First fundamental theorem of calculus

The area function is built by accumulation. Let $f$ be continuous on $[a,b]$. Define $A(x) = \int_a^x f(t)\,dt$ — the running total of signed area under $f$, swept from the fixed point $a$ to the moving point $x$. Because $f$ is continuous, this running total is a genuine, well-behaved function of $x$, and $A(a)=0$ automatically since $\int_a^a f(t)\,dt = 0$.

Differentiating the accumulation returns the height of the curve. This is the theorem itself:

where: $A(x)=\int_a^x f(t)\,dt$ is the area function, $A'(x)$ is its derivative with respect to the moving upper limit $x$, and $f(x)$ is the value (height) of the original curve at $x$. In words — the rate at which accumulated area grows, as the right-hand boundary slides to the right, equals the height of the curve exactly there. A tall curve sweeps out area fast; a curve near zero barely adds any.

This is why $A(x)$ qualifies as an anti-derivative. By definition, $F$ is an anti-derivative of $f$ precisely when $F'=f$. FTC1 says $A$ satisfies this condition, so $A$ belongs to the family of anti-derivatives of $f$ — specifically the one member of that family that vanishes at $x=a$, since $A(a)=0$. Every other anti-derivative $F(x)$ differs from $A(x)$ only by a constant, $F(x)=A(x)+C$.

The theorem also explains why "any" anti-derivative works when evaluating a definite integral. Elsewhere in the Integrals chapter you compute $\int_a^b f(x)\,dx$ as $F(b)-F(a)$ using whichever anti-derivative $F$ is convenient to find — not necessarily the area function $A$ itself. That step is only legal because FTC1 has already established that $A$ is an anti-derivative of $f$; since $F$ and $A$ differ by a constant $C$, the difference $F(b)-F(a)$ automatically cancels that constant and reproduces $A(b)-A(a)=A(b)$, the actual accumulated area. Without FTC1, there would be no guarantee that some unrelated anti-derivative $F$ gives the same number as the real area.

The chain-rule extension covers a moving upper limit that is itself a function. When the upper limit is $u(x)$ rather than plain $x$, treat $\int_a^{u(x)} f(t)\,dt$ as the composite function $A(u(x))$ and apply the chain rule:

where: $u(x)$ is the (differentiable) upper limit expressed as a function of $x$, and $u'(x)$ is its derivative. This single line is the version examiners actually test most often, since a plain $x$ upper limit is the easy special case $u(x)=x$, $u'(x)=1$. The figure below shows the mechanism in one picture: as the right boundary $x$ nudges forward by a small $dx$, the extra sliver of area added is approximately a thin rectangle of height $f(x)$ and width $dx$, so the extra area is $\approx f(x)\,dx$ — dividing by $dx$ and shrinking it to zero is exactly the definition of $A'(x)$, and it lands on $f(x)$.

Find $\dfrac{d}{dx}\displaystyle\int_0^{x^2} \cos t\,dt$, and check it by evaluating the integral explicitly first.

Step 1. Identify the pieces for the chain-rule extension. The integrand is $f(t)=\cos t$, and the upper limit is $u(x)=x^2$ rather than plain $x$, so this needs $\dfrac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\cdot u'(x)$, with $u'(x)=2x$.

Step 2. Substitute directly. $\dfrac{d}{dx}\displaystyle\int_0^{x^2}\cos t\,dt = \cos(x^2)\cdot 2x = 2x\cos(x^2)$.

Step 3. Verify by evaluating the integral first. An anti-derivative of $\cos t$ is $\sin t$, so $\int_0^{x^2}\cos t\,dt = \big[\sin t\big]_0^{x^2} = \sin(x^2) - \sin 0 = \sin(x^2)$. Differentiating $\sin(x^2)$ with the ordinary chain rule gives $\cos(x^2)\cdot 2x$ — exactly matching Step 2.

Answer: $\dfrac{d}{dx}\displaystyle\int_0^{x^2}\cos t\,dt = 2x\cos(x^2)$, and the two routes — the theorem directly, or evaluate-then-differentiate — agree, which is the whole point of FTC1: it is a shortcut, not a different answer.

Competitive stretch

Find $\dfrac{d}{dx}\displaystyle\int_x^{5} t^3\,dt$. Here the variable limit is the lower one, so first flip it using $\int_x^5 t^3\,dt = -\int_5^x t^3\,dt$, then apply FTC1 to the flipped form: $\dfrac{d}{dx}\left[-\int_5^x t^3\,dt\right] = -x^3$. The sign flip is essential — differentiating as if $x$ were an upper limit without flipping first gives the wrong sign, $x^3$ instead of $-x^3$.

• Forgetting the chain-rule factor $u'(x)$ when the upper limit is a function of $x$ — writing $\int_0^{x^2}\cos t\,dt \to \cos(x^2)$ and dropping the $2x$ is the single most common slip.
• Confusing the dummy variable $t$ with $x$ — the answer to $\dfrac{d}{dx}\int_a^x f(t)\,dt$ is $f(x)$, never $f(t)$; the dummy variable is a placeholder that never appears in the final answer.
• Applying FTC1 directly when the variable limit is the lower one — $\dfrac{d}{dx}\int_x^b f(t)\,dt = -f(x)$, because reversing the limits (property P1) introduces a minus sign before the theorem can be applied.
• Trying to reproduce a full proof in an exam answer — CBSE states FTC1 without proof, so only correct statement and application are expected, not a derivation from the limit definition.

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