Here we provide RD Sharma Class 12 Ex 21.4 Solutions Chapter 21 Areas of Bounded Regions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 21.4 Solutions Chapter 21 Areas of Bounded Regions book pdf download. Now you will get step-by-step solutions to each question.
| Textbook | NCERT |
| Class | Class 12th |
| Subject | Maths |
| Chapter | 21 |
| Exercise | 21.4 |
| Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 21.4 Solutions Chapter 21 Areas of Bounded Regions
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.svg "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.svg "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.svg "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.svg "Rendered by QuickLaTeX.com")
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