NCERT Questions for Class 12 Maths Chapter 8 – Application of Integrals

Important Questions with Solutions of Class 12 Maths Chapter 8 – Application of Integrals

1) Determine the area included within the ellipse

Ans – Given that the ellipse exhibits symmetry with respect to both the x-axis and the y-axis

∴ Area of ellipse = 4 × Area of OAB

we know that,

As OAB resides in the first quadrant, the value of y is positive.

It is of the form

Therefore, the required area is πab square units.

2) Determine the area of the smaller region enclosed by the

Ans – The specified ellipse is defined by the equation x²/a² + y²/b² = 1, whereas the line is characterised by the equation x/a + y/b = 1.

3) Using integration, determine the area of the region enclosed by the triangle with vertices at (-1,0), (1,3), and (3,2).

Ans –

We require the equations of the lines for segments AB, BC, and AC, given the coordinates (-1,0) and (1,3).

For line BC, the points are (1, 3) and (3, 2).
For line AC, the points are (-1, 0) and (3, 2).

The required area is the shaded region, which is defined as the sum of the area under line AB from x = -1 to x = 1, plus the area under BC from x = 1 to x = 3, minus the area under AC from x = -1 to x = 3.

4) Determine the area of the region defined by the conditions { (x, y) : x² + y² ≤ 1 and x + y ≥ 1 }.

Ans – The provided curves are {(x:y): x² + y² < 1 ≤ x + y}.

5) Determine the area bounded by the curve y = sin x, the x-axis, and the vertical lines x = 0 and x = 3π/2.

Ans –

Required area = Area of OAB + Area of BC

Area OAB

= [-cos π -cos 0]

= -[-1-1]

= -[-2]

= 2

Area BC

As area cannot be negative, Area BC equals 1.

Therefore, Area Required = Area OAB + Area BC

=2+1

=3