Examples

Chapter 7 Class 12 Integrals
Serial order wise

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Example 41 (Introduction) Evaluate β«_(β1)^(3/2)β|π₯ sinβ‘(π π₯) | ππ₯ To find sign of |π₯ sinβ‘(π π₯) | in the interval, let us check sign of x and sinβ‘γ (ππ₯) γseparately π₯ > 0 & π₯ sinβ‘γ (ππ₯) γ> 0 π₯ < 0 & π₯ sinβ‘γ (ππ₯) γ< 0 π πππβ‘γ(ππ) γ< 0 π₯ < 0 & π₯ sinβ‘γ (ππ₯) γ> 0 π₯ > 0 & π₯ sinβ‘γ (ππ₯) γ< 0 Sign of x We have Interval β1< π₯ < 3/2 Sign of π¬π’π§β‘γ(ππ)γ Here, β1β€π₯β€3/2 βπ β€ππ₯β€3π/2 Let π=ππ₯ β΄ βπ 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‘(ππ₯) in interval β1 β€ x β€ 3/2 is So, we can write |π₯ π ππ (Οπ₯)|={β(π₯ sinβ‘γ (ππ₯)γ β1 β€ π₯ β€ 1@&βπ₯ sinβ‘γ (ππ₯)γ 1 β€ π₯ β€ 3/2)β€ Example 41 Evaluate β«_(β1)^(3/2)β|π₯ sinβ‘(π π₯) | ππ₯ Hence, |π₯ π ππ (Οπ₯)|={β(π₯ sinβ‘γ (ππ₯)γ β1 β€ π₯ β€ 1@&βπ₯ sinβ‘γ (ππ₯)γ 1 β€ π₯ β€ 3/2)β€ Therefore, we can write β΄ β«_(β1)^(3/2)β|π₯ sinβ‘(ππ₯) | ππ₯ = β«_(β1)^1βγπ₯ sinβ‘γ (ππ₯) ππ₯ ββ«_1^(3/2)βγπ₯ π ππγγ γ(ππ₯) ππ₯ Solving β«1βγπ πππβ‘γ (ππ) ππγ γ separately β«1βγπ₯ sinβ‘γ (ππ₯) ππ₯γ γ = x β«1βsinβ‘γ(ππ₯)ββ«1β(π(π₯)/ππ₯ β«1βsinβ‘γ(ππ₯) γ ) γ ππ₯ = x ((βcosβ‘γ(ππ₯))γ)/πββ«1β1((βcosβ‘γ(ππ₯)γ)/π) ππ₯ = (β π₯ cosβ‘γ(ππ₯) γ)/π+β«1βcosβ‘γ(ππ₯)γ/π ππ₯ = (βπ₯ cosβ‘γ(ππ₯)γ)/π+ sinβ‘γ(ππ₯)γ/π^2 Using by parts β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = x & g(x) = sin (πx) Putting limits β«_(βπ)^πβγπ π¬π’π§β‘γππ ππγ γ = [(βπ₯ cosβ‘γ(ππ₯)γ)/π+sinβ‘γ(ππ₯)γ/π^2 ]_(β1)^1 = ((β1 cosβ‘π)/π+sinβ‘π/π^2 ) β((β(β1)cosβ‘γ(βπ)γ)/π+sinβ‘γ(βπ)γ/π^2 ) = ((β1 Γ (β1))/π+0/π^2 ) β(cosβ‘π/π+γβ sinγβ‘π/π^2 ) = (1/π+0)β((β1)/π+0) = 1/π+1/π = 2/π Putting limits β«_π^(π/π)βγπ π¬π’π§β‘γππ ππγ γ = [(βπ₯ cosβ‘γ(ππ₯)γ)/π+sinβ‘γ(ππ₯)γ/π^2 ]_1^(3/2) = (((β3)/2 cosβ‘(3π/2))/π+sinβ‘((3π )/2)/π^2 )β((β1 cosβ‘π)/π+sinβ‘π/π^2 ) = (0+((β1))/π^2 )β((β1 Γ β1)/π+0/π^2 ) = (β1)/π^2 β1/π Now, β΄ β«_(β1)^(3/2)β|π₯ sinβ‘(ππ₯) | ππ₯ = β«_(β1)^1βγπ₯ sinβ‘γ (ππ₯) ππ₯ ββ«_1^(3/2)βγπ₯ π ππγγ γ(ππ₯) ππ₯ = 2/πβ((β1)/π^2 β1/π) = 2/π+1/π^2 +1/π = π/π+π/π^π