JEE Main Definite Integrals and Areas questions
102 Mathematics questions on Definite Integrals and Areas as asked in JEE Main, newest first. 58 come with a worked solution.
- Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x^2 - y^2 + 64x + 4y + 44 = 0. Then the area of the region above the parabola x^2 = y + 4, below the transverse axis T and on the right of…
2023 · with solution
- Let A = \(x, y) ∈ R^2 : y ≥ 0, 2x ≤ y ≤ √(4-(x-1)^2)\ and B = \(x, y) ∈ R × R : 0 ≤ y ≤ min \2x, √(4-(x-1)^2) \\. Then the ratio of the area of A to the area of B is
2023
- Let Δ be the area of the region \(x,y) ∈ R^2 : x^2 + y^2 ≤ 21, y^2 ≤ 4x, x ≥ 1 \. Then 1/2 (Δ - 21sin^-1frac2√(7) ) is equal to
2023
- Let [x] denote the greatest integer ≤ x. Consider the function f(x) = max\x^2, 1+[x]\. Then the value of the integral int_0^2 f(x) dx is
2023 · with solution
- If f(x) = dfrac2^2x2^2x+2, x ∈ R, then f (1/2023 ) + f (2/2023 ) + … + f (2022/2023 ) is equal to
2023 · with solution
- The integral 16int_1^2 dx/(x^3(x^2+2)^2) is equal to
2023
- int_frac3√(2)4^frac3√(3)4 frac48√(9-4x^2) dx is equal to
2023 · with solution
- The area enclosed by the curves y^2 + 4x = 4 and y - 2x = 2 is :
2023
- Let f(x) = x + a/(π^2 - 4)sin x + b/(π^2 - 4)cos x, x ∈ R be a function which satisfies f(x) = x + displaystyleint_0^π/2 sin(x+y) f(y) dy. Then (a+b) is equal to
2023
- The minimum value of the function f(x) = int_0^2 e^|x-t| dt is :
2023
- The value of 8/(π) int_0^(π)/2 dfrac(cos x)^2023(sin x)^2023 + (cos x)^2023 dx is __________
2023 · with solution
- If int_1^e [log_e x] dx = m/n log_e ( (e^2)/c ), where m and n are coprime natural numbers, then m^2 + n^2 - 5 is equal to
2023
- The value of 12int_0^3 |x^2 - 3x + 2 | dx is ___________
2023
- If the area of the region bounded by the curves y^2 - 2y = -x, x + y = 0 is A, then 8A is equal to
2023
- The value of int_-π/2^π/2 dx/([x] + [sin x] + 4), where [t] denotes the greatest integer less than or equal to t, is
2019
- A value of α such that int_α^α+1 dx/((x+α)(x+α+1)) = log_e (9/8 ) is
2019 · with solution
- lim_n → ∞ ( frac(n+1)^1/3n^4/3 + frac(n+2)^1/3n^4/3 + ..... + frac(2n)^1/3n^4/3 ) is equal to
2019 · with solution
- If int_x^x f(t) dt = x^2 + int_x^1 t^2 f(t) dt, then f (1/2 ) is
2019
- If the area (in sq units) bounded by the parabola y^2 = 4λ x and the line y = λ x, λ > 0, is 1/9, then λ is equal to
2019 · with solution
- If the area enclosed between the curves y = kx^2 and x = ky^2, (k > 0), is 1 square unit. Then, k is
2019
- lim_n → ∞ ( n/(n^2+1^2) + n/(n^2+2^2) + n/(n^2+3^2) + … + 1/5n ) is equal to
2019
- If int_0^π/2 (cot x)/(cot x + cosec x) dx = m(π + n), then m · n is equal to
2019
- If the area (in sq units) of the region \(x, y): y^2 ≤ 4x, x + y ≤ 1, x ≥ 0, y ≥ 0\ is a√(2) + b, then a - b is equal to
2019
- The integral int_π/6^π/3 sec^2/3 x csc^4/3 x dx is equal to
2019
- The area (in sq units) of the region bounded by the curves y = 2^x and y = |x+1| in the first quadrant is
2019
- The value of ∫limits_0^2π [sin 2x(1 + cos 3x)] dx, where [t] denotes the greatest integer function, is
