68 free MCQs on Integrals with worked answers and explanations. Indefinite and definite integrals, areas under curves
Take the timed Integrals chapterwise test →Below are 68 practice questions on Integrals, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Integrals notes.
The exact area under the curve from a to b (shaded) is approximated by a few rectangles, the Riemann sum idea behind the definite integral.
Q1.
Integration is the reverse of:
Answer: B. Differentiation
Why: Integration is the inverse operation of differentiation. If d/dx(F(x)) = f(x), then integral of f(x) = F(x) + C.
Q2.
Integral of xⁿ dx = (n not equal to -1)
Answer: C. xⁿ⁺¹/(n+1) + C
Why: Power rule for integration: integral of xⁿ = xⁿ⁺¹/(n+1) + C.
Q3.
Integral of cos x dx =
Answer: A. sin x + C
Why: Integral of cos x = sin x + C (since d/dx(sin x) = cos x).
Q4.
Integral of sin x dx =
Answer: B. -cos x + C
Why: Integral of sin x = -cos x + C (since d/dx(-cos x) = sin x).
Q5.
Integral of eˣ dx =
Answer: A. eˣ + C
Why: Integral of eˣ = eˣ + C (exponential function is its own integral).
Q6.
Integral of 1/x dx =
Answer: C. ln|x| + C
Why: Integral of 1/x = ln|x| + C. This is the exception to the power rule (n = -1).
Q7.
What does C represent in integration?
Answer: A. A constant of integration (arbitrary constant)
Why: C is the constant of integration. Since d/dx(C) = 0, any constant is lost when differentiating, so we add it back when integrating.
Q8.
Evaluate: integral of 3x² dx
Answer: B. x³ + C
Why: Integral of 3x² = 3 × x³/3 = x³ + C.
Q9.
Integral of 2x dx =
Answer: B. x² + C
Why: Integral of 2x = 2 × x²/2 = x² + C.
Q10.
Evaluate definite integral: integral from 0 to 1 of x dx
Answer: B. 1/2
Why: Integral of x = x²/2 + C. Evaluate from 0 to 1: [1²/2] - [0²/2] = 1/2 - 0 = 1/2.
Q11.
Integral of sec²x dx =
Answer: A. tan x + C
Why: Integral of sec²x = tan x + C (since d/dx(tan x) = sec²x).
Q12.
Which symbol is used for integration?
Answer: C. ∫
Why: The integral symbol ∫ (elongated S for sum) represents integration.
Q13.
Integral of a constant k dx =
Answer: B. kx + C
Why: Integral of k (constant) = kx + C.
Q14.
Definite integral represents:
Answer: B. Area under the curve between two points
Why: A definite integral calculates the exact area under a curve between two x-values.
Q15.
Evaluate: integral from 0 to 3 of 2 dx
Answer: C. 6
Why: Integral of 2 = 2x. Evaluate from 0 to 3: 2(3) - 2(0) = 6 - 0 = 6.
Q16.
Integral of x⁰ dx =
Answer: B. x + C
Why: x⁰ = 1. Integral of 1 = x + C.
Q17.
Integral of √x dx =
Answer: C. (2/3)x<sup>3/2</sup> + C
Why: sqrt(x) = x<sup>1/2</sup>. Integral = x<sup>3/2</sup>/(3/2) = (2/3)x<sup>3/2</sup> + C.
Q18.
Integral of (3x² + 2x + 1) dx =
Answer: A. x³ + x² + x + C
Why: Integrate each term: x³ + x² + x + C.
Q19.
Evaluate: integral from 1 to 2 of x² dx
Answer: A. 7/3
Why: Integral of x² = x³/3. Evaluate: [8/3] - [1/3] = 7/3.
Q20.
The upper and lower limits in a definite integral are written:
Answer: B. At the top and bottom of the ∫ symbol
Why: In definite integral, limits appear at top (upper) and bottom (lower) of the integral symbol ∫.
Q21.
Integrate: (1+x)² dx
Answer: A. x + x² + x³/3 + C
Why: Expand: 1 + 2x + x². Integrate: x + x² + x³/3 + C.
Q22.
Integrate: sin²x dx using identity
Answer: A. x/2 - sin(2x)/4 + C
Why: sin²x = (1-cos2x)/2. Integrate: x/2 - sin(2x)/4 + C.
Q23.
Integrate: e<sup>3x</sup> dx
Answer: B. e<sup>3x</sup>/3 + C
Why: Integral of e<sup>ax</sup> = e<sup>ax</sup>/a + C. Here a=3: e<sup>3x</sup>/3 + C.
Q24.
