Find the point of intersection of the curves y=ln(x+2) and y=2+ln(2-x)
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Answer:
Correct option is
C
3−e
Given curves, y=lnx & y=(lnx)
2
Points of intersection of given curves are, (1,0) and (e,1)
Hence required area is,
A=∫
1
e
(lnx−(lnx)
2
)dx
On solving it by parts we get
A=[xlnx−x]
1
e
−[x(lnx)
2
−∫
1
e
2lnxdx]
A=[xlnx−x]
1
e
−[x(lnx)
2
−2xlnx+2x]
1
e
A=[xlnx−x−x(lnx)
2
+2xlnx−2x]
1
e
A=∣3x(lnx−1)∣
1
e
−∣x(lnx)
2
∣
1
e
=3−e
Step-by-step explanation:
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