2019
- The area (in sq units) of the region A = \(x, y): (y^2)/2 ≤ x ≤ y + 4 \ is
2019 · with solution
- The value of the integral ∫limits_0^1 xcot^-1(1-x^2+x^4)dx is
2019
- The area (in sq units) of the region A = \(x, y) : x^2 ≤ y ≤ x + 2\ is
2019 · with solution
- The value of int_0^π/2 (sin^3 x)/(sin x + cos x) dx is
2019
- Let S(α) = \(x, y) : y^2 ≤ x, 0 ≤ x ≤ α\ and A(α) is area of the region S(α). If for λ, 0 < λ < 4, A(λ) : A(4) = 2 : 5, then λ equals
2019 · with solution
- Let f(x) = ∫limits_0^x g(t)dt, where g is a non-zero even function. If f(x+5) = g(x), then ∫limits_0^x f(t)dt equals
2019
- If f(x) = (2 - xcos x)/(2 + xcos x) and g(x) = log_e x, (x > 0) then the value of the integral int_-π/4^π/4 g(f(x)) dx is
2019 · with solution
- The area (in sq units) of the region A = \(x, y) ∈ R × R mid 0 ≤ x ≤ 3, 0 ≤ y ≤ 4, y ≤ x^2 + 3x\ is
2019 · with solution
- Let f and g be continuous functions on [0, a] such that f(x) = f(a - x) and g(x) + g(a - x) = 4, then int_0^a f(x) g(x) dx is equal to
2019
- The area (in sq units) of the region bounded by the parabola, y = x^2 + 2 and the lines, y = x + 1, x = 0 and x = 3, is
2019
- If f(x) = ∫ (5x^8 + 7x^6)/((x^2 + 1 + 2x^7)^2) dx, \ (x ≥ 0), and f(0) = 0, then the value of f(1) is
2019 · with solution
- The integral int_1^e [ (x/e )^2x - (e/x )^x ]log_e x dx is equal to
2019
- The area (in sq units) in the first quadrant bounded by the parabola, y = x^2 + 1, the tangent to it at the point (2, 5) and the coordinate axes is
2019 · with solution
- The integral int_π/6^π/4 dx/(sin 2x(tan^5 x + cot^5 x)) equals
2019
- The area (in sq units) of the region bounded by the curve x^2 = 4y and the straight line x = 4y - 2 is
2019 · with solution
- The value of the integral int_-2^2 (sin^2 x)/( [x/(π) ] + 1/2) dx (where, [x] denotes the greatest integer less than or equal to x) is
2019 · with solution
- The area of the region A = \(x, y) 0 ≤ y ≤ |x| + 1 and -1 ≤ x ≤ 1\ in sq. units, is
2019
- The area (in sq units) bounded by the parabola y = x^2 - 1, the tangent at the point (2, 3) to it and the Y-axis is
2019
- If int_0^π/3 fractanθ√(2ksecθ) dθ = 1 - frac1√(2), (k > 0), then the value of k is
2019 · with solution
- The value of int_-π^π |cos x|^3 dx is
2019 · with solution
- Let g(x) = cos x^2, f(x) = √(x) and α, β (α < β) be the roots of the quadratic equation 18x^2 - 9π x + π^2 = 0. Then, the area (in sq units) bounded by the curve y = (gof)(x) and the lines x = α, x = β and y = 0, is
2018 · with solution
- The area (in sq units) of the region \(x, y) : x ≥ 0, x + y ≤ 3, x^2 ≤ 4y and y ≤ 1 + √(x)\ is
2017
- int_π/4^3π/4 dx/(1 + cos x) is equal to
2017 · with solution
- The area (in sq units) of the region \(x, y): y^2 ≥ 2x and x^2 + y^2 ≤ 4x, x ≥ 0, y ≥ 0\ is
2016 · with solution
- lim_n → ∞ ( frac(n+1)(n+2)·s 3nn^2n )^1/n is equal to
2016 · with solution
- Area of the region \(x, y) ∈ R^2 : y ≥ √(|x+3|), 5y ≤ (x+9) ≤ 15\ is equal to
2016
- The value of int_-π/2^π/2 (x^2 cos x)/(1 + e^x) dx is equal to
2016 · with solution
- The area (in sq units) of region described by \(x, y) y^2 ≤ 2x and y ≥ 4x - 1\ is
