Make the upper limit of a definite integral a variable x and move it, and the signed area collected so far becomes a function of x, F(x).
While x moves through a stretch where f is positive, area is added and F grows; where f is negative, area is taken away and F shrinks.
Differentiate this area function F and you get the original integrand f back, which is the fundamental theorem of calculus.
Move the upper limit x and the shaded area and the gold curve F change together: F falls, reaches its lowest point at x = 1, and then rises again.
Area Piles Up into a Function
seegongsik.com
F(x) = ∫₀ˣ (t − 1) dt: red area is subtracted, green area is added, and the gold curve is F
1.5
👀 What to read in the picture
①F(x) is the signed area under f(t) = t − 1 from 0 to x
②For 0 < x < 1, f is negative, so red area piles up and F decreases
③At x = 1 the sign of f changes from negative to positive, so F has a local minimum, F(1) = −1/2
④For x > 1, green area is added and F rises again; at x = 2 the two areas are equal and F(2) = 0
⑤In the top band, F′(x) = f(x) is the height of the blue point, and its sign decides whether the gold curve is rising or falling
Differentiate and the Integrand Comes Back
Fundamental theorem of calculus
ddx ∫ax f(t) dt = f(x)
With a constant lower limit a, differentiating with respect to the upper limit x gives f(x) itself. If the upper limit is g(x), the chain rule gives F′(x) = f(g(x)) · g′(x)
Value at the lower limit
F(a) = ∫aa f(t) dt = 0
When the interval has length 0, the area is 0
Where F Rises, Falls and Turns
Increasing or decreasing
F′(x) = f(x) ⇒ F increases where f > 0 and decreases where f < 0
F has an extremum where f(x) = 0 and f changes sign
Worked Examples
Example 1
Find the extreme value of F(x) = ∫0x (t − 2) dt.
1
By the fundamental theorem, differentiating F gives the integrand.
F′(x) = x − 2
2
F′(x) = 0 at x = 2, and the sign changes from negative to positive there, so it is a local minimum.
F(2) = ∫02 (t−2) dt = [t22 − 2t]02 = −2
▸ Local minimum −2
You can tell where F rises and falls from F′ = f without integrating first. Integrate only when you need the value at that x.
Example 2
If F(x) = ∫0x (t² − 3t + 2) dt, find F′(1).
1
By the fundamental theorem F′(x) = f(x), so replacing t with x in the integrand gives F′(x).
2
Substitute x = 1.
F′(1) = 1 − 3 + 2 = 0
▸ 0
There is no need to find F by integrating; just compute f(1).
Summary
Key facts
F(x) = ∫ax f(t) dt ⇒ F′(x) = f(x), F(a) = 0
Differentiating gives back the integrand, and the value at the lower limit is 0. Most problems start from these two facts
2021 KICE mock exam Math (Calculus) type, adapted
What is the local minimum value of F(x) = ∫0x (t² − 2t) dt?
①−4/3
②−2/3
③0
④4/3
⑤2
▸ ① −4/3
1
F′(x) = x² − 2x = x(x − 2) = 0 gives x = 0 and x = 2. Around x = 2 the sign changes from negative to positive, so it is a local minimum.
2
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
③F has extrema where f changes sign
④When asked for an extreme value, integrate up to that x to get it
⑤If the upper limit is g(x), the chain rule gives F′(x) = f(g(x)) · g′(x)