=∫_a^x f(t) dt ⇒ F'(x)=f(x), F(a)=0

Differentiate to get the integrand, 0 at the lower limit — the start of every problem

2021 KICE mock exam Math (Calculus) type, adapted

What is the local minimum of F(x)=∫₀^x (t²−2t) dt?

①−4/3

②−2/3

③0

④4/3

⑤2

▸ ① −4/3

F'(x)=x²−2x=x(x−2)=0 ⇒ x=0, x=2. At x=2 the sign is −→+, a local min.

Compute F(2)=∫₀² (t²−2t) dt = [t³3 − t²]₀².

F(2) = 83 − 4 = −43

🎯 Exam Points

①If F(x)=∫_a^x f(t)dt then F'(x)=f(x)

②F(a)=0 (zero area at the lower limit)

③Extrema of F are where f changes sign

④For extremum problems, integrate again at that x for the value

⑤If the upper limit is g(x), chain rule gives F'=f(g(x))·g'(x)

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Definite Integrals

Definite Integrals & Series

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