Order this chain of reasoning that explains why the Fundamental Theorem of Calculus works, for $A(x) = \int_a^x f(t)\,dt$ .

Conclude $A'(x) = f(x)$ : accumulation undoes differentiation
Define $A(x)$ as the accumulated area from $a$ to $x$
Nudge $x$ by a sliver $h$ : the area gains a thin strip
So $\frac{A(x+h) - A(x)}{h} \approx f(x)$ , exactly in the limit
That strip is nearly a rectangle: height $f(x)$ , width $h$

Hints

The theorem is a statement about how accumulated area responds to moving its right edge.
Everything hinges on one picture: a sliver of area is a rectangle of height $f(x)$ .

Show the answer

Define $A(x)$ as the accumulated area from $a$ to $x$
Nudge $x$ by a sliver $h$ : the area gains a thin strip
That strip is nearly a rectangle: height $f(x)$ , width $h$
So $\frac{A(x+h) - A(x)}{h} \approx f(x)$ , exactly in the limit
Conclude $A'(x) = f(x)$ : accumulation undoes differentiation

Why

The area function’s growth rate at $x$ is just the curve’s height there, one thin rectangle of insight. That is the bridge joining the two halves of calculus, and it is why antiderivatives compute areas.

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Order this chain of reasoning that explains why the Fundamental Theorem of Calculus works, for A(x)=∫axf(t) dtA(x) = \int_a^x f(t)\,dtA(x)=∫ax​f(t)dt.

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More questions on Fundamental theorem of calculus

Order this chain of reasoning that explains why the Fundamental Theorem of Calculus works, for $A(x) = \int_a^x f(t)\,dt$ .