NDA Area under Curves previous year questions with solutions

Answer: (b) (2e − 2) square units

Method. even function: area is twice the area on [0, 1]

e^|x| > 0 and is even, and |x| = 1 means the lines x = -1 and x = 1.
Area = ∫_-1¹ e^|x| dx = 2∫₀¹ e^x dx = 2[e^x]₀¹ = 2(e − 1) = (2e − 2) square units.

Answer: (a) 1/⁠3 square unit

Method. find the curve from its slope, then integrate

dydx = 2x ⇒ y = x² + C; the curve passes through (0, 0), so C = 0 and y = x².
Area = ∫₀¹ x² dx = [x³3]₀¹ = 13 square unit.

Answer: (b) 1 − k

Method. areas under sin x on [0, π4] and [π4, π2] add to 1

k = ∫₀^π/⁠4 sin x dx = [-cos x]₀^π/⁠4 = 1 − 1⁠√2.⁠
Required area = ∫_π/⁠4^π/⁠2 sin x dx = [-cos x]_π/⁠4^π/⁠2 = 1⁠√2.⁠
So the required area = 1 − k.

Shortcut. ∫₀^π/⁠2 sin x dx = 1, so the rest of the arch beyond [0, π4] is 1 − k.

Answer: (a) k

Method. cos x on [π4, π2] mirrors sin x on [0, π4]

∫_π/⁠4^π/⁠2 cos x dx = [sin x]_π/⁠4^π/⁠2 = 1 − 1⁠√2.⁠
From the previous item, k = 1 − 1⁠√2.⁠
So the required area = k.

Shortcut. cos x = sin(π2 − x), so the graph on [π4, π2] is the reflection of sin x on [0, π4].

Answer: (a) 23

Method. odd function: area is twice the area on [0, 1]

f(x) = x|x| = x² for x ≥ 0 and -x² for x < 0; it is odd, so the regions on [-1, 0] and [0, 1] are equal in area.
Area = 2∫₀¹ x² dx = 2 × 13 = 23.
(The signed integral ∫_-1¹ x|x| dx = 0 is not the area.)

Answer: (c) 0.6 square unit

Method. greatest integer function is constant between integers

For -1.8 ≤ x < -1.5 we have -2 ≤ x < -1, so [x] = -2.
The region is a rectangle of width -1.5 − (-1.8) = 0.3 and height |-2| = 2 (below the x-axis).
Area = 0.3 × 2 = 0.6 square unit.

Answer: (b) 2c

Method. one arch of the sine curve, ∫₀^π sin x dx = 2

y = c sin x meets the x-axis at x = 0, π, 2π, …; one loop lies over an interval of length π, say [0, π] (c > 0, as the options assume; in general the loop area is 2|c|).
Area = ∫₀^π c sin x dx = c[-cos x]₀^π = c(1 + 1) = 2c.
Every loop has the same area by periodicity.

Answer: (d) 8k²

Method. area of a rectangle

|x| ≤ 2k gives -2k ≤ x ≤ 2k: width 4k.
|y| ≤ k gives -k ≤ y ≤ k: height 2k.
Area = 4k × 2k = 8k².

Answer: (a) -12

Method. the region is a triangle with a vertical side on x = 2

The lines y = x and y = mx meet at the origin; x = 2 cuts them at (2, 2) and (2, 2m).
Since m < 0, the vertical side has length 2 − 2m, and the height from the origin is 2.
Area = 12 × 2 × (2 − 2m) = 2 − 2m = 3 ⇒ m = -12.

Answer: (c) 4

Method. the region is a triangle

x − |y| = 0 means x = |y|: the lines y = x and y = -x for x ≥ 0.
With x = 2 they form the triangle with vertices (0, 0), (2, 2), (2, -2).
Area = 12 × base × height = 12 × 4 × 2 = 4.

