68 free MCQs on Application of Integrals with worked answers and explanations. Using definite integrals to compute the area under a curve, the area between two curves, and the areas enclosed by standard curves like circles, parabolas, and ellipses.
Take the timed Application of Integrals chapterwise test →Below are 68 practice questions on Application of Integrals, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Application of Integrals notes.
The definite integral ∫ₐᵇf(x)dx computes the exact area of the shaded region bounded by the curve, the x-axis, and the vertical lines x=a and x=b - the same idea behind the Riemann sum, but evaluated exactly rather than approximated by rectangles.
Q1.
The area under the curve y = f(x), above the x-axis, between x = a and x = b is given by:
Answer: B. Integral from a to b of f(x) dx
Why: The area bounded by a non-negative curve, the x-axis, and the vertical lines x=a, x=b is the definite integral from a to b of f(x) dx.
Q2.
If the curve lies entirely below the x-axis over [a,b], the area is:
Answer: B. The absolute value of the definite integral (which is negative)
Why: When f(x) < 0 on [a,b], the definite integral is negative; the actual area is its absolute value.
Q3.
The area between two curves f(x) and g(x) (with f(x) >= g(x)) from x=a to x=b is:
Answer: B. Integral of [f(x) - g(x)] dx from a to b
Why: Area between two curves is the integral of (upper curve minus lower curve) over the interval.
Q4.
The area of a circle x<sup>2</sup> + y<sup>2</sup> = a<sup>2</sup> found using integration is:
Answer: C. pi a<sup>2</sup>
Why: Using integration (4 times the first-quadrant area), the area of the circle x<sup>2</sup>+y<sup>2</sup>=a<sup>2</sup> comes out to pi a<sup>2</sup>, matching the known formula.
Q5.
The area enclosed by the ellipse x<sup>2</sup>/a<sup>2</sup> + y<sup>2</sup>/b<sup>2</sup> = 1 is:
Answer: A. pi a b
Why: By integration over the ellipse (4 times the first-quadrant area), the total enclosed area is pi a b.
Q6.
Evaluate the area under y = x from x = 0 to x = 4.
Answer: B. 8
Why: Area = integral from 0 to 4 of x dx = [x<sup>2</sup>/2] from 0 to 4 = 16/2 - 0 = 8.
Q7.
Evaluate the area under y = 3 (a constant function) from x = 1 to x = 5.
Answer: B. 12
Why: Area = integral from 1 to 5 of 3 dx = 3(5-1) = 12, which is just the rectangle area.
Q8.
If a curve crosses the x-axis within [a,b], the correct method to find total area is to:
Answer: B. Split at the crossing point, take absolute values of each piece, and add
Why: Where the sign of f(x) changes, you must split the interval at the root, evaluate each piece, take absolute values, and sum them to get the correct total area.
Q9.
Area bounded by curve x = g(y), the y-axis, and lines y=c, y=d is given by:
Answer: A. Integral of x dy from c to d
Why: When integrating with respect to y, area = integral from c to d of x dy = integral from c to d of g(y) dy.
Q10.
Find the area under y = x<sup>2</sup> from x = 0 to x = 3.
Answer: C. 9
Why: Area = integral from 0 to 3 of x<sup>2</sup> dx = [x<sup>3</sup>/3] from 0 to 3 = 27/3 - 0 = 9.
Q11.
The first step in solving an area-between-curves problem should be to:
Answer: B. Find the points of intersection of the curves
Why: Finding the intersection points first establishes the correct limits of integration and shows which curve lies above the other in each region.
Q12.
Area under y = sqrt(x) from x = 0 to x = 4 is:
Answer: B. 16/3
Why: Integral of x<sup>1/2</sup> dx from 0 to 4 = [(2/3)x<sup>3/2</sup>] from 0 to 4 = (2/3)(8) - 0 = 16/3.
Q13.
