# NCERT Solutions class 12 Maths Exercise 7.8 (Ex 7.8) Chapter 7 Integrals

## NCERT Solutions for Class 12 Maths Exercise 7.8 hapter 7 Integrals – FREE PDF Download

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# NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.8) Exercise 7.8

Evaluate the following definite integrals as limit of sums:

1.

Ans. We know that

abf(x)dx=limh0[f(a)+f(a+h)+f(a+2h)+...+f(a+(n1)h)]∫abf(x)dx=limh→0⁡[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]

where

Here,  and

2.

Ans. We know that

abf(x)dx=limh0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n1)h)]∫abf(x)dx=limh→0⁡h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]

where

Here,  and

=05(x+1)dx=limh0h[1+(h+1)+(2h+1)+...+((n1)h+1)]=∫05(x+1)dx=limh→0⁡h[1+(h+1)+(2h+1)+…+((n−1)h+1)]

=limh0h[n+h(1+2+3+...+(n1))]=limh→0⁡h[n+h(1+2+3+…+(n−1))]

=limh0[nh+h2n(n1)2]=limh→0⁡[nh+h2n(n−1)2]

=limh0[nh+nh(nhh)2]=limh→0⁡[nh+nh(nh−h)2]

=

3.

Ans. We know that

abf(x)dx=limh0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n1)h)]∫abf(x)dx=limh→0⁡h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]

where

Here,  and

=23x2dx=limh0h[4+(4+4h+h2)+(4+8h+22h2)+...+(4+4(n1)h+(n1)2.h2)]=∫23x2dx=limh→0⁡h[4+(4+4h+h2)+(4+8h+22h2)+…+(4+4(n−1)h+(n−1)2.h2)]

=limh0h[(4+4+...n times)+4h(1+2+...+(n1))+h2(12+22+....+(n1)2)]=limh→0⁡h[(4+4+…n times)+4h(1+2+…+(n−1))+h2(12+22+….+(n−1)2)]

=limh0h[4n+4hn(n1)2+h2n(n1)(2n1)6]=limh→0⁡h[4n+4hn(n−1)2+h2n(n−1)(2n−1)6]

=limh0[4nh+2nh(nhh)+nh(nhh)(2nhh)6]=limh→0⁡[4nh+2nh(nh−h)+nh(nh−h)(2nh−h)6]

=

4.

Ans. We know that

abf(x)dx=limh0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n1)h)]∫abf(x)dx=limh→0⁡h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]

where

Here,  and

=14(x2x)dx=limh0h[0+h+h2+2h+22h2+.....+(n1)h+(n1)2.h2)]=∫14(x2−x)dx=limh→0⁡h[0+h+h2+2h+22h2+…..+(n−1)h+(n−1)2.h2)]

=limh0h[h(1+2+3+..+(n1))+h2(1++22+32.....+(n1)2)]=limh→0⁡h[h(1+2+3+..+(n−1))+h2(1++22+32…..+(n−1)2)]

=limh0h[hn(n1)2+h2n(n1)(2n1)6]=limh→0⁡h[hn(n−1)2+h2n(n−1)(2n−1)6]

=limh0[nh(nhh)2+nh(nhh)(2nhh)6]=limh→0⁡[nh(nh−h)2+nh(nh−h)(2nh−h)6]

=

5.

Ans. We know that

abf(x)dx=limh0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n1)h)]∫abf(x)dx

=limh→0⁡h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]

where

Here,  and

=11exdx=limh0h[e1+e1eh+e1e2h+........+e1e(n1)h]

=∫−11exdx=limh→0⁡h[e−1+e−1eh+e−1e2h+……..+e−1e(n−1)h]

[ The series within brackets is a G.P. and ]

So, We get

=limh0 he1[(eh)n1]eh1=limh→0⁡ he−1[(eh)n−1]eh−1

=limh0 he1(enh1)eh1=limh→0⁡ he−1(enh−1)eh−1

=limh0 he1(e21)eh1=limh→0⁡ he−1(e2−1)eh−1

=e1e=e−1e

6.

Ans. We know that

abf(x)dx=limh0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n1)h)]∫abf(x)dx

=limh→0⁡h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]

where

Here,  and

=04(x+e2x)dx=limh0h[1+(h+e2h)+(2h+e4h)+........+((n1)h+e2(n1)h]=∫04(x+e2x)dx

=limh→0⁡h[1+(h+e2h)+(2h+e4h)+……..+((n−1)h+e2(n−1)h]

=limh0h[(h+2h+3h+......+(n1)h)+(1+e2h+e4h+.....+e2(n1)h]

=limh→0⁡h[(h+2h+3h+……+(n−1)h)+(1+e2h+e4h+…..+e2(n−1)h]

=limh0h[h(1+2+3+......+(n1))+(1+e2h+e4h+.....+e2(n1)h]

=limh→0⁡h[h(1+2+3+……+(n−1))+(1+e2h+e4h+…..+e2(n−1)h]

[ The series within brackets is a G.P. and ]

=limh0h[hn(n1)2+1((e2h)n1)e2h1]=limh→0⁡h[hn(n−1)2+1((e2h)n−1)e2h−1]

=limh0[nh(nh1)2+he2nh1e2h1]=limh→0⁡[nh(nh−1)2+he2nh−1e2h−1]