Exams › JEE Main › Maths › Integrals
324 questions with worked solutions.
Answer: (1/a)(a+b)²
The expression for y can be minimized using calculus or the method of Lagrange multipliers, leading to the conclusion that the least value occurs at the critical points derived from the function's derivative. The correct option, (1/a)(a+b)², represents the minimum value of y when evaluated at the optimal point within the given interval.
Q2. Let f(x)=∫ from 2x to sin x of cos(t³) dt. Then the value of f''(x) is
Answer: cos(sin³ x) cos x − 2 cos(8x³)
With f(x)=int_{2x}^{sin x} cos(t^3) dt, the Leibniz rule gives f'(x)=cos(sin^3 x)*cos x - cos((2x)^3)*2 = cos(sin^3 x)cos x - 2cos(8x^3), which is option 1.
Q3. Evaluate the integral ∫_(√(ln 3))^(√(ln 2)) (xsin(x²))/(sin(x²)+sin(ln 6-x²)) dx.
Answer: (1)/(4)ln ((3)/(2))
Let u=x^2 (du=2x dx). The integral becomes (1/2)*Integral sin(u)/(sin(u)+sin(ln6-u)) du over [ln2,ln3]. Since ln2+ln3=ln6, adding the King-reflected integral gives (1/2)*(ln3-ln2)=(1/2)ln(3/2), so each half is (1/4)ln(3/2).
Q4. Evaluate the integral ∫ (e^x (1+xⁿ⁻¹-xⁿ))/((1-xⁿ)√(1-x²ⁿ)) dx.
Answer: (e^x√(1-x²ⁿ))/(1-xⁿ)+C
The integrand has the structure e^x[f(x)+f'(x)] with f(x)=sqrt(1-x^(2n))/(1-x^n). Therefore the integral is e^x*sqrt(1-x^(2n))/(1-x^n)+C.
Answer: K < 2/3 and J < 2
K=integral_0^1 (sin x)/sqrt(x) dx ~ 0.62 < 2/3, and J=integral_0^1 (cos x)/sqrt(x) dx ~ 1.81 < 2. Hence K < 2/3 and J < 2.
Answer: ∛(logₑ(3y-2))
The correct option is ∛(logₑ(3y-2)) because it correctly represents the inverse relationship derived from the original integral expression, ensuring that when y is substituted back, it satisfies the original curve's conditions.
Q7. Evaluate the definite integral ∫₀^(π/2) |sin x - cos x| dx.
Answer: 2(√(2)-1)
For 0..pi/4, cos>sin; for pi/4..pi/2, sin>cos. Integral = Integral(cos-sin) from 0 to pi/4 + Integral(sin-cos) from pi/4 to pi/2 = (sqrt2-1)+(sqrt2-1) = 2(sqrt2-1).
Answer: 1/2
Substitute x->1-x in I1; the argument x(1-x) is unchanged and the limits swap back, giving I1 = integral of (1-x) f = I2 - I1. Hence 2I1 = I2, so I1:I2 = 1/2.
Q9. Evaluate the integral ∫ e^(tan⁻¹x) (1+x+x²)/(1+x²) dx.
Answer: x e^(tan⁻¹x) + C
The integral evaluates to the product of the variable x and the exponential function of the inverse tangent, indicating that the function's growth is influenced by both x and the exponential term, which is consistent with the result of option A.
Answer: ∫₀^a f(x) dx
Let I=int_0^a f(x)g(x)dx. Substituting x->a-x and using f(a-x)=f(x), g(a-x)=2-g(x): I=int_0^a f(x)(2-g(x))dx = 2*int_0^a f dx - I. So 2I=2*int_0^a f dx, giving I=int_0^a f(x)dx.
Answer: 6, 34
Note (x-5)(x-4)=x^2-9x+20=Q. d/dx[A sqrt(Q)] = A(2x-9)/(2 sqrt Q) and d/dx[B ln|x-9/2+sqrt Q|] = B/sqrt Q. So 6x+7 = A(2x-9)/2 + B. Matching: A=6, then -9(6)/2 + B = 7 => B=34. Thus A=6, B=34.
Q12. Evaluate the definite integral ∫ₐ^b (|x|)/(x) dx, where a<b.
Answer: |b|-|a|
The integral ( rac{|x|}{x}) evaluates to 1 for positive values of x and -1 for negative values of x, leading to the result of (|b| - |a|) when integrating over the interval from a to b, accounting for the sign changes.
Q13. Let f:(0,∞)→ R and define F(x)=∫₀^x f(t) dt. If F(x²)=x²(1+x), then the value of f(4) is
Answer: 4
From F(x^2)=x^2+x^3, differentiate: F'(x^2)*2x = 2x+3x^2, and F'=f so f(x^2)=1+3x/2. Putting x^2=4 (x=2) gives f(4)=1+3=4.
