1.

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

Answer» Correct Answer - 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`


Discussion

No Comment Found

Related InterviewSolutions