Answer: (b) 80 square units

Method. area of a rectangle

|x| < 5 means -5 < x < 5: a strip of width 10.
The lines y = 0 and y = 8 cut it off at height 8.
The region is a 10 × 8 rectangle, so area = 80 square units.

Answer: (c) 2 square units

Method. |x| + |y| = 1 is a square (rhombus) with vertices on the axes

In each quadrant |x| + |y| = 1 is a straight line, e.g. x + y = 1 in the first quadrant.
The four lines join (1, 0), (0, 1), (-1, 0), (0, -1): a square whose diagonals both have length 2.
Area = 12 × d₁ × d₂ = 12 × 2 × 2 = 2 square units.

Shortcut. Area enclosed by |x| + |y| = a is 2a²; here a = 1.

Answer: (b) 3

Method. solve x² = 2|x|

x² = 2|x| ⇒ |x|² - 2|x| = 0 ⇒ |x|(|x| − 2) = 0 ⇒ |x| = 0 or 2.
x = 0, 2, -2 give the points (0, 0), (2, 4), (-2, 4).
Number of points of intersection = 3.

Answer: (b) 23 square unit

Method. area between two curves

For 0 ≤ x ≤ 1, 2|x| = 2x, and 2x ≥ x² (since x ≤ 2), so y = 2x is on top.
Area = ∫₀¹ (2x − x²) dx = [x² - x³3]₀¹ = 1 − 13 = 23 square unit.

Answer: (b) 14 square unit

Method. area between two curves

Intersection: x = x³ ⇒ x = 0, ±1; in the first quadrant the curves meet at x = 0 and x = 1.
For 0 < x < 1, x³ < x, so y = x is on top.
Area = ∫₀¹ (x − x³) dx = 12 − 14 = 14 square unit.

Answer: (c) 23

Method. area between a parabola and a line

Intersection: x² = 2x ⇒ x = 0 or x = 2, giving (0, 0) and (2, 2).
For 0 < x < 2 the upper branch y = ⁠√(2x)⁠ lies above y = x.
Area = ∫₀² (⁠√(2x)⁠ − x) dx = [2⁠√23⁠ x^3/⁠2 - x²2]₀² = 83 − 2 = 23.

Shortcut. Area between y² = 4ax and y = mx is 8a²3m³; here a = 12, m = 1 gives 23.

Answer: (a) 23 square units

Method. area between a parabola and a line

Intersection: x² = 2x ⇒ x = 0 or x = 2, giving (0, 0) and (2, 2).
For 0 < x < 2 the parabola's upper branch y = ⁠√(2x)⁠ lies above y = x (at x = 1: ⁠√2⁠ > 1).
Area = ∫₀² (⁠√(2x)⁠ − x) dx = [2⁠√23⁠ x^3/⁠2 - x²2]₀² = 2⁠√23⁠ × 2⁠√2⁠ − 2 = 83 − 2 = 23 square units.

Shortcut. Area between y² = 4ax and y = mx is 8a²3m³; here a = 12, m = 1 gives 23.

Answer: (c) 4⁠√63⁠

Method. area between two curves, integrating along y

Intersection: 6(x − 1) = 3x ⇒ x = 2, y² = 6 ⇒ y = ±⁠√6.⁠
For |y| < ⁠√6⁠ the right boundary is y² = 6(x − 1), i.e. x = 1 + y²6, and the left boundary is y² = 3x, i.e. x = y²3.
Area = ∫_-⁠√6^√6⁠ [(1 + y²6) − y²3] dy = 2∫₀^⁠√6⁠ (1 − y²6) dy
= 2[y − y³18]₀^⁠√6⁠ = 2(⁠√6⁠ − 6⁠√618)⁠ = 2 × 2⁠√63⁠ = 4⁠√63.⁠

Answer: (b) 1

Method. area of a parabolic segment cut off by a vertical line

y² = kx is symmetric about the x-axis, with y = ±⁠√(kx)⁠.
Area = 2∫₀^k ⁠√(kx)⁠ dx = 2⁠√k⁠ [23 x^3/⁠2]₀^k = 43 k².
43 k² = 43 ⇒ k² = 1 ⇒ k = 1 (k > 0).