2015 · with solution
- The integral int_2^4 (log x^2)/(log x^2 + log(36 - 12x + x^2)) dx is equal to
2015 · with solution
- Let f:[0,2] → R be a function which is continuous on [0,2] and is differentiable on (0,2) with f(0)=1. Let F(x) = int_0^x^2 f(√(t)) dt, for x ∈ [0,2]. If F'(x) = f'(x), forall x ∈ (0,2), then F(2) equals
2014
- The common tangents to the circle x^2 + y^2 = 2 and the parabola y^2 = 8x touch the circle at the points P, Q and the parabola at the points R, S. Then, the area (in sq units) of the quadrilateral PQRS is
2014
- The area (in sq units) of the region described by A = \(x, y) : x^2 + y^2 ≤ 1 and y^2 ≤ 1 - x\ is
2014
- The area enclosed by the curves y = sin x + cos x and y = |cos x - sin x| over the interval [0, (π)/2 ] is
2014 · with solution
- The integral int_π/4^π/2 (2cosec x)^17 dx is equal to
2014 · with solution
- The integral ∫limits_0^(π)/2 √(1 + 4sin^2x/2 - 4sinx/2) dx is equal to
2014
- The intercepts on X-axis made by tangents to the curve, y = int_0^x |t| dt, x ∈ R, which are parallel to the line y = 2x, are equal to
2013
- Let f : [1/2, 1] → R (the set of all real numbers) be a positive, non-constant and differentiable function such that f'(x) < 2f(x) and f(1/2) = 1. Then, the value of int_1/2^1 f(x) dx lies in the interval
2013 · with solution
- The area (in sq units) bounded by the curves y = √(x), 2y - x + 3 = 0, X-axis and lying in the first quadrant, is
2013 · with solution
- The value of the integral int_-π/2^π/2 (x^2 + log(π - x)/(π + x) ) cos x dx is
2012 · with solution
- Let f: [-1, 2] → [0, ∞) be a continuous function such that f(x) = f(1-x), forall x ∈ [-1, 2]. If R_1 = int_-1^2 x f(x) dx and R_2 are the area of the region bounded by y = f(x), x = -1, x = 2 and the x-axis. Then,
2011
- If the straight line x = b divide the area enclosed by y = (1-x)^2, y = 0 and x = 0 into two parts R_1(0 ≤ x ≤ b) and R_2(b ≤ x ≤ 1) such that R_1 - R_2 = 1/4. Then, b equals
2011
- The value of int_√(log 2)^√(log 3) (x sin x^2)/(sin x^2 + sin(log 6 - x^2)) dx is \hspace{1em}
2011 · with solution
- The area of the region between the curves y = √((1+sin x)/(cos x)) and y = √((1-sin x)/(cos x)) and bounded by the lines x = 0 and x = (π)/4 is
2008 · with solution
- The area of the equilateral triangle, in which three coins of radius 1 cm are placed, as shown in the figure, is [figure]
2005 · with solution
- If int_sin x^1 t^2 f(t) dt = 1 - sin x, forall x ∈ (0, π/2), then f (frac1√(3) ) is
2005 · with solution
- The value of int_-1^0 [x^3 + 3x^2 + 3x + 3 + (x+1)cos(x+1)] dx is
2005
- The area bounded by the curves y = (x-1)^2, y = (x+1)^2 and y = 1/4 is
2005 · with solution
- If f(x) is differentiable and int_0^t^2 x f(x) dx = 2/5t^5, then f (4/25 ) equals
2004 · with solution
- The value of the integral int_0^1 √((1-x)/(1+x)) dx is
2004 · with solution
- The area enclosed between the curves y = ax^2 and x = ay^2 (a > 0) is 1 sq unit. Then, the value of a is
2004
- The area of the quadrilateral formed by the tangents at the end points of latusrectum to the ellipse (x^2)/9 + (y^2)/5 = 1, is
