Class 12 RD Sharma Solutions- Chapter 21 Areas of Bounded Regions – Exercise 21.4
Question 1. Find the area of the region between the parabola x = 4y − y2 and the line x = 2y − 3.
Solution:
Area of the bounded region
![=\int_{-1}^{3}(4y-y^2-2y+3)dy\\ =[2\frac{y^2}{2}-\frac{y^3}{3}+3y]_{-1}^{3}\\ =9-9+9-1-\frac{1}{3}+3-\frac{(16a)^3}{48a}\\ =\frac{32}{3}sq.\ units](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-fe38a5f4699d156a306362885d245ff3_l3.png "Rendered by QuickLaTeX.com")
Question 2. Find the area bounded by the parabola x = 8 + 2y − y2; the y-axis and the lines y = −1 and y = 3.
Solution:
Area of the bounded region
![=\int_{-1}^{3}(5-0)dy+\int_{-1}^{3}8+2y-y^2-5\ dy\\ =[5y]_{-1}^{3}+[3y+y^2-\frac{y^3}{3}]_{-1}^{3}\\ =15+5+9+9-\frac{27}{3}+3-1-\frac{1}{3}\\ =\frac{92}{3}sq.\ units](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-4270a694de3595cc3456db501d61c48f_l3.png "Rendered by QuickLaTeX.com")
Question 3. Find the area bounded by the parabola y2 = 4x and the line y = 2x − 4.
(i) By using horizontal strips
(ii) By using vertical strips
Solution:
Area of the bounded region
![=\int_{-2}^{4}(\frac{y+4}{2}-\frac{y^2}{4})\ dy\\ =[\frac{y^2}{4}+2y-\frac{y^3}{12}]_{-2}^{4}\\ =4+8-\frac{16}{3}-1+4-\frac{2}{3}\\ =9\ sq.\ units](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-9f0448a9caf4a3074bc860802d420f8a_l3.png "Rendered by QuickLaTeX.com")
Question 4. Find the area of the region bounded the parabola y2 = 2x and straight line x − y = 4.
Solution:
Area of the bounded region
![=\int_{-2}^{4}(y+4-\frac{y^2}{2})\ dy\\ =[\frac{y^2}{2}+4y-\frac{y^3}{6}]_{-2}^{4}\\ =8+16-\frac{32}{3}-2+8-\frac{4}{3}\\ =18\ sq.\ units](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-61b96ca97f017f461b8dc1408f7873ab_l3.png "Rendered by QuickLaTeX.com")
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