When set up correctly, area is always expressed as a:
Answer: B. Non-negative quantity
Why: Even though a definite integral can be negative, the area itself is defined as a non-negative quantity, so absolute values are taken where needed.
Q14.
The area enclosed by a semicircle of radius r above the x-axis (y = sqrt(r<sup>2</sup>-x<sup>2</sup>)) is:
Answer: B. (pi r<sup>2</sup>)/2
Why: A semicircle is half of a full circle, so its area is (pi r<sup>2</sup>)/2.
Q15.
The definite integral of a function between two limits gives the ___ under the curve:
Answer: A. area
Why: A definite integral evaluates the area bounded by the curve and the x-axis.
Q16.
Integration is the reverse process of:
Answer: A. differentiation
Why: Integration (antidifferentiation) undoes differentiation.
Q17.
The area under a curve y = f(x) from x = a to x = b is given by:
Answer: A. ∫ f(x) dx from a to b
Why: Area = ∫ₐᵇ f(x) dx.
Q18.
Physical area is always taken to be a ___ quantity:
Answer: A. a strictly non-negative value
Why: Area is measured as a non-negative amount, so we use absolute values where a curve dips below the axis.
Q19.
The symbol ∫ represents:
Answer: A. integration
Why: The elongated S symbol ∫ denotes integration.
Q20.
In a definite integral, the numbers a and b are called the ___ of integration:
Answer: A. limits
Why: a and b are the lower and upper limits of integration.
Q21.
Find the area enclosed between y = x and y = x<sup>2</sup> for x in [0,1].
Answer: A. 1/6
Why: Since x >= x<sup>2</sup> on [0,1], Area = integral from 0 to 1 of (x - x<sup>2</sup>) dx = [x<sup>2</sup>/2 - x<sup>3</sup>/3] from 0 to 1 = 1/2 - 1/3 = 1/6.
Q22.
Find the area bounded by y = x<sup>2</sup> and y = 4 (the horizontal line), between their intersection points.
Answer: B. 32/3
Why: Intersections at x = -2, 2. Area = integral from -2 to 2 of (4 - x<sup>2</sup>) dx = 2 times integral from 0 to 2 of (4-x<sup>2</sup>) dx = 2[4x - x<sup>3</sup>/3] from 0 to 2 = 2(8 - 8/3) = 2(16/3) = 32/3.
Q23.
Find the area under y = sin x from x = 0 to x = pi.
Answer: C. 2
Why: Area = integral from 0 to pi of sin x dx = [-cos x] from 0 to pi = -cos(pi) - (-cos(0)) = 1 + 1 = 2 (sin x is non-negative on [0,pi], so the integral directly gives area).
Q24.
Find the area under y = cos x from x = 0 to x = pi (careful with sign change at pi/2).
Answer: C. 2
Why: cos x is positive on [0,pi/2] and negative on [pi/2,pi]. Area = |integral 0 to pi/2 of cos x dx| + |integral pi/2 to pi of cos x dx| = |1| + |-1| = 2.
Q25.
Find the area of the region bounded by y<sup>2</sup> = 4x and the line x = 4 (right half of the parabola's enclosed area, upper part only).
Answer: A. 32/3
Why: Upper half: y = 2sqrt(x). Area (upper half only) = integral from 0 to 4 of 2sqrt(x) dx = 2 times [(2/3)x<sup>3/2</sup>] from 0 to 4 = 2 times (16/3) = 32/3.
Q26.
Find the area between the curves y = x<sup>3</sup> and y = x for x in [0,1].
Answer: A. 1/4
Why: On [0,1], x >= x<sup>3</sup>, so Area = integral from 0 to 1 of (x - x<sup>3</sup>) dx = [x<sup>2</sup>/2 - x<sup>4</sup>/4] from 0 to 1 = 1/2 - 1/4 = 1/4.
Q27.
Find the area enclosed by the circle x<sup>2</sup> + y<sup>2</sup> = 9 in the first quadrant only.