2003 · with solution
- If f(x) = int_x^2^x^2+1 e^-t^2 dt, then f(x) increases in
2003 · with solution
- If I(m,n) = int_0^1 t^m(1+t)^n dt, then the expression for I(m,n) in terms of I(m+1, n-1) is
2003 · with solution
- Let T > 0 be a fixed real number. Suppose, f is a continuous function such that for all x ∈ R, f(x + T) = f(x). If I = int_0^T f(x) dx, then the value of int_3^3+3T f(2x) dx is
2002 · with solution
- The area (in sq units) bounded by the curves y = |x| - 1 and y = -|x| + 1 is
2002 · with solution
- Let f(x) = int_1^x √(2 - t^2) dt. Then, the real roots of the equation x^2 - f'(x) = 0 are
2002 · with solution
- The integral int_-1/2^1/2 [ [x] + log ((1+x)/(1-x) ) ] dx equals
2002 · with solution
- The triangle formed by the tangent to the curve f(x) = x^2 + bx - b at the point (1,1) and the coordinate axes, lies in the first quadrant. If its area is 2 sq units, then the value of b is
2001
- Let f:(0,∞) → R and F(x) = int_0^x f(t) dt. If F(x^2) = x^2(1+x), then f(4) equals
2001 · with solution
- The value of int_-π^π (cos^2 x)/(1 + a^x) dx, a > 0, is
2001
- If f(x) = begincases e^cos x sin x & , for |x| ≤ 2 \ 2 & , otherwise endcases then int_-2^3 f(x) dx is equal to
2000 · with solution
- The value of the integral int_e^-1^e^2 | (log_e x)/x | dx is
2000 · with solution
- If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the integral int_π/2^3π/2 [2sin x] dx is
1999
- int_π/4^3π/4 dx/(1 + cos x) is equal to
1999 · with solution
- If int_0^x f(t) dt = x + int_x^1 t f(t) dt, then the value of f(1) is
1998 · with solution
- If g(x) = int_0^x cos^4 t dt, then g(x + π) equals
1997 · with solution
- Let f be a positive function. If I_1 = int_1-k^k x f [x(1-x) ] dx and I_2 = int_1-k^k f [x(1-x) ] dx, where 2k - 1 > 0. Then, I_1/I_2 is
1997 · with solution
- The value of int_0^2π [2sin x] dx, where [·] represents the greatest integral functions, is
1995
- If f(x) = Asin ((π x)/2 ) + B, f' (1/2 ) = √(2) and int_0^1 f(x) dx = 2A/(π), then constants A and B are
1995 · with solution
- The value of int_0^π/2 dx/(1+tan^3 x) is
1993 · with solution
- Let f: R → R be a differentiable function and f(1) = 4. Then, the value of lim_x → 1 int_4^f(x) 2t/(x-1) dt is
1990 · with solution
- Let f: R → R and g: R → R be continuous functions. Then, the value of the integral int_-π/2^π/2 [f(x) + f(-x)][g(x) - g(-x)] dx
1990
- For any integer n, the integral int_0^π e^cos^2 x cos^3(2n+1)x dx has the value
1985
- The value of the integral int_0^π/2 dfrac√(cot x)√(cot x) + √(tan x) dx is
1983 · with solution
- The value of the definite integral int_0^1 (1 + e^-x^2) dx is
1981 · with solution
- The slope of tangent to a curve y = f(x) at [x, f(x)] is 2x + 1. If the curve passes through the point (1, 2), then the area bounded by the curve, the X-axis and the line x = 1 is
with solution
Practise Definite Integrals and Areas the way it is asked
A timed set of real JEE Main Definite Integrals and Areas questions, graded as the exam grades. No sign-up.
Start a Definite Integrals and Areas test