Integral Calculus: JEE Advanced PYQ, 2007–2026

Answer: (C)

∫ dx/(1 + x²) = tan⁻¹x. Between 0 and √3 this is tan⁻¹√3 − tan⁻¹0 = π/3.

Answer: (C)

The curve meets y = e² at x = 2, and at x = 0 it meets the y-axis at y = 1. For 0 ≤ x ≤ 2 the line y = e² is above the curve, so Area = ∫₀²(e² − eˣ)dx = [e²x − eˣ]₀² = (2e² − e²) − (0 − 1) = e² + 1. Check by horizontal strips: ∫₁^{e²} ln y dy = [y ln y − y]₁^{e²} = e² + 1.

Answer: (A)

The factors x, ln x and ln(ln x) in the denominator suggest a nested logarithmic substitution. Put w = ln(ln x), so dw = dx/(x ln x). The integral becomes ∫dw/w. At x = e^e, ln x = e, so w = ln e = 1. At x = e^(e²), ln x = e², so w = ln(e²) = 2. Hence the value is [ln w]₁² = ln 2 − ln 1 = ln 2.

Answer: (A)

Choose u = x (algebraic) and dv = sec²x dx, so v = tan x. Then ∫x sec²x dx = x tan x − ∫tan x dx. Now ∫tan x dx = −ln|cos x| + C, so the result is x tan x + ln|cos x| + C. Differentiating gives tan x + x sec²x − tan x = x sec²x, which confirms it.

Answer: (A)

Divide numerator and denominator by x²: the integrand becomes (1 + 1/x²)/(x² + 1/x²). Put t = x − 1/x, so dt = (1 + 1/x²)dx. Also x² + 1/x² = t² + 2. The integral becomes ∫dt/(t² + 2) = (1/√2) tan⁻¹(t/√2) + C. Substituting back, t/√2 = (x² − 1)/(√2 x).

Answer: (A)

Let u = √x. Then du = dx/(2√x), so dx/√x = 2 du. The integral becomes 2∫e^u du = 2e^u + C = 2e^(√x) + C. Differentiating back gives 2·e^(√x)·1/(2√x) = e^(√x)/√x, which confirms the result.

Answer: (A)

f is strictly increasing with f(0) = 0 and f(1) = 2. So g(0) = 0 and g(2) = 1, and the region is the area under y = g(x) for 0 ≤ x ≤ 2. Reflect it in y = x. The region under the graph of g, read as x as a function of y, lies between the curve y = f(x) and the y-axis. Equivalently, the area under g on [0, 2] plus the area under f on [0, 1] equals the rectangle 1 × 2 = 2. So ∫₀² g(x) dx = 2 − ∫₀¹ (x³ + x) dx = 2 − (1/4 + 1/2) = 5/4. No formula for g is needed.

Answer: (A)

Differentiate both sides with respect to x using the first fundamental theorem and the chain rule: f(x²+x³)·(2x+3x²) = 1. We need the argument to equal 2, so solve x³ + x² = 2, which gives x = 1. At x = 1 the equation becomes f(2)·(2+3) = 1, so f(2) = 1/5.

Answer: (A)

Let 3 sin x + 2 cos x = A(3 cos x + 2 sin x) + B(−3 sin x + 2 cos x). Comparing cos x: 3A + 2B = 2. Comparing sin x: 2A − 3B = 3. Solving gives A = 12/13 and B = −5/13. Then the integral is Ax + B ln|3 cos x + 2 sin x| + C, so a = 12/13 and b = −5/13. Hence 13(a − b) = 12 + 5 = 17.

Answer: (A)

By the Leibniz rule, d/dx ∫_{u(x)}^{v(x)} g(t)dt = g(v)·v′ − g(u)·u′. Here u = x, v = x², g(t) = ln t. So F′(x) = ln(x²)·2x − ln x·1. At x = e, ln(e²) = 2, so F′(e) = 2·2e − 1 = 4e − 1.