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

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

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.8

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.8

As can be seen in the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.8, Chapter 7 focuses on the idea of the definite integral. This chapter is particularly tricky because, if one follows the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.8, they will observe that there is serious incorporation of concepts from different chapters in it. Exercise 7.8 specifically focuses on questions that are based on definite integrals when they are the limit of a sum, and the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.8 helps them get a better grasp of the topic, making them more self-reliant.

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Students all over the country have benefited from the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.8

Q.1

Evaluate the following definite integrals as limit of sums.∫abx dx

Ans

By definition∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)],where, h=b−anHere,  f(x)=x,  h=b−an∴∫abx dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)]          =(b−a)limn→∞1n[a+a+h+...+a+(n−1)h]          =(b−a)limn→∞1n[na+h+2h+...+(n−1)h]          =(b−a)limn→∞1n[na+{1+2+...+(n−1)}h]          =(b−a)limn→∞1n[na+(n−1)n2h]        [∵∑n=n(n+1)2]          =(b−a)limn→∞[a+(n−1)2(b−an)]          =(b−a)limn→∞[a+(b−a)2(1−1n)]          =(b−a)[a+(b−a)2(1−0)]          =(b−a)(2a+b−a2)          =(b−a)(b+a)2  =b2−a22

 Q.2

Evaluate the following definite integrals as limit of sums.∫05(x+1) dx

Ans

We have,  ∫05(x+1) dxBy definition∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)],where, h=b−anHere,  a=0,  b=5,  f(x)=x+1,  h=5−0n=5n∴∫05(x+1) dx=(b−a)limn→∞1n[f(0)+f(0+h)+...+f(0+(n−1)h)]                            =5limn→∞1n[(0+1)+(h+1)+...+{(n−1)h+1}]                            =5limn→∞1n[n+h+2h+...+(n−1)h]                            =5limn→∞1n[n+{1+2+...+(n−1)}h]                            =5limn→∞1n[n+(n−1)n2h][∵∑n=n(n+1)2]                            =5limn→∞[1+(n−1)2(5n)]                            =5limn→∞[1+52(1−1n)]                            =5[1+52(1−0)]                            =5×72  =352

 Q.3

Evaluate the following definite integrals as limit of sums.∫23x2 dx

Ans

We have, ∫23x2 dx∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)],where, h=b−anHere,  a=2,  b=3,  f(x)=x2,  h=3−2n=1n∴ ∫23x2 dx          =(3−2)limn→∞1n[f(2)+f(2+h)+...+f(2+(n−1)h)]          =limn→∞1n[(2)2+(2+h)2+...+{2+(n−1)h}2]          =limn→∞1n[4+(4+4h+h2)+...+{4+4(n−1)h+(n−1)2h2}]          =limn→∞1n[4n+4{1+2+...+(n−1)}h+{1+22+32+...+(n−1)2}h2]          =limn→∞1n[4n+4(n−1)n2h+(n−1)n(2n−2+1)6h2]          =limn→∞1n[4n+4(n−1)n2.1n+(n−1)n(2n−2+1)6.1n2]                                        [∵∑n=n(n+1)2,∑n2=n(n+1)(2n+1)6]          =limn→∞[4+4(1−1n)2+(1−1n)(2−1n)6]          =[4+4(1−0)2+(1−0)(2−0)6]          =4+2+13          =193

 Q.4

Evaluate the following definite integrals as limit of sums.∫14(x2−x) dx

Ans

We have, ∫14(x2−x) dx=∫14x2dx−∫14x dx                           =I1−I2    (Let)∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)]where, h=b−anHere,  a=1,  b=4,  f(x)=x2,  h=4−1n=3n∴    I1=∫14x2dx       =(4−1)limn→∞1n[f(1)+f(1+h)+...+f(1+(n−1)h)]           =3limn→∞1n[(1)2+(1+h)2+...+{1+(n−1)h}2]           =3limn→∞1n[1+(1+2h+h2)+...+{1+2(n−1)h+(n−1)2h2}]            =3limn→∞1n[n+{1+2+...+(n−1)}2h+{1+22+32+...+(n−1)2}h2]           =3limn→∞1n[n+(n−1)n22h+(n−1)n(2n−2+1)6h2]           =3limn→∞1n[n+(n−1)n.3n+(n−1)n(2n−2+1)6.9n2]                   ∵∑n=nn+12,∑n2=nn+12n+16           =3limn→∞[1+(1−1n)×3+(1−1n)(2−1n)6×9]

