UPSEE 2017 Solved Paper

Given, f(x)=

2

ex+1

We know, if f(x) is continuous function on [a,b],
then
Mean value of the function,
M=

1

b−a

‌

b

∫

a

f(x)dx
∴ Mean value of the given function
M=

1

2−0

‌

2

∫

0

2

ex+1

dx
=

2

∫

0

e−x

1+e−x

dx
Let 1+e−x=t
On differentiating
e−x(−1)dx=dt
∴ M=−

1+ 1 e2

∫

2

dt

t

=

∫

1+ 1 e2

dt

t

[By changing limit of integration]
=[log|t|]21+

1

e2

=log‌2−log|1+

1

e2

|
=log‌2−log(1+

1

e2

)
[Here, 1+

1

e2

is always positive. So, we can omit modulus size]
=log‌2−log(

1+e2

e2

)
=log‌2−log(1+e2)+log‌e2
=log‌2−log(1+e2)+2
=2+log(

2

1+e2

)