Integral of 1/(x²+a²) dx =
Answer: A. (1/a)tan⁻¹(x/a) + C
Why: Standard result: integral of 1/(x²+a²) dx = (1/a)tan⁻¹(x/a) + C.
Q25.
Evaluate: integral from 0 to pi of sin x dx
Answer: C. 2
Why: [-cos x] from 0 to pi = -cos(pi) - (-cos 0) = -(-1) + 1 = 2.
Q26.
Integrate using substitution: integral of 2x(x²+1)⁵ dx
Answer: A. (x²+1)⁶/6 + C
Why: Let u = x²+1, du = 2x dx. Integral = u⁵ du = u⁶/6 = (x²+1)⁶/6 + C.
Q27.
Integrate by parts: integral of x eˣ dx
Answer: B. eˣ(x-1) + C
Why: By parts: u=x, dv=eˣdx. uv - integral(v du) = x eˣ - integral(eˣ dx) = x eˣ - eˣ + C = eˣ(x-1) + C.
Q28.
Integral of 1/√(1-x²) dx =
Answer: A. sin⁻¹x + C
Why: Standard integral: integral of 1/√(1-x²) = sin⁻¹x + C.
Q29.
Evaluate: integral from 0 to 1 of eˣ dx
Answer: C. e-1
Why: [eˣ] from 0 to 1 = e¹ - e⁰ = e - 1.
Q30.
Integral of tan x dx =
Answer: D. B and C are equivalent
Why: tan x = sin x / cos x. Integral = -ln|cos x| + C = ln|sec x| + C. Both B and C are correct.
Q31.
The area bounded by y = x², x-axis, x = 0 to x = 3:
Answer: C. 9
Why: Area = integral from 0 to 3 of x² dx = [x³/3] = 27/3 - 0 = 9.
Q32.
Integrate: (3x² + 2x) / (x³ + x²) dx
Answer: A. ln|x³ + x²| + C
Why: Numerator is derivative of denominator: d/dx(x³+x²) = 3x²+2x. So integral = ln|x³+x²| + C.
Q33.
Evaluate: integral from 0 to pi/2 of cos x dx
Answer: B. 1
Why: [sin x] from 0 to pi/2 = sin(pi/2) - sin(0) = 1 - 0 = 1.
Q34.
Integral of sec x tan x dx =
Answer: A. sec x + C
Why: d/dx(sec x) = sec x tan x. So integral of sec x tan x = sec x + C.
Q35.
Integrate: integral of x ln x dx (by parts)
Answer: A. x²/2 × ln x - x²/4 + C
Why: u = lnx, dv = x dx. v = x²/2. uv - integral(v du) = (x²/2)lnx - integral(x²/2 × 1/x dx) = (x²/2)lnx - x²/4 + C.
Q36.
Using partial fractions, integrate: 1/((x-1)(x+1)) dx
Answer: A. (1/2)ln|(x-1)/(x+1)| + C
Why: 1/((x-1)(x+1)) = 1/2 × [1/(x-1) - 1/(x+1)]. Integrate: (1/2)[ln|x-1| - ln|x+1|] = (1/2)ln|(x-1)/(x+1)| + C.
Q37.
Evaluate: integral from 1 to e of (1/x) dx
Answer: B. 1
Why: [ln x] from 1 to e = ln e - ln 1 = 1 - 0 = 1.
Q38.
Integral of cosec²x dx =
Answer: A. -cot x + C
Why: d/dx(-cot x) = cosec²x. So integral of cosec²x = -cot x + C.
Q39.
The area between y = x and y = x² from x = 0 to x = 1 is:
Answer: A. 1/6
Why: Area = integral from 0 to 1 of (x - x²) dx = [x²/2 - x³/3] = 1/2 - 1/3 = 1/6.
Q40.
Integrate: (1 + tan²x) dx
Answer: B. tan x + C
Why: 1 + tan²x = sec²x. Integral of sec²x = tan x + C.
Q41.
Evaluate: integral of x²/(1+x³) dx
Answer: A. (1/3)ln|1+x³| + C
Why: Substitution u = 1+x³, so du = 3x² dx, i.e. x² dx = du/3. Integral becomes ∫(1/u)(du/3) = (1/3)∫du/u = (1/3)ln|u| + C = (1/3)ln|1+x³| + C.
Q42.
Integral from 0 to π/2 of sin<sup>n</sup>(x)/(sin<sup>n</sup>(x)+cos<sup>n</sup>(x)) dx =
Answer: B. π/4
Why: King's rule: let I = ∫₀<sup>π/2</sup> sinⁿx/(sinⁿx+cosⁿx)dx. Replace x→(π/2−x): get ∫cosⁿx/(cosⁿx+sinⁿx)dx. Adding both: 2I = ∫₀<sup>π/2</sup>1 dx = π/2. So I = π/4.
