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=limh→0[f(a)+f(a+h)+f(a+2h)+...+f(a+(n−1)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=limh→0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n−1)h)]∫abf(x)dx=limh→0h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]
where
Here, 
and 
=∫05(x+1)dx=limh→0h[1+(h+1)+(2h+1)+...+((n−1)h+1)]=∫05(x+1)dx=limh→0h[1+(h+1)+(2h+1)+…+((n−1)h+1)]
=limh→0h[n+h(1+2+3+...+(n−1))]=limh→0h[n+h(1+2+3+…+(n−1))]
=limh→0[nh+h2n(n−1)2]=limh→0[nh+h2n(n−1)2]
=limh→0[nh+nh(nh−h)2]=limh→0[nh+nh(nh−h)2]
= 
= 
= 
= 
3. 
Ans. We know that
∫abf(x)dx=limh→0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n−1)h)]∫abf(x)dx=limh→0h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]
where
Here, 
and 
=∫23x2dx=limh→0h[4+(4+4h+h2)+(4+8h+22h2)+...+(4+4(n−1)h+(n−1)2.h2)]=∫23x2dx=limh→0h[4+(4+4h+h2)+(4+8h+22h2)+…+(4+4(n−1)h+(n−1)2.h2)]
=limh→0h[(4+4+...n times)+4h(1+2+...+(n−1))+h2(12+22+....+(n−1)2)]=limh→0h[(4+4+…n times)+4h(1+2+…+(n−1))+h2(12+22+….+(n−1)2)]
=limh→0h[4n+4hn(n−1)2+h2n(n−1)(2n−1)6]=limh→0h[4n+4hn(n−1)2+h2n(n−1)(2n−1)6]
=limh→0[4nh+2nh(nh−h)+nh(nh−h)(2nh−h)6]=limh→0[4nh+2nh(nh−h)+nh(nh−h)(2nh−h)6]
= 
= 
= 
= 
4. 
Ans. We know that
∫abf(x)dx=limh→0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n−1)h)]∫abf(x)dx=limh→0h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]
where
Here, 
and 
=∫14(x2−x)dx=limh→0h[0+h+h2+2h+22h2+.....+(n−1)h+(n−1)2.h2)]=∫14(x2−x)dx=limh→0h[0+h+h2+2h+22h2+…..+(n−1)h+(n−1)2.h2)]
=limh→0h[h(1+2+3+..+(n−1))+h2(1++22+32.....+(n−1)2)]=limh→0h[h(1+2+3+..+(n−1))+h2(1++22+32…..+(n−1)2)]
=limh→0h[hn(n−1)2+h2n(n−1)(2n−1)6]=limh→0h[hn(n−1)2+h2n(n−1)(2n−1)6]
=limh→0[nh(nh−h)2+nh(nh−h)(2nh−h)6]=limh→0[nh(nh−h)2+nh(nh−h)(2nh−h)6]
= 
= 
= 
= 
5. 
Ans. We know that
∫abf(x)dx=limh→0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n−1)h)]∫abf(x)dx
=limh→0h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]
where
Here, 
and 
=∫−11exdx=limh→0h[e−1+e−1eh+e−1e2h+........+e−1e(n−1)h]
=∫−11exdx=limh→0h[e−1+e−1eh+e−1e2h+……..+e−1e(n−1)h]
[
The series within brackets is a G.P. and 
]
So, We get
=limh→0 he−1[(eh)n−1]eh−1=limh→0 he−1[(eh)n−1]eh−1
=limh→0 he−1(enh−1)eh−1=limh→0 he−1(enh−1)eh−1
=limh→0 he−1(e2−1)eh−1=limh→0 he−1(e2−1)eh−1
= 
= 
=e−1e=e−1e
6. 
Ans. We know that
∫abf(x)dx=limh→0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n−1)h)]∫abf(x)dx
=limh→0h[f(a)+f(a+h)+f(a+2h)+…+f(a+(n−1)h)]
where
Here, 
and 
=∫04(x+e2x)dx=limh→0h[1+(h+e2h)+(2h+e4h)+........+((n−1)h+e2(n−1)h]=∫04(x+e2x)dx
=limh→0h[1+(h+e2h)+(2h+e4h)+……..+((n−1)h+e2(n−1)h]
=limh→0h[(h+2h+3h+......+(n−1)h)+(1+e2h+e4h+.....+e2(n−1)h]
=limh→0h[(h+2h+3h+……+(n−1)h)+(1+e2h+e4h+…..+e2(n−1)h]
=limh→0h[h(1+2+3+......+(n−1))+(1+e2h+e4h+.....+e2(n−1)h]
=limh→0h[h(1+2+3+……+(n−1))+(1+e2h+e4h+…..+e2(n−1)h]
[ The series within brackets is a G.P. and ]
=limh→0h[hn(n−1)2+1((e2h)n−1)e2h−1]=limh→0h[hn(n−1)2+1((e2h)n−1)e2h−1]
=limh→0[nh(nh−1)2+he2nh−1e2h−1]=limh→0[nh(nh−1)2+he2nh−1e2h−1]
= 
= 
= 
= 