Answer: (a) 1/⁠6 square unit

Method. area cut off by the latus rectum of y² = 4ax is 8a²3

With k = 1 the parabola is y² = x = 4 × 14 × x, so a = 14 and the latus rectum is x = 14.
Area = 2∫₀^1/⁠4 ⁠√x⁠ dx = 2 × 23 (14)^3/⁠2 = 43 × 18 = 16 square unit.

Answer: (c) 3

Method. area cut off by the latus rectum of y² = 4ax is 8a²3

The latus rectum of y² = 4kx is the line x = k.
Area = 2∫₀^k ⁠√(4kx)⁠ dx = 4⁠√k⁠ [23 x^3/⁠2]₀^k = 8k²3.
8k²3 = 24 ⇒ k² = 9 ⇒ k = 3 (k > 0).

Answer: (d) 3π + 24

Method. segment areas when the chord subtends a right angle

x² + y² - 2x = 0 ⇒ (x − 1)² + y² = 1: centre C(1, 0), radius 1.
y = x meets it where 2x² - 2x = 0 ⇒ O(0, 0) and P(1, 1); CO = (-1, 0) ⊥ CP = (0, 1), so the chord subtends π2 at C.
A₂ = sector − triangle = π4 − 12 = π − 24.
A₁ = π − A₂ = 3π + 24.

Answer: (a) π

Method. substitute the segment areas

From the previous item, A₁ = 3π + 24, A₂ = π − 24.
A₁ + A₂ = π (the whole circle).
A₁ - 3A₂ = 3π + 2 − 3π + 64 = 2.
2(A₁ + A₂)A₁ - 3A₂ = 2π2 = π.

Answer: (a) (π − 2)a²4

Method. segment area = sector − triangle, chord subtends a right angle

Circle: centre C(a, 0), radius a. Put y = x: (x − a)² + x² = a² ⇒ 2x² - 2ax = 0 ⇒ x = 0 or a.
The chord joins O(0, 0) and P(a, a); CO = (-a, 0) and CP = (0, a) are perpendicular, so the chord subtends π2 at C.
Minor segment = sector − triangle OCP = 12 a² × π2 − 12 × a × a = (π − 2)a²4.

Answer: (b) (3π + 2)a²4

Method. major segment = circle − minor segment

From the previous item, the minor segment = (π − 2)a²4.
Major segment = πa² - (π − 2)a²4 = (4π − π + 2)a²4 = (3π + 2)a²4.

Answer: (c) π8

Method. the region is a circular sector

y = ⁠√(1⁠ −⁠ x²) is the upper half of the unit circle; y = x meets it at (1⁠√2,⁠ 1⁠√2),⁠ at angle π4.
The region bounded by y = 0, the arc and y = x runs from angle 0 to angle π4: a sector of the unit circle.
Area = 12 r² θ = 12 × 1 × π4 = π8.

Answer: (b) π³4 − 2

Method. quarter circle minus the area under one arch of sin x

x² + y² = π² in the first quadrant is a quarter circle of radius π, meeting the x-axis at (π, 0); y = sin x also meets it at (π, 0).
For 0 < x < π, sin x lies below the circle, so the region lies between the circle and y = sin x for 0 ≤ x ≤ π.
Area = (quarter circle) - ∫₀^π sin x dx = π × π²4 − 2 = π³4 − 2.

Answer: (b) 8π square units

Method. upper half of a circle

y = ⁠√(16⁠ −⁠ x²) gives y ≥ 0 and x² + y² = 16: the upper half of the circle of radius 4.
With the x-axis it encloses a semicircle.
Area = 12 π (4)² = 8π square units.