Fundamental Theorem of Calculus: JEE Main Guide

The Fundamental Theorem of Calculus (FTC) connects differentiation and integration — two operations that appear to be inverses of each other. In JEE Main, the FTC appears in questions that ask you to differentiate an integral with variable limits, apply Leibniz's rule, or evaluate definite integrals using antiderivatives. These questions are worth mastering because they arise in both the integration section and the differential equations section.

FTC Part 1: Differentiation of an Integral

If F(x) = ∫_a^x f(t) dt, then F'(x) = f(x). The derivative of a definite integral with a variable upper limit is simply the integrand evaluated at that upper limit (with the constant lower limit having no effect). This is the "antiderivative" theorem — it tells us that every continuous function has an antiderivative. Extension: if F(x) = ∫_a^{g(x)} f(t) dt, then by chain rule: F'(x) = f(g(x)) × g'(x). For a variable lower limit: if F(x) = ∫_{h(x)}^b f(t) dt = −∫_b^{h(x)} f(t) dt, then F'(x) = −f(h(x)) × h'(x).

General Leibniz rule: if F(x) = ∫_{h(x)}^{g(x)} f(t) dt, then F'(x) = f(g(x)) × g'(x) − f(h(x)) × h'(x). JEE question type: "find d/dx [∫_{x}^{x²} sin(t²) dt]." Apply Leibniz: sin(x²×x²) × 2x − sin(x²) × 1 = 2x sin(x⁴) − sin(x²). Take a free calculus mock. See our definite integration guide.

FTC Part 2: Evaluating Definite Integrals

∫_a^b f(x) dx = F(b) − F(a) where F is any antiderivative of f. This is Newton-Leibniz formula — the cornerstone of calculus. Key: you need an antiderivative (indefinite integral). Once you have F, the definite integral is a simple subtraction. JEE tests this through: (1) straightforward evaluation (find ∫₀^π sin x dx = [−cos x]₀^π = −cos π + cos 0 = 1 + 1 = 2); (2) choosing the right antiderivative — any F works since the constant cancels in F(b)−F(a); (3) improper integrals (one or both limits infinite, or integrand unbounded). For improper integrals: ∫_1^∞ 1/x² dx = [−1/x]_1^∞ = 0 − (−1) = 1 (convergent); ∫_1^∞ 1/x dx = [ln x]_1^∞ = ∞ (divergent).

Differentiation Under the Integral Sign

If F(α) = ∫_a^b f(x, α) dx (where the integrand depends on parameter α), then under regularity conditions: dF/dα = ∫_a^b (∂f/∂α) dx. This technique (Feynman's trick) evaluates difficult integrals by differentiating with respect to a parameter. JEE Main rarely tests this directly, but the technique appears in the form: "find ∫_0^1 x^n dx in terms of n" (answer: 1/(n+1)), and "show that ∫_0^π log(1+a cosθ)/cosθ dθ = π log((1+√(1-a²))/2)" — this latter is a JEE Advanced technique. At Main level, recognise that some difficult integrals become easy when a parameter is differentiated.

Applications: Area and Average Value

Area under curve from a to b: A = ∫_a^b |f(x)| dx (take absolute value to count area below x-axis as positive). Average value of f on [a,b]: f_avg = (1/(b-a)) ∫_a^b f(x) dx. Mean Value Theorem for Integrals: there exists c ∈ [a,b] such that f(c) = f_avg = (1/(b-a)) ∫_a^b f(x) dx. JEE uses MVT for integrals in bounding integral inequalities: if m ≤ f(x) ≤ M on [a,b], then m(b-a) ≤ ∫_a^b f(x) dx ≤ M(b-a). This sandwiching technique answers comparison questions without computing the integral. For the complete integration strategy see our integration techniques guide and our area under curves guide.