           =3[1+3(1−0)+3(1−0)(2−0)2]           =3(1+3+3)  =21For​   I2=∫14x dx∴∫14x dx=(4−1)limn→∞1n[f(1)+f(1+h)+...+f(1+(n−1)h)]                   =3limn→∞1n[1+1+h+...+1+(n−1)h]                   =3limn→∞1n[n+h+2h+...+(n−1)h]                   =3limn→∞1n[n+{1+2+...+(n−1)}h]                   =3limn→∞1n[n+(n−1)n2h][∵∑n=n(n+1)2]                   =3limn→∞[1+(n−1)2(3n)]                    =3limn→∞[1+32(1−1n)]                   =3[1+32(1−0)]                    =3(52)=152So,  ∫14(x2−x) dx                    =I1−I2 =21−152 =272

 Q.5

Evaluate the following definite integrals as limit of sums.∫–11ex dx

Ans

We  have, ∫–11ex dx∵∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)],where, h=b−anHere,  a=−1,  b=1,  f(x)=ex,  h=1−(−1)n=2n∴∫–11ex dx=(1+1)limn→∞1n[f(−1)+f(−1+h)+...+f(−1+(n−1)h)]           =2limn→∞1n[e−1+e−1+h+...+e−1+(n−1)h]Using the sum to n terms of a G.P., where a=e−1, r=eh, we have∴∫–11ex dx=2limn→∞1n[e−1(enh−1)(eh−1)]           =2limn→∞1n[e−1(en.2n−1)(e2n−1)]           =2limn→∞1n[e−1(e2−1)(e2n−1)]           =2e−1(e2−1)limn→∞[e2n−12n].2           =2e−1(e2−1)2           =e−1e           ∵limx→0eh−1h=1

 Q.6

Evaluate the following definite integrals as limit of sums.∫04(x+e2x) dx

Ans

We have, ∫04(x+e2x) dx=∫04x dx+∫04e2x dx                           =I1+I2   (Let)∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)]where, h=b−anHere,  a=0,  b=4,  f(x)=x,  h=4−0n=4n        I1=∫04x dx∴∫04x dx=(4−0)limn→∞1n[f(0)+f(0+h)+...+f(0+(n−1)h)]             =4limn→∞1n[0+h+...+(n−1)h]             =4limn→∞1n[h+2h+...+(n−1)h]             =4limn→∞1n[{1+2+...+(n−1)}h]             =4limn→∞1n[(n−1)n2h][∵∑n=n(n+1)2]             =4limn→∞[(n−1)2(4n)]             =4limn→∞[42(1−1n)]             =4[2(1−0)]             =4(2)=8        I2=∫04e2x dx             =(4−0)limn→∞1n[f(0)+f(0+h)+...+f(0+(n−1)h)]             =4limn→∞1n[e0+e2h+...+e2(n−1)h]             =4limn→∞1n[1+e2h+...+e2(n−1)h]Using the sum to n terms of a G.P., where a=1, r=e2h, we have

∴∫04e2x dx=4limn→∞1n[1.(e2nh−1)(e2h−1)]           =4limn→∞1n[(e2n.4n−1)(e8n−1)]           =4limn→∞1n[(e8−1)(e8n−1)]           =4(e8−1)limn→∞[e8n−18n].8           =4(e8−1)8         [∵limx→0eh−1h=1]           =e8−12So,  ∫04(x+e2x) dx          =I1+I2           =8+e8−12          =15+e82

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