Q10Short answer
– x I= ∫ dx ...(1) 2 x + 10 – x 10 – (10 – x) = ∫ 10 – x + 10 – (10 – x) dx by (P3) ⇒ I=∫ dx (2) 2 10 – x + x Adding (1) and (2), we get 2I = ∫ 1dx = 8 – 2 = 6 Hence I=3 π
– x I= ∫ dx ...(1) 2 x + 10 – x 10 – (10 – x) = ∫ 10 – x + 10 – (10 – x) dx by (P3) ⇒ I=∫ dx (2) 2 10 – x + x Adding (1) and (2), we get 2I = ∫ 1dx = 8 – 2 = 6 Hence I=3 π