# Example 15 - Chapter 8 Class 12 Application of Integrals

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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≤ 𝒙𝟐+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= 𝑥2+1 & 𝑦=𝑥+1 𝑥2+1=𝑥+1 𝑥2−𝑥+1−1=0 𝑥2−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 = 01𝑦 𝑑𝑥 Here, 𝑦→ equation of Parabola PQ 𝑦= 𝑥2+1 ∴ Area OPQT = 01 𝑥2+1 = 𝑥33+𝑥01 = 133+1− 033+0 = 13+1 = 43 Area QRST Area QRST= 12𝑦 𝑑𝑥 Here, 𝑦→ equation of line QP 𝑦=𝑥 + 1 ∴ Area QRST= 12 𝑥+1 𝑑𝑥 = 𝑥22+𝑥12 = 222+2− 122+1 =2+2− 12+1 =4− 32 = 52 Thus, Area Required = Area OPQT + Area QPST = 43+ 52 = 8 + 156 = 𝟐𝟑𝟔

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.