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If n is a positive odd integer, then int...

If n is a positive odd integer, then `int |x^n| dx=`

A

`|(x^(n+1))/(n+1)|+C`

B

`(x^(n+1))/(n+1)+C`

C

`(|x^(n)|)/(n+1)+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
c
We have the following cases :
CASE I When `xge0`
In this case , we have , `int|x^(n)|dx=int|x|^(n)dx`
`rArrint|x^(n)|dx=intx^(n)dx " " [because|x|=x]`
`rArrint|x^(n)|dx=(x^(n+1))/(n+1)+C`
`rArrint|x^(n)|dx=(|x|^(n)x)/(n+1)+C " " [becausexge0therefore|x|=x]`
CASE II When ` xle0`
In this case , we have
`|x| =-x`
`thereforeint|x^(n)|dx= int|x|^(n)dx=int (-x)^(n)dx`
`rArrint|x^(n)|dx=-intx^(n)dx " " [because n is odd"]`
`rArrint|x^(n)|dx=-(x^(n)+1)/(n+1)+C`
`rArr int|x^(n)|dx=((-x)^(n)x)/(n+1)+C " " [{:(because " n is odd",),(therefore(-x)^(n)=-x^(n),):}]`
`rArrint|x^(n)|dx=(|x|^(n)x)/(n+1)+C`
Hence , `int|x^(n)|dx=(|x|^(n)x)/(n+1)+C`
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