Q14. Evaluate the definite integral ∫₀¹⁰⁰⁰ e^(x-x) dx.
Answer: 1000(e-1)
The exponent is the fractional part x-[x], periodic with period 1. Over each unit interval integral of e^(x-[x]) equals integral_0^1 e^t dt = e-1. Over [0,1000] there are 1000 such intervals, so the value is 1000(e-1).
Answer: Statement 1 is true, Statement 2 is true, and Statement 2 correctly explains Statement 1
From int g dx=g we get g'=g. Then d/dx[g f]=g'f+g f'=g(f+f'), proving Statement 2. Writing h=f-f', note h+h'=f-f'', so int g(f-f'')dx=int g(h+h')dx=g h+C=g(f-f')+C, proving Statement 1 directly from Statement 2. Hence both are true and Statement 2 correctly explains Statement 1.
Answer: 1 square unit
Area = |integral_0^1 ln x dx| = |[x ln x - x]_0^1| = |0 - (-1)| = 1 square unit.
Q17. Find the area enclosed by the curves y = x e^x, y = x e^(-x), and the vertical line x = 1.
Answer: 2/e
On [0,1], xe^x >= xe^{-x}. Area = integral_0^1 (xe^x - xe^{-x}) dx. With integral xe^x dx = (x-1)e^x and integral xe^{-x} dx = -(x+1)e^{-x}, the first part = 1 and the second = 1 - 2/e. Area = 1 - (1 - 2/e) = 2/e.
Q18. Find the area enclosed by the graph of y = ln(x) and the straight lines y = 0, y = ln(3), and x = 0.
Answer: 2
The region is bounded left by x=0 and right by x=e^y, for y from 0 to ln3. Area = integral_0^(ln3) e^y dy = e^(ln3)-1 = 3-1 = 2.
Q19. Find the indefinite integral of 8/((x + 2)(x² + 4)) with respect to x.
Answer: ln|x + 2| − 1/2 ln(x² + 4) + tan^(-1)(x/2) + C
The correct option accurately combines the natural logarithm of the linear term and the inverse tangent function, which arises from the integration of the rational function involving a quadratic term, ensuring the proper application of integration techniques such as partial fraction decomposition.
Q20. Define aₙ = ∫₀^(π/2) l(1-sinⁿ t r)sin 2t dt. Then the value of lim_(n→∞) (1)/(n)∑_(i=1)ⁿ a_i is
Answer: 1
a_n = int_0^(pi/2)(1-sin^n t)sin2t dt = 1 - 2/(n+2) using int sin^n t sin2t dt = 2/(n+2). Since a_n -> 1, the average (1/n)sum a_i also -> 1.
Q21. Evaluate the integral I = ∫₀¹ x(1−x)ⁿ dx.
Answer: 1/(n+1) − 1/(n+2)
The correct option is derived from the integration of the function x(1−x)ⁿ, which can be solved using integration by parts or the beta function. The result simplifies to 1/(n+1) - 1/(n+2), reflecting the contributions from both the polynomial and the limits of integration.
Q22. Evaluate the limit as n approaches infinity of the sum ∑_(r=1)ⁿ (1)/(n) e^(r/n).
Answer: e - 1
The limit can be interpreted as a Riemann sum for the function e^x over the interval [0, 1]. As n approaches infinity, the sum converges to the integral of e^x from 0 to 1, which evaluates to e - 1.
Q23. Evaluate the definite integral ∫₋₂³ |1-x²| dx.
Answer: (28)/(3)
The integral evaluates the area under the curve of the function |1 - x²| from -2 to 3. The function changes behavior at the points where 1 - x² = 0, leading to two distinct regions to integrate, resulting in a total area of (28)/(3).
Q24. Evaluate the integral I = ∫ from 0 to π/2 of ((sin x + cos x)² / √(1 + sin 2x)) dx.
Answer: 2
The integral simplifies to a form that can be evaluated using trigonometric identities and symmetry properties of sine and cosine functions, ultimately leading to the result of 2.
Q25. If ∫₀^(π) x f(sin x) dx = A ∫₀^(π/2) f(sin x) dx, then the value of A is
Answer: π
The integral on the left can be simplified using the symmetry of the sine function and the properties of definite integrals, leading to the conclusion that the factor A must equal C0 to balance the two sides.
Q26. Evaluate the definite integral ∫_(1/3)⁶ (√(x))/(√(9 - x + √(x))) dx.
