Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 7.8

Ex 7.8, 1

Ex 7.8, 2

Ex 7.8, 3

Ex 7.8, 4 Important

Ex 7.8, 5

Ex 7.8, 6

Ex 7.8, 7

Ex 7.8, 8 Important

Ex 7.8, 9

Ex 7.8, 10

Ex 7.8, 11 Important

Ex 7.8, 12

Ex 7.8, 13

Ex 7.8, 14 Important You are here

Ex 7.8, 15

Ex 7.8, 16 Important

Ex 7.8, 17 Important

Ex 7.8, 18

Ex 7.8, 19 Important

Ex 7.8, 20 Important

Ex 7.8, 21 (MCQ) Important

Ex 7.8, 22 (MCQ)

Last updated at June 13, 2023 by Teachoo

Ex 7.8, 14 ∫_0^1▒(2𝑥 + 3)/(5𝑥2 + 1) dx Let F(𝒙)=∫1▒(2𝑥 + 3)/(5𝑥^2 + 1) 𝑑𝑥 =∫1▒2𝑥/(5𝑥^2 + 1) 𝑑𝑥+∫1▒3/(5𝑥^2 + 1) 𝑑𝑥 Solving ∫1▒𝟐𝒙/(𝟓𝒙^𝟐 + 𝟏) 𝒅𝒙 Put 𝑥^2=𝑡 Differentiating w.r.t.𝑥 2𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/2𝑥 Hence ∫1▒〖(2𝑥 )/(5𝑥^2 + 1) 𝑑𝑥=∫1▒〖2𝑥/(5𝑡+1) 𝑑𝑡/2𝑥〗〗 =∫1▒𝑑𝑡/(5𝑡+1) =1/5 𝑙𝑜𝑔|5𝑡+1| =𝟏/𝟓 𝒍𝒐𝒈|𝟓𝒙^𝟐+𝟏| Integrating ∫1▒〖𝟑/(𝟓𝒙^𝟐+𝟏) 𝒅𝒙〗 ∫1▒〖3/(5𝑥^2+1) 𝑑𝑥〗 =3∫1▒𝑑𝑥/(5𝑥^2+1) =3/5 ∫1▒𝑑𝑥/(𝑥^2 + 1/5) =3/5 ∫1▒𝑑𝑥/(𝑥^2 +(1/√5)^2 ) =3/5×(1/1)/√5 tan^(−1) ((𝑥/1)/√5) =3/5×√5 〖𝑡𝑎𝑛〗^(−1) (√5 𝑥) =𝟑/𝟓 〖𝒕𝒂𝒏〗^(−𝟏) (√𝟓 𝒙) Hence, F(𝒙)=∫1▒2𝑥/(5𝑥^2 + 1) 𝑑𝑥+∫1▒3/(5𝑥^2 + 1) 𝑑𝑥 =𝟏/𝟓 𝒍𝒐𝒈|𝟓𝒙^𝟐+𝟏|+𝟑/√𝟓 〖𝒕𝒂𝒏〗^(−𝟏) (√𝟓 𝒙) Now, ∫_0^1▒〖(2𝑥 + 3)/(〖5𝑥〗^2+ 1) 𝑑x=𝐅(𝟏)−𝐅(𝟎)〗 =1/5 𝑙𝑜𝑔|5〖×1〗^2+1|+3/√5 〖𝑡𝑎𝑛〗^(−1) (√5 𝑥) −[1/5 𝑙𝑜𝑔|5×0+1|+3/√5 〖𝑡𝑎𝑛〗^(−1) (√5×0)] =1/5 |6|+3/√5 〖𝑡𝑎𝑛〗^(−1) √5−1/5 𝑙𝑜𝑔1+3/5 〖𝑡𝑎𝑛〗^(−1) 0 =1/5 𝑙𝑜𝑔 6+3/√5 〖𝑡𝑎𝑛〗^(−1) √5−1/5×0+3/5×0 =𝟏/𝟓 𝒍𝒐𝒈 𝟔+𝟑/√𝟓 〖𝒕𝒂𝒏〗^(−𝟏) √𝟓