Question 30 - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at Oct. 24, 2019 by Teachoo

Evaluate β« |x
^{
2
}
- 2x| Β dx from 1 to 3

Note
: - This
is similar to
Example 30 of NCERT β Chapter 7 Class 12

Check the answer here
https://www.teachoo.com/4811/727/Example-30---Evaluate-integral--1----2--x3---x--dx/category/Examples/

Transcript
Question 30 Evaluate β« 3 1 |π₯^2β2π₯| dx
|π₯^2β2π₯|=|π₯(π₯β2)|
=|π₯| |π₯β2|
Thus, π₯=0, π₯=2
Since our integration is from 1 to 3, we ignore x = 0
β΄ |π₯^2β2π₯|= {(π₯Γβ(π₯β2) ππ 1β€π₯<2π₯Γ(π₯β2) ππ 2β€π₯<3)β€
|π₯^2β2π₯|= {(β(π₯^2β2π₯) ππ 1β€π₯<2(π₯^2β2π₯) ππ 2β€π₯<3)β€
Now,
β«_1^3 |π₯^2β2π₯| dx
= ββ«_1^2β(π₯^2β2π₯) ππ₯+β«_2^3β(π₯^2β2π₯) ππ₯
= ββ«_1^2βπ₯^2 ππ₯+β«_1^2β2π₯ ππ₯+β«_2^3βπ₯^2 ππ₯ββ«_2^3β2π₯ ππ₯
= ββ«_1^2βπ₯^2 ππ₯+β«_2^3βπ₯^2 ππ₯+β«_1^2β2π₯ ππ₯ββ«_2^3β2π₯ ππ₯
= β[π₯^3/3]_1^2+[π₯^3/3]_2^3+[π₯^2 ]_1^2β[π₯^2 ]_2^3
= β[2^3/3β1^3/3]+[3^3/3β2^3/3]+[2^2β1^2 ]β[3^2β2^2 ]
= β[8/3β1/3]+[27/3β8/3]+[4β1]β[9β4]
= β[7/3]+[19/3]+[3]β[5]
= 12/3β2
= 4β2
= 2

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