Q43.
Evaluate integral of √(1+x²) dx.
Answer: A. x√(1+x²)/2 + (1/2)sinh⁻¹x + C
Why: Trig sub x = tanθ, dx = sec²θ dθ, √(1+x²) = secθ. ∫sec³θ dθ = (secθ tanθ)/2 + (1/2)ln|secθ+tanθ| + C. Back-substituting: x√(1+x²)/2 + (1/2)sinh⁻¹x + C.
Q44.
Integral of (x+1)/((x+2)(x+3)) dx using partial fractions:
Answer: A. -ln|x+2| + 2ln|x+3| + C
Why: Partial fractions: (x+1)/[(x+2)(x+3)] = A/(x+2)+B/(x+3). Setting x=−2: A=−1; x=−3: B=2. Integral = −∫dx/(x+2)+2∫dx/(x+3) = −ln|x+2|+2ln|x+3|+C.
Q45.
Evaluate: integral from 0 to π of x sinx dx.
Answer: A. π
Why: Integration by parts: u=x, dv=sinx dx → du=dx, v=−cosx. [−x cosx]₀^π + ∫₀^π cosx dx = (−π cosπ+0) + [sinx]₀^π = π + 0 = π.
Q46.
Reduction formula for integral of sin<sup>n</sup>(x) dx when n is positive integer involves:
Answer: C. Recursion: I<sub>n</sub> = -(sin<sup>n-1</sup>x cosx)/n + (n-1)/n × I_(n-2)
Why: Integration by parts: u=sinⁿ⁻¹x, dv=sinx dx. After differentiating and substituting cos²x=1−sin²x, the recursion Iₙ = −sinⁿ⁻¹x cosx/n + (n−1)/n·Iₙ₋₂ emerges, reducing the power by 2 each step.
Q47.
Integral of 1/(a + b cosx) dx uses the substitution:
Answer: B. t = tan(x/2) (Weierstrass substitution)
Why: Weierstrass substitution t=tan(x/2): cosx=(1−t²)/(1+t²), dx=2dt/(1+t²). Denominator a+b·(1−t²)/(1+t²) becomes rational in t, converting the trigonometric integral into a standard rational integral.
Q48.
The Beta function B(m,n) = integral from 0 to 1 of x<sup>m-1</sup>(1-x)<sup>n-1</sup> dx is related to Gamma by:
Answer: B. B(m,n) = Gamma(m)Gamma(n)/Gamma(m+n)
Why: Using the Gamma convolution: Γ(m)Γ(n) = ∫₀^∞ u<sup>m−1</sup>e<sup>−u</sup>du · ∫₀^∞ v<sup>n−1</sup>e<sup>−v</sup>dv. Substituting u=t·s, v=t(1−s) and integrating out t gives B(m,n) = Γ(m)Γ(n)/Γ(m+n).
Q49.
Evaluate: integral from 0 to ∞ of e<sup>-x²</sup> dx.
Answer: C. √π/2
Why: Gaussian integral: let I=∫₀^∞ e<sup>−x²</sup>dx. Then (2I)²=∫∫e<sup>−x²−y²</sup>dx dy over the plane. Converting to polar: 2π∫₀^∞ re<sup>−r²</sup>dr = π. So (2I)²=π, giving I=√π/2.
Q50.
Using shell method, volume of solid of revolution of y=x² (0 to 1) around y-axis:
Answer: A. π/2
Why: Shell method: V = 2π∫₀¹ x·f(x)dx = 2π∫₀¹ x·x² dx = 2π∫₀¹ x³ dx = 2π[x⁴/4]₀¹ = 2π·(1/4) = π/2 cubic units.
Q51.
integral of sin<sup>3</sup>(x) dx =
Answer: A. -cos x + cos³x/3 + C
Why: Write sin³x = sinx·(1−cos²x). Substitute u=cosx, du=−sinx dx. Integral = −∫(1−u²)du = −u+u³/3+C. Back-substituting: −cosx+cos³x/3+C.
Q52.
Evaluate: integral from 0 to 1 of ln(x) dx.
Answer: A. -1
Why: Integration by parts: u=ln x, dv=dx → du=dx/x, v=x. = [x ln x]₀¹ − ∫₀¹ dx = (1·0 − lim_{x→0⁺} x ln x) − 1 = 0 − 0 − 1 = −1. Note lim x ln x=0 by L'Hôpital.
Q53.