Answer: A. 9 pi/4
Why: Total circle area = pi(3)<sup>2</sup> = 9pi. The first quadrant is one-fourth of the circle: 9pi/4.
Q28.
Find the area bounded by y = x<sup>2</sup> - 4 and the x-axis between x = -2 and x = 2.
Answer: A. 32/3
Why: y = x<sup>2</sup>-4 is below the x-axis on [-2,2] (since x<sup>2</sup><=4). Area = |integral from -2 to 2 of (x<sup>2</sup>-4) dx| = |[x<sup>3</sup>/3 - 4x] from -2 to 2| = |(8/3-8) - (-8/3+8)| = |-16/3 - 16/3| = 32/3.
Q29.
Find the area of the region in the first quadrant bounded by y = 4 - x<sup>2</sup>, the x-axis, and the y-axis.
Answer: B. 16/3
Why: The curve meets the x-axis at x=2. Area = integral from 0 to 2 of (4-x<sup>2</sup>) dx = [4x - x<sup>3</sup>/3] from 0 to 2 = 8 - 8/3 = 16/3.
Q30.
Find the area enclosed between the line y = x + 2 and the parabola y = x<sup>2.</sup>
Answer: A. 9/2
Why: Intersections: x<sup>2</sup> = x+2 gives x=-1,2. Area = integral from -1 to 2 of [(x+2)-x<sup>2</sup>] dx = [x<sup>2</sup>/2+2x-x<sup>3</sup>/3] from -1 to 2 = (2+4-8/3)-(0.5-2+1/3) = (10/3) - (-7/6) = 27/6 = 9/2.
Q31.
Find the area bounded by y = 1/x, x = 1, x = e, and the x-axis.
Answer: A. 1
Why: Area = integral from 1 to e of (1/x) dx = [ln x] from 1 to e = ln(e) - ln(1) = 1 - 0 = 1.
Q32.
Find the area between the curve y = x<sup>2</sup> + 1 and the x-axis from x = 0 to x = 2.
Answer: A. 14/3
Why: Area = integral 0 to 2 of (x<sup>2</sup>+1) dx = [x<sup>3</sup>/3 + x] from 0 to 2 = 8/3 + 2 = 14/3.
Q33.
The area bounded by y = x², the x-axis and x = 0 to x = 3 is ∫₀³ x² dx =:
Answer: A. 9
Why: ∫x² dx = x³/3; evaluated from 0 to 3 gives 27/3 = 9.
Q34.
Found by integration, the area of a full circle of radius r is:
Answer: A. πr²
Why: Integrating the circle’s equation gives the enclosed area πr².
Q35.
The area between an upper curve f(x) and a lower curve g(x) is ∫[f(x) − g(x)] dx taken over their:
Answer: A. points of intersection
Why: The limits are the x-values where the two curves intersect.
Q36.
The area under y = x from x = 0 to x = 2 is:
Answer: A. 2
Why: ∫x dx = x²/2; from 0 to 2 gives 4/2 = 2.
Q37.
If a curve lies below the x-axis, its definite integral gives a ___ value:
Answer: A. negative
Why: Below the axis the signed area is negative, so the raw integral is negative.
Q38.
The value of ∫₀¹ x dx is:
Answer: A. 1/2
Why: ∫x dx = x²/2; from 0 to 1 gives 1/2.
Q39.
The area under y = sin x from x = 0 to x = π is:
Answer: A. 2
Why: ∫sin x dx = −cos x; from 0 to π gives (−cos π) − (−cos 0) = 1 + 1 = 2.
Q40.
To find an area with respect to the y-axis, we integrate x with respect to:
Answer: A. y
Why: Horizontal strips require integrating x as a function of y, with respect to y.
Q41.
Find the area of the region bounded by the parabola y<sup>2</sup> = 4ax and its latus rectum x = a (full region, both above and below the x-axis).