Answer: 3/2
The integral evaluates to 3/2 because the substitution and simplification of the integrand lead to a manageable expression that, when integrated over the specified limits, yields this result.
Q27. ∫₀^π x f(sin x) dx is equal to
Answer: π ∫₀^(π/2) f(cos x) dx
By x -> pi - x, ∫0^pi x f(sin x)dx = (pi/2)∫0^pi f(sin x)dx = pi∫0^(pi/2) f(sin x)dx. Since ∫0^(pi/2) f(sin x)dx = ∫0^(pi/2) f(cos x)dx, this equals pi∫0^(pi/2) f(cos x)dx.
Q28. ∫_(-3π/2)^(-π/2) [(x + π)³ + cos²(x + 3π)] dx is equal to
Answer: π/2
Let t=x+pi. Limits become -pi/2 to pi/2 and cos^2(x+3pi)=cos^2(t). Integral = int_{-pi/2}^{pi/2}(t^3+cos^2 t)dt. The t^3 term is odd -> 0, and int_{-pi/2}^{pi/2}cos^2 t dt = pi/2. Value = pi/2.
Q29. The value of ∫₁^a [x] f'(x) dx, a > 1 where [x] denotes the greatest integer not exceeding x is
Answer: [a]f(a) - f(1) + f(2) +..........f([a])
Split the integral over [1,2],[2,3],...,[[a],a] using [x]=k on (k,k+1): sum_k k(f(k+1)-f(k)) + [a](f(a)-f([a])). Collecting terms gives [a]f(a) - (f(1)+f(2)+...+f([a])).
Q30. ∫ dx / (cos x + √3 sin x) equals
Answer: (1/2) log tan(x/2 + π/12) + C
The integral simplifies to a form that involves the tangent function, and the presence of the factor of 1/2 indicates that the argument of the logarithm must be adjusted accordingly, leading to the correct option involving tan(x/2 + π/12).
Q31. Let F(x) = f(x) + f(1/x), where f(x) = ∫₁^x (log t)/(1 + t) dt. Then F(e) equals
Answer: 1/2
To find F(e), we evaluate f(e) and f(1/e). The integral f(x) is symmetric around x=1, leading to the result that f(e) + f(1/e) equals 1/2, thus making F(e) equal to 1/2.
Q32. The solution for x of the equation ∫_(√2)^x dt / √(t² - 1) = π/2 is
Answer: None of these
The integral from √2 to x of dt / √(t² - 1) evaluates to arcsinh(x), and setting this equal to π/2 leads to x = sinh(π/2), which is not among the provided options.
Q33. Let I = ∫₀¹ (sin x)/√x dx and J = ∫₀¹ (cos x)/√x dx. Then which one of the following is true?
Answer: I < 2/3 and J < 2
Since sin x < x on (0,1), sin x/sqrt(x) < sqrt(x), so I < integral of sqrt(x) = 2/3. Since cos x < 1, cos x/sqrt(x) < 1/sqrt(x), so J < integral of 1/sqrt(x) = 2. Hence I < 2/3 and J < 2.
Q34. The value of √2 ∫ [sin x dx] / [sin(x - π/4)] is
Answer: x + log|sin(x - π/4)| + c
The integrand becomes 1+(sinx+cosx)/(sinx-cosx). Integrating with u=sinx-cosx gives x+log|sinx-cosx|+c = x+log|sin(x-pi/4)|+c (constant absorbed).
Q35. ∫₀^π [cot x] dx, where [.] denotes the greatest integer function, is equal to:
Answer: -π/2
The integral of cot(x) over the interval from 0 to π results in a value of -π/2, as the cotangent function approaches infinity at the endpoints and has a negative area under the curve in this range.
Q36. Evaluate the definite integral ∫₀¹ (8log(1+x))/(1+x²) dx.
Answer: πlog 2
The integral evaluates to ( rac{ ext{numerical constant}}{1+x²}) which, when integrated, yields a result involving ( rac{ ext{constant}}{8}) and the logarithm of 2, ultimately leading to the correct answer of ( rac{ ext{constant}}{8}) multiplied by ( ext{constant}) from the integration process.
Q37. Evaluate the integral ∫ (1+x-(1)/(x))e^(x+1/x) dx.
Answer: x e^(x+1/x)+c
The integral can be solved using integration by parts, where the derivative of the exponent, which is a function of both x and 1/x, simplifies the expression. This leads to the result of the integral being x multiplied by the exponential function, plus a constant of integration.
Q38. Evaluate the definite integral ∫₀^(π) √(1+4sin² ((x)/(2))-4sin ((x)/(2))) dx:
Answer: 4√(3)-4-(π)/(3)
The correct option is derived from evaluating the integral by simplifying the integrand using trigonometric identities and then applying integration techniques, which ultimately leads to the result that includes both a constant term and a term involving pi, confirming the complexity of the integral.
