Fundamental theorem of calculus

What are the fundamental theorems of calculus?

In plain words: the signed area under a curve changes, as x moves right, at a rate equal to the height of the curve there. So the area function is an anti derivative of f, and any anti derivative gives the signed area by subtraction. For nonnegative f and increasing limits, this equals ordinary geometric area. This is the link between the derivative and the integral. NCERT states both theorems without proof, as the proofs are beyond the scope of the book (page 267). NCERT writes the area function as ∫_a^x f(x) dx; we write t inside, so that the variable of integration is not confused with the upper limit x.

NCERT slip (page 267): Section 7.8.1 begins “We have defined ∫_ₐ^ᵇ f(x) dx as the area of the region bounded by the curve y = f(x), the ordinates x = a and x = b and x-axis”, but the 2026–27 book has not defined it that way: Section 7.7 says only that it can be introduced as the limit of a sum or as F(b) − F(a). Read the sentence as a fact for f(x) ≥ 0: the integral equals that area.

The area function

For f(t) = 2t, a = 0 and x > 0, the region under the line from 0 to x is a triangle with base x and height 2x, so A(x) = ½ × x × 2x = x². Its derivative is 2x = f(x), as Theorem 1 says.

Beyond the NCERT text (JEE): when the upper limit is itself a function g(x), for continuous f and differentiable g, Theorem 1 and the chain rule give

For example, d/dx ∫₀^x2 cos t dt = cos (x²) × 2x.

Beyond the NCERT text (JEE): if the differentiable lower limit h(x) varies too, the derivative is f(g(x)) g′(x) minus f(h(x)) h′(x).

Evaluating a definite integral

NCERT’s steps (page 268):

1. Find the indefinite integral ∫f(x) dx = F(x). No constant is needed: with F(x) + C, the C cancels in [F(b) + C] − [F(a) + C].
2. Evaluate F(b) − F(a) = [F(x)]_a^b.

To apply NCERT’s theorem as stated, check that the integrand is defined and continuous on [a, b] (NCERT’s Remark (iv), page 268). For instance, ∫₋₂³ x(x² − 1)^1/2 dx makes no sense, because x(x² − 1)^1/2 is not defined for −1 < x < 1.

Beyond the NCERT text: a break at an end point only, as with 1/√(1 − x²) at 1 (NCERT Exercise 7.8, Q9), still gives F(b) − F(a) when F has a finite limit there. This is an improper integral, evaluated by a one-sided limit, assuming continuity on the interior. An interior singularity requires splitting there and testing the two improper integrals separately. Continuity is a sufficient condition for NCERT’s theorem, not a necessary condition for every definite integral to exist. A break inside the interval cannot simply be skipped. Used blindly, F(x) = −1/x would give ∫₋₁¹ dx/x² = −2, which is impossible for a positive integrand.

NCERT Example 25 (pages 269–270):

• ∫₂³ x² dx = 27/3 − 8/3 = 19/3.
• ∫₄⁹ √x/(30 − x^3/2)² dx: with t = 30 − x^3/2, F(x) = (2/3)/(30 − x^3/2), and F(9) − F(4) = (2/3)(1/3 − 1/22) = 19/99.
• ∫₁² x/((x + 1)(x + 2)) dx: partial fractions give F(x) = −log |x + 1| + 2 log |x + 2|, and the value is log (32/27).
• ∫₀^π/4 sin³ 2t cos 2t dt: with u = sin 2t, F(t) = (1/8) sin⁴ 2t, and the value is 1/8.

Solved examples

Common mistakes to avoid

Quick revision summary

• The area function A(x) = ∫_a^x f(t) dt has A′(x) = f(x) when f is continuous (first theorem).
• If F is any anti derivative of a continuous f on [a, b], then ∫_a^b f(x) dx = F(b) − F(a) (second theorem); no constant is needed.
• Check that the integrand is defined and continuous on [a, b], apart at most from an end point where F has a finite limit, before using F(b) − F(a).

Questions students ask

What is the difference between the first and second fundamental theorems of calculus?

