Exercise 8.1 answers: Application of Integrals

Open NCERT p. 296Matches NCERT’s answer

1. Compare x²/16 + y²/9 = 1 with the standard form x²/a² + y²/b² = 1.
2. So a² = 16 and b² = 9, which gives a = 4 and b = 3.
3. The area of a full ellipse is π × a × b.
4. π × 4 × 3 = 12π square units.

Answer12π square units

Watch this explained “Make the formula prove itself”, 17:44 into Recovering the areas of a circle and an ellipse by integration, using their own symmetry

Open NCERT p. 296Matches NCERT’s answer

1. Compare x²/4 + y²/9 = 1 with the standard form x²/a² + y²/b² = 1.
2. So a² = 4 and b² = 9, which gives a = 2 and b = 3.
3. The area of a full ellipse is π × a × b, whichever semi-axis is bigger.
4. π × 2 × 3 = 6π square units.

Answer6π square units

Watch this explained “Make the formula prove itself”, 17:44 into Recovering the areas of a circle and an ellipse by integration, using their own symmetry

Open NCERT p. 296Checked by computer

1. x² + y² = 4 is a circle of radius 2, centred at the origin.
2. The region in the first quadrant between x = 0 and x = 2, under this circle, is exactly one quarter of the whole circular disc.
3. Area of the full circle = πr² = π × 2² = 4π.
4. Area of one quarter = 4π ÷ 4 = π.

Answerπ square units, option (A).

Watch this explained “The same argument, as a question”, 11:40 into Recovering the areas of a circle and an ellipse by integration, using their own symmetry

Open NCERT p. 296Checked by computer

1. y² = 4x means x = y²/4.
2. The region lies between the y-axis and the curve, from y = 0 to y = 3, so use strips lying flat against the y-axis.
3. Area = ∫ from 0 to 3 of (y²/4) dy.
4. ∫ y²/4 dy = y³/12; putting in the limits gives 27/12 − 0 = 9/4.

Answer9/4 square units, option (B).

Watch this explained “The one where reading is the work”, 12:18 into Choosing vertical or horizontal strips, and integrating in the matching variable