Q39. Evaluate the integral ∫ (dx)/(x² (x⁴+1)^(3/4)):
Answer: - ((x⁴+1)/(x⁴))^(1/4)+c
Factor x^4: integrand = 1/(x^5 (1+x^-4)^(3/4)). Let u=1+x^-4, du=-4x^-5 dx, giving -(1/4) integral u^(-3/4) du = -u^(1/4)+c = -((x^4+1)/x^4)^(1/4)+c.
Q40. Evaluate the indefinite integral ∫ (2x¹²+5x⁹)/((x⁵+x³+1)³) dx. It is equal to:
Answer: (x¹⁰)/(2(x⁵+x³+1)²)+C
Dividing numerator and denominator by x^15 and letting t=(x^5+x^3+1)/x^5 gives dt=-(2x^-3+5x^-6)dx, so the integral = ∫ -dt/t^3 = 1/(2t^2) = x^10/(2(x^5+x^3+1)^2) + C.
Q41. Evaluate the definite integral ∫_(π/4)^(3π/4) (dx)/(1+cos x). What is its value?
Answer: 2
1+cos x = 2cos^2(x/2), so the integrand is (1/2)sec^2(x/2) and the antiderivative is tan(x/2). Evaluating from pi/4 to 3pi/4 gives tan(3pi/8)-tan(pi/8) = (1+sqrt(2))-(sqrt(2)-1) = 2.
Answer: (1/5, 0)
The integral of tanⁿ x can be expressed in terms of lower powers of tan x and x, and through integration by parts or reduction formulas, it can be shown that the combination I₄ + I₆ results in a term proportional to tan⁵ x with a coefficient of 1/5, while the x⁵ term does not appear, leading to the ordered pair (1/5, 0).
Answer: −1 / [3(1 + tan³ x)] + C
With denominator (sin^3 x+cos^3 x)^2, d/dx[-1/(3(1+tan^3 x))] = tan^2 x sec^2 x/(1+tan^3 x)^2 = sin^2 x cos^2 x/(sin^3 x+cos^3 x)^2, which is the integrand. Hence the integral equals -1/[3(1+tan^3 x)] + C.
Q44. Evaluate the definite integral ∫_(-π/2)^(π/2) (sin² x)/(1+2^x) dx.
Answer: π/4
The integral evaluates to frac{ ext{Area under the curve of } rac{ ext{sin}² x}{1+2^x}}{ ext{symmetry and properties of definite integrals}} over the interval from - rac{ ext{pi}}{2} to rac{ ext{pi}}{2}, resulting in frac{ ext{pi}}{4}.
Q45. Evaluate the definite integral ∫₀^(π) |cos x|³ dx.
Answer: 4/3
By symmetry of |cos x|, integral_0^pi |cos x|^3 dx = 2 integral_0^(pi/2) cos^3 x dx = 2*(2/3) = 4/3.
Q46. The integral ∫ sec^(2/3) x cosec^(4/3) x dx is equal to:
Answer: -3 tan^(-1/3) x + C
The integral evaluates to -3 tan^(-1/3) x + C because it involves the integration of sec^(2/3) x and cosec^(4/3) x, which can be simplified using trigonometric identities and substitution methods that lead to this specific antiderivative.
Q47. The value of ∫[0 to π/2] (sin³ x)/(sin x + cos x) dx is:
Answer: (π - 1)/4
The integral evaluates to (π - 1)/4 due to the symmetry and periodic properties of the sine and cosine functions, which allows for simplification and substitution that leads to this specific result.
Answer: ∫[a−1 to b−1] f(x + 1) dx
The given functional equation indicates that the function f is symmetric about the point (a + b + 1)/2, which allows us to transform the integral by shifting the variable. This transformation leads to the conclusion that the integral of f(x + 1) over the interval [a−1, b−1] matches the original integral's structure, confirming that option C is the correct choice.
Q49. Find the area bounded by the graph of y = ln(x + e) and the two coordinate axes.
Answer: 1
The area under the curve y = ln(x + e) from x = 0 to x = 1 is calculated using integration, resulting in an area of 1. This is because the natural logarithm function increases at a decreasing rate, and the specific limits yield this exact area.
Q50. The integral ∫ (1 + x - 1/x) e^(x + 1/x) dx is equal to
Answer: (3) x e^(x + 1/x) + c
The correct option is right because when differentiating the expression x e^(x + 1/x), the product rule and the chain rule yield the integrand (1 + x - 1/x) e^(x + 1/x), confirming that it is indeed the antiderivative.