Answer: A. (8/3)a<sup>2</sup>
Why: By symmetry, total area = 2 times the upper-half area. Upper-half area = integral from 0 to a of 2sqrt(ax) dx = 2sqrt(a)[(2/3)x<sup>3/2</sup>] from 0 to a = 2sqrt(a)(2/3)a<sup>3/2</sup> = (4/3)a<sup>2.</sup> Doubling for both halves gives 2 times (4/3)a<sup>2</sup> = (8/3)a<sup>2.</sup>
Q42.
Find the area common to the circle x<sup>2</sup>+y<sup>2</sup> = 4 and the line x = 1 (area of the circular region to the right of x=1).
Answer: A. (4pi/3) - sqrt(3)
Why: Area to the right of x=1 inside x<sup>2</sup>+y<sup>2</sup>=4 is 2 times integral from 1 to 2 of sqrt(4-x<sup>2</sup>) dx, which evaluates using the standard formula integral sqrt(a<sup>2</sup>-x<sup>2</sup>)dx = (x/2)sqrt(a<sup>2</sup>-x<sup>2</sup>)+(a<sup>2</sup>/2)sin-1(x/a), giving total area = (4pi/3) - sqrt(3).
Q43.
Find the area bounded by y = |x - 1| and the x-axis between x = 0 and x = 3.
Answer: B. 2.5
Why: Split at x=1: integral from 0 to 1 of (1-x) dx = [x - x<sup>2</sup>/2] from 0 to 1 = 1 - 0.5 = 0.5. Integral from 1 to 3 of (x-1) dx = [x<sup>2</sup>/2 - x] from 1 to 3 = (4.5-3) - (0.5-1) = 1.5 + 0.5 = 2. Total area = 0.5 + 2 = 2.5.
Q44.
Find the area enclosed between the parabola y = x<sup>2</sup> and the line y = 4x.
Answer: A. 32/3
Why: Intersections: x<sup>2</sup> = 4x gives x=0,4. Since 4x >= x<sup>2</sup> on [0,4], Area = integral from 0 to 4 of (4x-x<sup>2</sup>) dx = [2x<sup>2</sup> - x<sup>3</sup>/3] from 0 to 4 = 32 - 64/3 = 96/3-64/3 = 32/3.
Q45.
Find the area of the smaller region bounded by the ellipse x<sup>2</sup>/9 + y<sup>2</sup>/4 = 1 and the line x/3 + y/2 = 1.
Answer: A. 3(pi - 2)/2
Why: Using the standard NCERT technique (integrating the ellipse boundary minus the line, from x=0 to x=3), the smaller region's area simplifies to 3(pi-2)/2.
Q46.
Find the area bounded by the curves y = x<sup>2</sup> and y = |x|.
Answer: A. 1/3
Why: By symmetry, consider x>=0: y=x and y=x<sup>2</sup> intersect at x=0,1. Area (x>=0 part) = integral 0 to 1 of (x-x<sup>2</sup>) dx = 1/6. Doubling for symmetry (x<=0 mirrors it) gives total area = 1/3.
Q47.
Find the area bounded by y = sin x and y = cos x between x = 0 and x = pi/2.
Answer: A. 2(sqrt(2)-1)
Why: They intersect at x=pi/4. Area = integral 0 to pi/4 (cos x - sin x)dx + integral pi/4 to pi/2 (sin x - cos x)dx. Each piece evaluates to (sqrt2-1), so total = 2(sqrt(2)-1).
Q48.
Find the area of the region bounded by y = x<sup>3</sup> - x and the x-axis between x = -1 and x = 1.
Answer: A. 1/2
Why: x<sup>3</sup>-x = x(x-1)(x+1), positive on (-1,0), negative on (0,1). Area = |integral -1 to 0| + |integral 0 to 1| = 1/4 + 1/4 = 1/2 (by symmetry of the odd function about the origin in absolute terms).
Q49.
Find the area bounded by the curve y = 4 - x<sup>2</sup> and the lines x = -1 and x = 2 (curve stays above axis throughout).
