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Question

Let S be the area of the region enclosed by y=ex2,y=0,x=0 and x=1. Then

A
S14(1+1e)
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B
S11e
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C
S1e
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D
S12+1e(11e)
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Solution

The correct option is D S12+1e(11e)
x[0,1]x2xx2xex2ex
S=10ex2dx10exdx=[ex]10=11e
i.e., S11e option (2) is correct.

S is the area under the curve y=ex2
Hence, S area of the rectangle OPQS
i.e., S1×1e=1e
Hence, option (1) is correct.
S=10ex2dx
=120ex2dx+112ex2dx

Area A+ Area B
Now, Area A=1201.dx=12
Area B=1121edx=1e(112)
Hence, S12+1e(112)
Hence, option (4) is correct
Now, consider the integral e1dxx3/2=2[11e]
Substitute t=lnx
e1dxx3/2=10et/2dt
Now, for 0<x<1
ex2>ex/2
10ex2dx>2[112]
Now, 2[11e]14(1+1e)=7494e
=7e94e>0
So, option (3) is not correct.


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