Integral of e<sup>x</sup> cos(x) dx =
Answer: A. e<sup>x</sup>(sinx + cosx)/2 + C
Why: Integration by parts twice: I=∫eˣcosx dx. First: u=cosx, dv=eˣdx → I=eˣcosx+∫eˣsinx dx. Second: ∫eˣsinx dx=eˣsinx−I. So 2I=eˣ(cosx+sinx), giving I=eˣ(sinx+cosx)/2+C.
Q54.
The area of region bounded by y = |x| and y = 1 is:
Answer: A. 1
Why: y=|x| meets y=1 at x=±1. Region lies between the two curves. By symmetry: Area = 2∫₀¹(1−x)dx = 2[x−x²/2]₀¹ = 2(1−1/2) = 2·(1/2) = 1 square unit.
Q55.
Integral from a to b of f(x) dx = - integral from b to a of f(x) dx. This is because:
Answer: A. Reversing limits changes sign
Why: By the Fundamental Theorem: ∫ₐᵇf(x)dx = F(b)−F(a). Reversing limits: ∫ᵦₐf(x)dx = F(a)−F(b) = −[F(b)−F(a)]. Reversing integration limits negates the value; this property is independent of whether f is odd or even.
Q56.
Wallis product: (π/2) = product of [2n/(2n-1)] × [2n/(2n+1)] for n=1,2,3,...
Answer: B. π/2
Why: Wallis derived this from ∫₀<sup>π/2</sup>sin<sup>2n</sup>x dx evaluated two ways. The infinite product (2/1)(2/3)(4/3)(4/5)(6/5)(6/7)... converges to π/2, giving a beautiful product representation of π.
Q57.
Evaluate: integral of (sin 2x)/(1+sin²x) dx.
Answer: A. ln(1+sin²x) + C
Why: Note sin2x = 2sinx cosx = d(sin²x)/dx. Substitution u=1+sin²x, du=sin2x dx. Integral = ∫du/u = ln|u|+C = ln(1+sin²x)+C. Since 1+sin²x≥1>0, absolute value is unnecessary.
Q58.
Second Mean Value Theorem for integrals: integral from a to b of f(x)g(x) dx =
Answer: A. f(c) × integral from a to b of g(x) dx for some c in [a,b]
Why: Second MVT for integrals: if f is monotone and g is integrable, then integral fg = f(c) integral g for some c. Generalization allows Bonnet form.
Q59.
The volume of solid formed by rotating y=√x (from 0 to 4) around x-axis:
Answer: A. 8π
Why: Disk method: V = π∫₀⁴[√x]²dx = π∫₀⁴ x dx = π[x²/2]₀⁴ = π(16/2−0) = 8π cubic units.
Q60.
The value of ∫ x² dx is:
Answer: A. x³/3 + C
Why: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, so ∫ x² dx = x³/3 + C.
Q61.
The value of the definite integral of sin²x from 0 to π/2 is:
Answer: B. π/4
Why: ∫ sin²x dx over [0, π/2] = π/4 by symmetry of sin² and cos².
Q62.
The value of the definite integral of x/(1 + x²) from 0 to 1 is:
Answer: B. (1/2)ln 2
Why: The antiderivative is (1/2)ln(1 + x²); evaluating from 0 to 1 gives (1/2)ln 2.
Q63.
The value of the definite integral of x·sinx from 0 to π is:
Answer: B. π
Why: ∫ x sinx dx = −x cosx + sinx. Evaluated from 0 to π this gives π.
Q64.
The value of the definite integral of 1/(1 + tanx) from 0 to π/2 is:
Answer: B. π/4
Why: Using the King property I = ∫ 1/(1+cotx), and adding gives 2I = π/2, so I = π/4.
Q65.
The value of the definite integral of lnx from 1 to e is:
Answer: B. 1
Why: ∫ lnx dx = x lnx − x; from 1 to e this is (e − e) − (0 − 1) = 1.
Q66.
The value of the definite integral of the greatest integer function [x] from 0 to 2 is:
Answer: B. 1
Why: [x] = 0 on [0,1) and 1 on [1,2), so the integral is 0·1 + 1·1 = 1.
Q67.
The value of the definite integral of x³·e<sup>x²</sup> from −1 to 1 is:
Answer: A. 0
Why: The integrand is an odd function on a symmetric interval, so the integral is 0.
Q68.
The indefinite integral of 1/(x² + 4) is:
Answer: B. (1/2)arctan(x/2) + C
Why: ∫ dx/(x² + a²) = (1/a)arctan(x/a) + C; with a = 2 this is (1/2)arctan(x/2) + C.