Area bounded by curve and horizontal or vertical line

Chapter 8 Class 12 Application of Integrals
Concept wise

### Transcript

Example 3 Find the area of region bounded by the line 𝑦=3𝑥+2, the 𝑥−𝑎𝑥𝑖𝑠 and the ordinates 𝑥=−1 and 𝑥=1 First Plotting 𝑦=3𝑥+2 In graph Now, Area Required = Area ACB + Area ADE Area ACB Area ACB = ∫_(−1)^((−2)/( 3))▒〖𝑦 𝑑𝑥〗 𝑦→ equation of line Area ACB = ∫_(−𝟏)^((−𝟐)/( 𝟑))▒〖(𝟑𝒙+𝟐) 𝒅𝒙〗 Since Area ACB is below x-axis, it will come negative , Hence, we take modulus Area ACB = |∫_(−1)^((−2)/( 3))▒〖(3𝑥+2) 𝑑𝑥〗| = |[𝟑 𝒙^𝟐/𝟐+𝟐𝒙]_(−𝟏)^((−𝟐)/𝟑) | = |" " [3/2 ((−2)/3)^2+2×−2/3]| − [3/2 (−1)^2+2(−1)] = |" " [3/2×4/9−4/3]−[3/2−2]| = |(−2)/3−(−1/2)| = |(−2)/3+1/2| = |(−𝟏)/𝟔| = 𝟏/𝟔 square units Area ADE Area ADE = ∫1_((−𝟐)/𝟑)^𝟏▒〖𝒚 𝒅𝒙〗 y → equation of line = ∫1_((−𝟐)/𝟑)^𝟏▒(𝟑𝒙+𝟐)𝒅𝒙 = [(3𝑥^2)/2+2𝑥]_((−2)/3)^1 =[(3〖(1)〗^2)/2+2×1] − [3/2 ((−2)/3)^2+2×((−2)/3)] = [3/2+2] − [2/3−4/3] = 7/2+2/3 = 𝟐𝟓/𝟔 square units Thus, Required Area = Area ACB + Area ADE = 1/6 + 25/6 = 26/6 = 𝟏𝟑/𝟑 square units

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.