Example 15 - Chapter 8 Class 12 Application of Integrals

Transcript

Example 15 Find the area of the region {(𝑥, 𝑦) : 0 ≤ 𝑦 ≤ 𝑥2 + 1, 0 ≤ 𝑦 ≤ 𝑥 + 1, 0 ≤ 𝑥 ≤ 2} Step 1: 𝟎≤𝒚≤ 𝒙﷮𝟐﷯+𝟏 𝟎≤𝒚≤𝒙+𝟏 Also 𝟎≤𝒙≤𝟐 Since 𝑦≥0 & 0≤𝑥≤2 So, our figure is Finding point of intersection P & Q Solving 𝑦= 𝑥﷮2﷯+1 & 𝑦=𝑥+1 𝑥﷮2﷯+1=𝑥+1 𝑥﷮2﷯−𝑥+1−1=0 𝑥﷮2﷯−𝑥+0=0 𝑥 𝑥−1﷯=0 So, 𝑥=0 , 𝑥=1 Finding area Area required = Area OPQRST Area OPQRST = Area OPQT + Area QRST Area OPQT Area OPQT = 0﷮1﷮𝑦 𝑑𝑥﷯ Here, 𝑦→ equation of Parabola PQ 𝑦= 𝑥﷮2﷯+1 ∴ Area OPQT = 0﷮1﷮ 𝑥﷮2﷯+1﷯﷯ = 𝑥﷮3﷯﷮3﷯+𝑥﷯﷮0﷮1﷯ = 1﷮3﷯﷮3﷯+1− 0﷮3﷯﷮3﷯+0﷯ = 1﷮3﷯+1 = 4﷮3﷯ Area QRST Area QRST= 1﷮2﷮𝑦 𝑑𝑥﷯ Here, 𝑦→ equation of line QP 𝑦=𝑥 + 1 ∴ Area QRST= 1﷮2﷮ 𝑥+1﷯﷯ 𝑑𝑥 = 𝑥﷮2﷯﷮2﷯+𝑥﷯﷮1﷮2﷯ = 2﷮2﷯﷮2﷯+2− 1﷮2﷯﷮2﷯+1﷯ =2+2− 1﷮2﷯+1﷯ =4− 3﷮2﷯ = 5﷮2﷯ Thus, Area Required = Area OPQT + Area QPST = 4﷮3﷯+ 5﷮2﷯ = 8 + 15﷮6﷯ = 𝟐𝟑﷮𝟔﷯