Exercise 8.1 — Application of Integrals

1. 13 marksNCERT Exercise

   Find the area of the region bounded by the ellipse x^2/16+y^2/9=1.

   Hint. Identify a and b from the denominators, then apply Area=pi.a.b directly.

   Here a^2=16 so a=4, and b^2=9 so b=3. Since the area of any ellipse in standard position is pi.a.b, substituting gives the result immediately.

   ✦ 12*pi square units

2. 23 marksNCERT Exercise

   Find the area of the region bounded by the ellipse x^2/4+y^2/9=1.

   Hint. Identify a and b from the denominators — note this ellipse is taller than it is wide, but the area formula is unaffected.

   Here a^2=4 so a=2, and b^2=9 so b=3; even though b is larger than a here (the ellipse is taller than wide), the formula Area=pi.a.b still applies directly.

   ✦ 6*pi square units

3. 33 marksNCERT Exercise

   Choose the correct answer: area lying in the first quadrant and bounded by the circle x^2+y^2=4 and the lines x=0 and x=2 is (A) pi (B) pi/2 (C) pi/3 (D) pi/4

   Hint. This region is exactly one quarter of the full circle of radius 2, bounded by the two axes and the arc.

   Since the circle has radius r=2, its full area is pi.r^2=4pi; the region described is exactly one quadrant of the circle, so dividing by 4 gives the answer.

   ✦ (A) pi

4. 43 marksNCERT Exercise

   Choose the correct answer: area of the region bounded by the curve y^2=4x, y-axis, and the line y=3 is (A) 2 (B) 9/4 (C) 9/3 (D) 9/2

   Hint. Since the curve is naturally x in terms of y here, use the y-axis area formula: Area = integral of x dy.

   Solving for x gives x=y^2/4, so the area is integral from 0 to 3 of (y^2/4)dy = [y^3/12] from 0 to 3 = 27/12, which simplifies to the answer.

   ✦ (B) 9/4

Solutions written by the tuition.in editorial team and checked against lemh202.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.