Fundamental theorem of calculus

The area function. Let $f$ be continuous on $[a,b]$. Define

where: $A(x)$ is the signed area under the graph of $f$ from the fixed point $a$ up to the variable point $x$, and $t$ is a dummy integration variable (it disappears once the integral is evaluated). As $x$ slides rightward, $A(x)$ accumulates more (or less, if $f$ is negative) area.

First FTC — differentiating the area function. $A(x)$ is differentiable at every point of $(a,b)$, and

where: $A'(x)$ is the rate at which accumulated area grows as $x$ increases. In words, the derivative of the area function gives back the original function — this is the precise sense in which $A(x)$ is an anti-derivative of $f$, built directly out of integration rather than guessed.

Second FTC — evaluation using any anti-derivative. If $F$ is any function with $F'(x) = f(x)$ on $[a,b]$, then

where: $F$ is any convenient anti-derivative found by substitution, parts, or partial fractions, and $F(b)-F(a)$ is shorthand often written $\big[F(x)\big]_a^b$. This is the everyday evaluation tool. The figure below makes the link concrete: the shaded region is exactly $A(x)$, and the theorem says the rate at which that shaded area grows, as the right edge $x$ sweeps rightward, equals the height $f(x)$ of the curve at that edge.

Together, the two parts explain why integration and differentiation are inverse operations (first part) and give a fast way to compute any definite integral (second part), replacing an infinite sum of thin strips with a single anti-differentiation. It is worth pausing on why this is remarkable. Without the theorem, evaluating $\int_a^b f(x)\,dx$ would mean going back to the underlying definition — an infinite sum of infinitesimally thin rectangular strips under the curve, then a limit as the strip width shrinks to zero. That limiting process is conceptually important (it is how the definite integral is defined in the first place) but painfully slow to carry out by hand for anything beyond the simplest functions, such as constants or low-degree polynomials. The theorem's real power is that it replaces this limiting process with ordinary anti-differentiation, a skill already built up through substitution, integration by parts, and partial fractions earlier in the chapter. This is precisely why CBSE asks students to use the FTC confidently while explicitly not requiring its proof: the proof belongs to a more theoretical treatment of calculus, while the statement and its application are what solve real problems on definite integrals and, later, on areas bounded by curves.

Use the second FTC to evaluate $\displaystyle\int_0^{\pi/2}\cos x\,dx$, then verify the first FTC on the related area function.

Step 1. Find any anti-derivative of $f(x)=\cos x$. $F(x)=\sin x$ works, since $F'(x)=\cos x$.

Step 2. Apply the second FTC. $\displaystyle\int_0^{\pi/2} \cos x\,dx = F\!\left(\frac{\pi}{2}\right) - F(0) = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = \mathbf{1}$.

Step 3. Confirm the constant of integration is irrelevant. Try $F(x)=\sin x + 5$ instead: $\big(\sin\frac{\pi}{2}+5\big) - \big(\sin 0 + 5\big) = (1+5)-(0+5)=1$, the same final number — exactly as Step 3 of the derivation above predicts, since the constant $C=5$ cancels.

Step 4. Now check the first FTC on a related area function. Let $A(x) = \displaystyle\int_2^x (3t^2-4)\,dt$. By the first FTC, $A'(x)=f(x)=3x^2-4$ directly, so $A'(3) = 3(3)^2-4 = 27-4=23$ with no integration needed.

Step 5. Verify by direct computation. An anti-derivative of $3t^2-4$ is $F(t)=t^3-4t$, so $A(x) = F(x)-F(2) = (x^3-4x)-(8-8) = x^3-4x$. Differentiating ordinarily gives $A'(x)=3x^2-4$, and $A'(3)=3(9)-4=\mathbf{23}$ — matching the first-FTC shortcut exactly.

Competitive stretch

For $A(x) = \displaystyle\int_1^x \frac{1}{t}\,dt$ ($x>0$), the first FTC gives $A'(x) = \dfrac{1}{x}$ immediately, without recognising $A(x)=\ln x$ first — a fast way to plant the derivative of $\ln x$ if that result has not yet been proved separately. The same trick works for any integral with a variable upper limit that is itself a function of $x$, say $A(x)=\int_a^{g(x)} f(t)\,dt$: the first FTC combined with the chain rule gives $A'(x) = f\big(g(x)\big)\cdot g'(x)$, a pattern that appears occasionally in JEE-level problems even though it is a small step beyond the plain NCERT statement.

• Thinking the FTC is a mere shortcut formula rather than a genuine bridge between two operations — the "why" (first part, $A'(x)=f(x)$) and the "how to compute" (second part, $F(b)-F(a)$) are both the theorem.
• Reversing the limits: $F(b)-F(a)$ is always upper minus lower, never $F(a)-F(b)$.
• Worrying about "which" anti-derivative is correct for the second FTC — any one works, because the constant of integration always cancels in a definite integral.
• Trying to reproduce a proof in an exam answer — CBSE states the FTC without proof; only the statement and its use are examinable (the derivation above is for understanding, not for reproduction verbatim).

Ask the tutor: want a fresh example evaluated with both FTC parts side by side, or curious why the constant of integration always cancels — ask the AI tutor to walk it through.