Answer: B. 27/3 = 9
Why: Area = integral from -1 to 2 of (4-x<sup>2</sup>) dx = [4x - x<sup>3</sup>/3] from -1 to 2 = (8-8/3) - (-4+1/3) = (16/3) - (-11/3) = 27/3 = 9.
Q50.
Find the area bounded by the curve y = sqrt(x), the x-axis, and the lines x = 0 and x = 4.
Answer: C. 16/3
Why: Area = ∫₀⁴√x dx = ∫₀⁴ x<sup>1/2</sup> dx = [(2/3)x<sup>3/2</sup>]₀⁴ = (2/3)(4<sup>3/2</sup>)−0 = (2/3)(8) = 16/3 square units.
Q51.
The area of the region bounded by the parabola y² = 4ax and its latus rectum is:
Answer: A. (8/3)a²
Why: This standard result gives the enclosed area as 8a²/3.
Q52.
The area enclosed by the ellipse x²/a² + y²/b² = 1 is:
Answer: A. πab
Why: The area of an ellipse is πab, reducing to πr² when a = b = r.
Q53.
The area under y = x² from x = 0 to x = 2 is:
Answer: A. 8/3
Why: ∫x² dx = x³/3; from 0 to 2 gives 8/3.
Q54.
The area enclosed between y = x and y = x² from x = 0 to x = 1 is:
Answer: A. 1/6
Why: ∫(x − x²) dx from 0 to 1 = 1/2 − 1/3 = 1/6.
Q55.
The value of ∫₀<sup>π/2</sup> cos x dx is:
Answer: A. 1
Why: ∫cos x dx = sin x; from 0 to π/2 gives 1 − 0 = 1.
Q56.
The area of the circle x² + y² = a², obtained by integration, equals:
Answer: A. πa²
Why: The enclosed area of a circle of radius a is πa².
Q57.
The total area between y = x³ and the x-axis from x = −1 to x = 1 (adding absolute areas) is:
Answer: A. 1/2
Why: Each half contributes 1/4, so the total (unsigned) area is 1/2.
Q58.
The value of ∫₁² (1/x) dx is:
Answer: A. ln 2
Why: ∫(1/x) dx = ln|x|; from 1 to 2 gives ln 2 − ln 1 = ln 2.
Q59.
The area under a curve, being a two-dimensional measure, has units of:
Answer: A. square units
Why: Area is measured in square units.
Q60.
The area enclosed by one arch of y = sin x (from 0 to π) is:
Answer: A. 2
Why: ∫₀^π sin x dx = 2, the area of one full arch.
Q61.
The area bounded by y = x², the x-axis and the lines x = 0 and x = 3 is:
Answer: B. 9
Why: ∫₀³ x² dx = 27/3 = 9.
Q62.
The area enclosed between the curves y = x² and y = x is:
Answer: A. 1/6
Why: ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.
Q63.
The area enclosed between y² = 4x and x² = 4y is:
Answer: B. 16/3
Why: The two parabolas meet at (0,0) and (4,4); the enclosed area is 16/3.
Q64.
The area enclosed by the ellipse x²/9 + y²/4 = 1 is:
Answer: A. 6π
Why: Area = πab = π·3·2 = 6π.
Q65.
The area under one arch of y = sinx from 0 to π is:
Answer: B. 2
Why: ∫₀^π sinx dx = [−cosx]₀^π = 1 + 1 = 2.
Q66.
The area bounded by y = x³, the x-axis and the lines x = 0 and x = 2 is:
Answer: B. 4
Why: ∫₀² x³ dx = 16/4 = 4.
Q67.
The area between y = √x and y = x from x = 0 to x = 1 is:
Answer: A. 1/6
Why: ∫₀¹ (√x − x) dx = 2/3 − 1/2 = 1/6.
Q68.
The area of the part of the disc x² + y² = 16 lying in the first quadrant is:
Answer: B. 4π
Why: One quarter of the disc of area 16π is 4π.