Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Example 41 - Chapter 7 Class 12 Integrals
Last updated at May 29, 2023 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii)
Example 2 (iii) Important
Example 3 (i)
Example 3 (ii) Important
Example 3 (iii)
Example 4
Example 5 (i)
Example 5 (ii)
Example 5 (iii) Important
Example 5 (iv) Important
Example 6 (i)
Example 6 (ii) Important
Example 6 (iii) Important
Example 7 (i)
Example 7 (ii) Important
Example 7 (iii)
Example 8 (i)
Example 8 (ii) Important
Example 9 (i)
Example 9 (ii) Important
Example 9 (iii) Important
Example 10 (i)
Example 10 (ii) Important
Example 11
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22 Important
Example 23
Example 24
Example 25 (i)
Example 25 (ii) Important
Example 25 (iii)
Example 25 (iv) Important
Example 26
Example 27
Example 28 Important
Example 29
Example 30
Example 31
Example 32 Important
Example 33 Important
Example 34 Important
Example 35
Example 36 Important
Example 37 Important
Example 38 Important
Example 39 Important
Example 40 Important
Example 41 Important You are here
Example 42 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 41 (Introduction) Evaluate β«_(β1)^(3/2)β|π₯ sinβ‘(π π₯) | ππ₯ To find sign of |π₯ sinβ‘(π π₯) | in the interval, let us check sign of x and sinβ‘γ (ππ₯) γseparately π₯ > 0 & π₯ sinβ‘γ (ππ₯) γ> 0 π₯ < 0 & π₯ sinβ‘γ (ππ₯) γ< 0 π πππβ‘γ(π π) γ< 0 π₯ < 0 & π₯ sinβ‘γ (ππ₯) γ> 0 π₯ > 0 & π₯ sinβ‘γ (ππ₯) γ< 0 Sign of x We have Interval β1< π₯ < 3/2 Sign of π¬π’π§β‘γ(π π)γ Here, β1β€π₯β€3/2 βπ β€ππ₯β€3π/2 Let π=ππ₯ β΄ βπ β€ΞΈβ€3π/2 From graph it is seen, Thus, Sign for π¬π’π§β‘γ(π π)γ is Now, Sign for π₯ sinβ‘(ππ₯) in interval β1 β€ x β€ 3/2 is So, we can write |π₯ π ππ (Οπ₯)|={β(π₯ sinβ‘γ (ππ₯)γ β1 β€ π₯ β€ 1@&βπ₯ sinβ‘γ (ππ₯)γ 1 β€ π₯ β€ 3/2)β€ Example 41 Evaluate β«_(β1)^(3/2)β|π₯ sinβ‘(π π₯) | ππ₯ Hence, |π₯ π ππ (Οπ₯)|={β(π₯ sinβ‘γ (ππ₯)γ β1 β€ π₯ β€ 1@&βπ₯ sinβ‘γ (ππ₯)γ 1 β€ π₯ β€ 3/2)β€ Therefore, we can write β΄ β«_(β1)^(3/2)β|π₯ sinβ‘(ππ₯) | ππ₯ = β«_(β1)^1βγπ₯ sinβ‘γ (ππ₯) ππ₯ ββ«_1^(3/2)βγπ₯ π ππγγ γ(ππ₯) ππ₯ Solving β«1βγπ πππβ‘γ (π π) π πγ γ separately β«1βγπ₯ sinβ‘γ (ππ₯) ππ₯γ γ = x β«1βsinβ‘γ(ππ₯)ββ«1β(π(π₯)/ππ₯ β«1βsinβ‘γ(ππ₯) γ ) γ ππ₯ = x ((βcosβ‘γ(ππ₯))γ)/πββ«1β1((βcosβ‘γ(ππ₯)γ)/π) ππ₯ = (β π₯ cosβ‘γ(ππ₯) γ)/π+β«1βcosβ‘γ(ππ₯)γ/π ππ₯ = (βπ₯ cosβ‘γ(ππ₯)γ)/π+ sinβ‘γ(ππ₯)γ/π^2 Using by parts β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = x & g(x) = sin (πx) Putting limits β«_(βπ)^πβγπ π¬π’π§β‘γπ π π πγ γ = [(βπ₯ cosβ‘γ(ππ₯)γ)/π+sinβ‘γ(ππ₯)γ/π^2 ]_(β1)^1 = ((β1 cosβ‘π)/π+sinβ‘π/π^2 ) β((β(β1)cosβ‘γ(βπ)γ)/π+sinβ‘γ(βπ)γ/π^2 ) = ((β1 Γ (β1))/π+0/π^2 ) β(cosβ‘π/π+γβ sinγβ‘π/π^2 ) = (1/π+0)β((β1)/π+0) = 1/π+1/π = 2/π Putting limits β«_π^(π/π)βγπ π¬π’π§β‘γπ π π πγ γ = [(βπ₯ cosβ‘γ(ππ₯)γ)/π+sinβ‘γ(ππ₯)γ/π^2 ]_1^(3/2) = (((β3)/2 cosβ‘(3π/2))/π+sinβ‘((3π )/2)/π^2 )β((β1 cosβ‘π)/π+sinβ‘π/π^2 ) = (0+((β1))/π^2 )β((β1 Γ β1)/π+0/π^2 ) = (β1)/π^2 β1/π Now, β΄ β«_(β1)^(3/2)β|π₯ sinβ‘(ππ₯) | ππ₯ = β«_(β1)^1βγπ₯ sinβ‘γ (ππ₯) ππ₯ ββ«_1^(3/2)βγπ₯ π ππγγ γ(ππ₯) ππ₯ = 2/πβ((β1)/π^2 β1/π) = 2/π+1/π^2 +1/π = π/π +π/π ^π
