Definite Integrals Part 2

  Let y=

lim

n→∞

(Un)

−4

n2

Taking log on both sides, we get
ln‌y=

lim

n→∞

‌

n

∑

‌r=1

[

−4

n2

.r‌ln(1+

r2

n2

)]
Now, replace

lim

n→∞

Σ→∫

r

n

→x,

1

n

→dx
Lower limit = 0
Upper limit = 1
∴ln‌y=∫01−4x‌ln(1+x2)‌d‌x
Let 1+x2=t
⇒xdx=

dt

2

When x→ 0, t → 1
and x → 1, t → 2
∴ln‌y=∫12−2‌ln‌t‌d‌t
=−2(t‌ln‌t−t)12
=−2(2‌ln‌2−2+1)
=−2(2‌ln‌2−1)
⇒ln‌y=ln‌

1

16

+2
⇒y=

1

16

e2