Letting a definite integral's upper bound move turns it into a genuine function, A(x) = \int_a^x f(t)\,dt. This is a running total of signed area: it grows wherever f is positive and shrinks wherever f is negative. That "running total" picture, more than the mechanics of any one antiderivative, is the idea double and triple integrals generalize next. They accumulate a quantity over a region instead of over an interval.
Think of A(x) as a running total on a receipt. It updates continuously as x scans rightward, adding whatever is under the curve between a and the current x. Where f dips negative, the running total ticks down. That's the same way a refund lowers a receipt's balance instead of raising it. FTC Part 1 just says the receipt's own rate of change, at any instant, equals whatever value f currently reads. In other words, how fast the total is moving right now has to match the current input.
The definite integral as accumulation is what results when a definite integral's upper bound is allowed to move. A(x) = \int_a^x f(t)\,dt becomes a genuine function of x. It is a running total of signed area. A(x) is built from a limit of Riemann sums the same way any single definite integral is. The only new feature is that the upper bound x is now a variable instead of a fixed number. That makes A itself a function, one that can be differentiated, evaluated, and graphed like any other.
FTC Part 1 says that differentiating gives back the integrand: A'(x)=f(x). This is the fact that makes area computable by antidifferentiation in the first place. It's also the fact a double integral leans on directly: an iterated double integral's inner integral is an accumulation function evaluated at the outer variable's current value.
Theorem
Fundamental Theorem of Calculus, Part 1
also known as: FTC1, the accumulation-function derivative theorem
Recalled from Calculus I — Differential & Integral Calculus · FTC Part 1: the accumulation function — first taught there.
Suppose f is continuous on [a, b]. Define A(x) = \int_a^x f(t)\,dt for x in [a, b]. Then A is differentiable on (a, b) and
A'(x) = f(x).Fix x in (a,b) and form the difference quotient defining A'(x): \frac{A(x+h)-A(x)}{h} = \frac{1}{h}\left[\int_a^{x+h}f(t)\,dt -\int_a^x f(t)\,dt\right] = \frac{1}{h}\int_x^{x+h} f(t)\,dt, the last step by additivity of the definite integral over adjacent intervals. The remaining integral is f's net signed area over a sliver of width h -- and by the Mean Value Theorem for Integrals (f continuous on [x,x+h]), that sliver equals f(c)\cdot h for SOME c between x and x+h, so \frac{A(x+h)-A(x)}{h} = f(c) \quad\text{for some $c$ strictly between $x$ and $x+h$.} As h\to0, c is squeezed between x and x+h and is forced to x too, and f's continuity at x then forces f(c)\to f(x). Taking the limit, A'(x) = \lim_{h\to0}\frac{A(x+h)-A(x)}{h} = \lim_{h\to0} f(c) = f(x).
The same conclusion falls out of the Riemann-sum definition directly, without invoking the Mean Value Theorem for Integrals: for small h, \int_x^{x+h} f(t)\,dt is itself approximated by a single thin rectangle of width h and height approximately f(x) (since f barely changes over an interval that short, by continuity), so \frac{1}{h}\int_x^{x+h}f(t)\,dt\approx f(x), with the approximation becoming exact in the limit h\to0 -- the same "thin slice, height \approx f(x)" idea Chapter 7 built Riemann sums from in the first place, just run over a single infinitesimally short interval instead of the whole [a,b].
Continuity of f is load-bearing in both arguments: it is what supplies the Mean Value Theorem for Integrals' own hypothesis, and it is what forces f(c)\to f(x) as the squeezed point c\to x -- a discontinuous f can make A fail to be differentiable exactly at the discontinuity, even though A itself (an accumulated area) still exists there.
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A running total that can go negative
Drag x past where f(t) dips below the axis — watch A(x) start decreasing instead of increasing.
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A(x) is a limit of exactly these rectangles
Drag n up and flip between left/right/midpoint — every rule converges to the same accumulated total as n grows.
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Worked example
Evaluating and differentiating an accumulation function
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1
For A(x) = \displaystyle\int_0^x (2t+1)\,dt, find A(3) and A'(3).
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2
A(3) is an ordinary definite integral: A(3) = \displaystyle\int_0^3 (2t+1)\,dt = \left[t^2+t\right]_0^3 = 9+3.
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3
A'(x) = f(x) by FTC Part 1. No antidifferentiation is needed to get the derivative at all; just substitute x=3 into the original integrand 2t+1.
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4
\boxed{A(3) = 12, \quad A'(3) = 2(3)+1 = 7}
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Where this goes next
Chapter 7: Double integrals
\displaystyle\int_c^d\!\left(\int_a^b f(x,y)\,dx\right)dy computes the inner integral first, at each fixed value of y. That inner integral is an accumulation function evaluated one y at a time, then integrated again over the outer variable. Fluency with A(x) here is fluency with the inner half of every iterated integral ahead.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
For A(x) = \int_0^x (2t+1)\,dt, find A(3).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\left[t^2+t\right]_0^3 = 12-0=12.
Correct answer
Also accepted:
Why
Find an antiderivative of 2t+1, then evaluate it at the two bounds and subtract.
From Calculus II — Integral Calculus & Infinite Series
In A(x) = \int_a^x f(t)\,dt, why is the integration variable written as t rather than x?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
There's a concrete reason — it isn't arbitrary styling.
t here is just a placeholder integration variable — it doesn't have to represent time at all.
f is the same function regardless of which letter names its input — the letter choice is about avoiding a collision with the bound, not about f itself.
Why
x already names the moving bound; a second use of x as the integration variable would be ambiguous.
x is already doing a job in this expression -- naming the upper (moving) bound.
From Calculus II — Integral Calculus & Infinite Series
For A(x) = \int_1^x \frac{1}{t}\,dt, find A(e) (where e is Euler's number).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\left[\ln t\right]_1^e = \ln e - \ln 1 = 1-0=1.
Correct answer
Also accepted:
Why
Find an antiderivative of 1/t, then evaluate it at e and at 1 and subtract.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus I — Differential & Integral Calculus
For f(t) = t^2 + 1 and A(x) = integral_0^x f(t) dt, find A'(3).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
By FTC Part 1, A'(x)=f(x)=x^2+1, so A'(3)=3^2+1=10.
Correct answer
Also accepted:
Why
By FTC Part 1, A'(x) = f(x) directly -- no antiderivative needs to be found first. Just substitute x=3 into f.
From Calculus I — Differential & Integral Calculus
If A(x) = integral_a^x f(t) dt and f is continuous, what does FTC Part 1 say A'(x) equals?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
A'(x) matches f evaluated at the RUNNING upper limit x, not the fixed lower limit a.
FTC Part 1 differentiates A itself -- it never asks for the derivative of f.
The average value of f on [a,x] is integral_a^x f(t) dt divided by (x-a), a different quantity from A'(x).
Why
FTC Part 1 says the accumulation function A(x)=integral_a^x f(t) dt satisfies A'(x)=f(x) -- differentiating an accumulation undoes the accumulation and returns the original integrand, evaluated at the boundary x.
From Calculus I — Differential & Integral Calculus
Try a different f -- for f(t) = 4t - 1 and A(x) = integral_0^x f(t) dt, find A'(5).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
By FTC Part 1, A'(x)=f(x)=4x-1, so A'(5)=4(5)-1=19.
Correct answer
Also accepted:
Why
By FTC Part 1, A'(x)=f(x) directly -- substitute x=5 into f.