Important Questions: Applications of Integrals | Mathematics (Maths) Class 12 - JEE PDF Download

Q1: Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is.
Ans: Given,
Equation of circle is x²+ y² = 4……….(i)
x²+ y² = 2²
y = √(2² – x²) …………(ii)
Equation of a lines is x + y = 2 ………(iii)
y = 2 – x

Therefore, the graph of equation (iii) is the straight line joining the points (0, 2) and (2, 0).
From the graph of a circle (i) and straight-line (iii), it is clear that points of intersections of circle
(i) and the straight line (iii) is A (2, 0) and B (0.2).
Area of OACB, bounded by the circle and the coordinate axes is

= [ 1 × √0 + 2 sin^-1(1) – 0√4 – 2 × 0]
= 2 sin^-1(1)
= 2 × π/2
= π sq. units
Area of triangle OAB, bounded by the straight line and the coordinate axes is

= 4 – 2 – 0 + 0
= 2 sq.units
Hence, the required area = Area of OACB – Area of triangle OAB
= (π – 2) sq.units

Q2: Find the area of the curve y = sin x between 0 and π.
Ans: Given,
y = sin x

Area of OAB

= – [cos π – cos 0]
= -(-1 -1)
= 2 sq. units

Q3: Find the area enclosed by the ellipse x²/a² + y²/b² =1.
Ans: Given,


We know that,
Ellipse is symmetrical about both x-axis and y-axis.

Area of ellipse = 4 × Area of AOB

Substituting the positive value of y in the above expression since OAB lies in the first quadrant.

= 2ab × sin^-1(1)
= 2ab × π/2
= πab
Hence, the required area is πab sq.units.

Q4: Find the area of the region bounded by the two parabolas y = x² and y² = x.
Ans: Given two parabolas are y = x² and y² = x.
The point of intersection of these two parabolas is O (0, 0) and A (1, 1) as shown in the below figure.

Now,
y² = x
y = √x = f(x)
y = x² = g(x), where, f (x) ≥ g (x) in [0, 1].
Area of the shaded region

= (⅔) – (⅓)
= ⅓
Hence, the required area is ⅓ sq. units.

Q3: Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Ans: We can draw the figure of y² = 9x; x = 2, x = 4 and the x-axis in the first curve as below.

y² = 9x
y = ±√(9x)
y = ±3√x
We can consider the positive value of y since the required area is in the first quadrant.
The required area is the shaded region enclosed by ABCD.

= 2 [(2)³ – (√2)³]
= 2[8 – 2√2]
= 16 – 4√2
Hence, the required area is 16 – 4√2 sq.units.