Miscellaneous

Chapter 7 Class 12 Integrals
Serial order wise

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Misc 33 Evaluate the definite integral β«_1^4β[|π₯β1|+|π₯β2|+|π₯β3|] ππ₯ I=β«_1^4β[|π₯β1|+|π₯β2|+|π₯β3|] ππ₯ I=β«_1^4β|π₯β1| ππ₯+β«_1^4β|π₯β2| ππ₯+β«_1^4β|π₯β3| ππ₯ Solving ππ I1=β«_1^4β|π₯β1| ππ₯ We kow that |π₯β1|= {β( (π₯β1) πππ π₯β₯[email protected]β(π₯β1) πππ π₯<1)β€ Therefore, I1=β«_1^4β|π₯β1| ππ₯ I1=β«_1^4β(π₯β1) ππ₯ I1=β«_1^4βπ₯ ππ₯ββ«_1^4β1 ππ₯ I1=[π₯^2/2]_1^4β[π₯]_1^4 I1=((4)^2 β (1)^2)/2 β [4β1] I1=(16 β 1)/2 β [3] I1=15/2 β3 I1=(15 β 6)/2 I1=9/2 Solving ππ I2=β«_1^4β|π₯β2| ππ₯ We know that |π₯β2|= {β( (π₯β2) πππ π₯β₯[email protected]β(π₯β2) πππ π₯<2)β€ Therefore I2=β«_1^4β|π₯β2| ππ₯ I2=β«_1^2βγβ(π₯β2) γ ππ₯+β«_2^4β(π₯β2) ππ₯ I2=β«_1^2β(βπ₯+2) ππ₯+β«_2^4β(π₯β2) ππ₯ I2=β«_1^2βγβπ₯γ ππ₯+β«_1^2β2 ππ₯+β«_2^4βπ₯ ππ₯ββ«_2^4β2 ππ₯ I2=β[π₯^2/2]_1^2+2[π₯]_1^2+[π₯^2/2]_2^4β2[π₯]_2^4 I2=β[(4 β 1)/2]+2[2β1]+[(16 β 4)/2]β2[4β2] I2=β[3/2]+2[1]+12/2β2[2] I2= (β 3)/2 + 2+6β4 I2= (β3)/2 +8β4 I2= (β3)/2 +4 I2= (β 3 + 8)/2 I2= 5/2 Solving ππ I3=β«_1^4β|π₯β3| ππ₯ We know |π₯β3|= {β( (π₯β3) πππ π₯β₯[email protected]β(π₯β3) πππ π₯<3)β€ Therefore, I3=β«_1^4β|π₯β3| ππ₯ I3=β«_1^3βγβ(π₯β3) γ ππ₯+β«_3^4β(π₯β3) ππ₯ I3=β«_1^3β(βπ₯+3) ππ₯+β«_3^4β(π₯β3) ππ₯ I3=β«_1^3βγβπ₯γ ππ₯+β«_1^3β3 ππ₯+β«_3^4βπ₯ ππ₯ββ«_3^4β3 ππ₯ I3=β[π₯^2/2]_1^3+3[π₯]_1^3+[π₯^2/2]_3^4β3[π₯]_3^4 I3=β[(9 β 1)/2]+3[3 β1]+[(16 β 9)/2]β3[4β3] I3=(β 8)/2 +3[2]+ 7/2 β 3[1] I3=β4 +6+ 7/2 β 3 I3=β7 +6+ 7/2 I3=β1+ 7/2 I3= (β2 + 7)/2 I3= 5/2 Putting the values of I1 , I2 , I3 in (1) I=9/2 + 5/2 + 5/2 I = ππ/π