The first fundamental theorem differentiates an integral and the second evaluates one: for a continuous f, the first says that the area function A(x), the integral of f(t) from a to x, has A′(x) = f(x), and the second that any anti derivative F gives the integral from a to b as F(b) − F(a). So the first guarantees that a function continuous on an interval has an anti derivative, A itself, even where no formula gives one, while the second turns a known anti derivative into a number. Numbering differs: NCERT’s Theorem 1 and Theorem 2 (pages 267–268) follow the usual order, but some books, such as the open textbook Active Calculus, call A′(x) = f(x) the second.

Why does the fundamental theorem of calculus work?

The fundamental theorem works because a thin strip of signed area under a continuous curve is nearly a rectangle (a sketch beyond the NCERT text, which states the theorems without proof, page 267): A(x + h) − A(x), the strip from x to x + h, is close to f(x) × h, so [A(x + h) − A(x)]/h tends to f(x) as h → 0, which is A′(x) = f(x). For f(t) = t³ and a = 0, A(x) = x⁴/4, and the strip from 2 to 2.01 is A(2.01) − A(2), about 0.0806, against f(2) × 0.01 = 0.08. Any other anti derivative F is A + C, and the C cancels in F(b) − F(a), which leaves A(b), as A(a) = 0: that is the second theorem.

What is the Leibniz rule for differentiating an integral, and what if x is inside the integrand?

The Leibniz rule (beyond the NCERT text) is this page’s rule for variable limits: the derivative of the integral of f(t) from h(x) to g(x) is f(g(x)) g′(x) minus f(h(x)) h′(x). It needs an integrand in t alone, so first take out any x that can be separated: ∫₀^x (x − t) cos t dt is x sin x minus ∫₀^x t cos t dt, and the product rule and the first theorem give its derivative as sin x plus x cos x minus x cos x, that is sin x. Integrating first agrees: the integral is 1 − cos x.

How is integration by parts used in a definite integral?

In a definite integral, integration by parts carries the limits along (reasoned from the product rule and the second theorem): the integral of uv′ from a to b is [uv]_a^b minus the integral of u′v from a to b, so the uv part is evaluated at once. For ∫₀¹ xe^2x dx, u = x and v = ½e^2x give ½e² minus ½∫₀¹ e^2x dx, which is ¼(e² − 1), so the value is (e² + 1)/4, about 2.097. Choose the first function as for an indefinite integral.

Who discovered the fundamental theorem of calculus?

The fundamental theorem of calculus is credited to Isaac Newton and Gottfried Wilhelm Leibniz, who worked independently between the 1660s and the 1680s; NCERT’s Historical Note to Chapter 8 (page 299) quotes the view that “The discovery that differentiation and integration are inverse operations belongs to Newton and Leibnitz”. Beyond the NCERT text, David Bressoud’s 2011 article in the American Mathematical Monthly recounts that geometric forms of the theorem also appeared in the seventeenth century, in James Gregory’s Geometriae pars universalis (1668) and in the Lectiones geometricae (1670) of Isaac Barrow, Newton’s predecessor as Lucasian professor at Cambridge.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Evaluate ∫₀^π/4 tan x dx (NCERT Exercise 7.8, Q7, page 270).

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   An anti derivative is log sec x, so the value is log √2 − log 1 = ½ log 2.

2. Q2. Evaluate ∫₂³ dx/(x² − 1) (NCERT Exercise 7.8, Q11, page 270).

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   An anti derivative is ½ log |(x − 1)/(x + 1)|. The value is ½[log (2/4) − log (1/3)] = ½ log (3/2).

3. Q3. Evaluate ∫₂³ x dx/(x² + 1) (NCERT Exercise 7.8, Q13, page 271).

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   An anti derivative is ½ log (x² + 1). The value is ½(log 10 − log 5) = ½ log 2.

4. Q4. Evaluate ∫₀^π/4 (2 sec² x + x³ + 2) dx (NCERT Exercise 7.8, Q17, page 271).

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   An anti derivative is 2 tan x + x⁴/4 + 2x. At π/4: 2 + π⁴/1024 + π/2; at 0 it is 0. The value is π⁴/1024 + π/2 + 2.