Definite Integrals Part 1

(b) ‌ Let ‌L‌‌=

lim

n→∞

[‌

1

n

+‌

n

(n+1)2

+‌

n

(n+2)2

+...]
‌ or ‌L‌‌=

lim

n→∞

[‌

n

(2n−1)2

]‌

n

(n+0)2

+‌

n

(n+1)2

+...+
‌‌=

lim

n→∞

[‌

n

(n+n−1)2

+‌

n

(n+n)2

−‌

n

(n+n)2

]
‌‌−

lim

n→∞

[‌

n

(n+n)2

]
‌=

lim

n→∞

‌

n

∑

r=0

‌

n

(n+1)2

+...+‌

n

(n+n)2

]
‌‌=

lim

n→∞

‌

n

∑

r=0

‌

n

(n+r)2

−0(since,

lim

n→∞

‌

1

n

=0)
Now, for solving limit summation, we integrate it using some replacement.
L=

lim

n→∞

‌

n

∑

r=0

‌

1

r(1+r∕n)

Take ‌

r

n

as x and ‌

1

n

as dx.
Lower limit is obtained by putting r=0 in ‌

r

n

, we get Lower limit =0
Upper limit is obtained by putting r=n in ‌

r

n

, we get
Upper limit = 1
∴‌‌L=

1

∫

0

‌

1

(1+x)2

dx=‌

−1

(1+x)

]01=−(‌

1

2

−1)=‌

1

2

∴‌‌L=‌

1

2