Fix the left end of an interval and let the right end move. The area accumulated so far is a new function of $x$. The Fundamental Theorem says its rate of change is the height of the curve at the moving edge. This ties integrals to derivatives, and the The Chain Rule extends it.
Key idea
Fundamental Theorem of Calculus, Part 1. If $f$ is continuous and $A(x) = \displaystyle\int_a^x f(t)\,dt$, then $A'(x) = f(x)$. With a function as the upper limit: $$\dfrac{d}{dx}\displaystyle\int_a^{u(x)} f(t)\,dt = f\big(u(x)\big)\cdot u'(x)$$.
Why it works
Increase $x$ by a little, $h$. The extra area is a thin strip from $x$ to $x + h$, almost a rectangle of height $f(x)$ and width $h$: $A(x + h) - A(x) \approx f(x)\,h$. So $\frac{A(x + h) - A(x)}{h} \approx f(x)$, and the approximation becomes exact as $h \to 0$ because $f$ is continuous. Area grows at a rate equal to the height of the leading edge.
Worked example: Basic
Find $\dfrac{d}{dx}\displaystyle\int_1^x \sqrt{1 + t^3}\,dt$.
- FTC Part 1: replace $t$ with $x$ in the integrand.
- No antiderivative is needed (and $\sqrt{1 + t^3}$ has none among familiar functions).
Answer: $\sqrt{1 + x^3}$
Worked example: A function in the upper limit
Find $\dfrac{d}{dx}\displaystyle\int_0^{x^2}\cos t\,dt$.
- The upper limit is $u = x^2$. Chain rule: $\cos(u)\cdot u'$.
- $= \cos(x^2)\cdot 2x$.
- Check: the integral equals $\sin(x^2)$, whose derivative is $2x\cos(x^2)$.
Answer: $2x\cos(x^2)$
Worked example: Variable in the lower limit
Find $\dfrac{d}{dx}\displaystyle\int_x^5 e^{t^2}\,dt$.
- Flip the limits: $-\int_5^x e^{t^2}\,dt$.
- FTC Part 1: $-e^{x^2}$.
Answer: $-e^{x^2}$
Common mistake
Forgetting the chain rule factor when the upper limit is a function of $x$: writing $\cos(x^2)$ instead of $2x\cos(x^2)$ in the second example. Check: if the upper limit is not plain $x$, there must be a factor $u'(x)$ in your answer.