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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 (Ex 7.8)

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.8 (Ex 7.8) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 7 Integrals Exercise 7.8 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 12 Subject: Class 12 Maths Chapter Name: Chapter 7 - Integrals Exercise: Exercise - 7.8 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
Competitive Exams after 12th Science

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

Exercise 7.8

Refer to page 94-100 for Exercise 7.8 in the PDF.

1. Integrate the following $\int\limits_a^b {x.dx}$.

Ans: Given:

$\int\limits_a^b {x.dx}$

As we know that $f\left( x \right)$ is continuous in $\left[ {a,b} \right]$.

Then we have,

$\int\limits_a^b {f(x)dx = \mathop {\lim }\limits_{n \to \infty } h\sum\limits_{r = 0}^{n - 1} {f(a + rh),} }$ where $h = \dfrac{{b - a}}{n}$

By substituting the value of $h$ in the above expression we get

$\int\limits_a^b {(x)dx = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{b - a}}{n}} \right)} \sum\limits_{r = 0}^{n - 1} {f\left( {a + \dfrac{{(b - a)r}}{n}} \right)}$

Since $f\left( a \right) = a$,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{b - a}}{n}} \right)\sum\limits_{r = 0}^{n - 1} {\left( {\dfrac{{(b - a)r}}{n}} \right)} + a$

Now, By expanding the summation we get,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{b - a}}{n}} \right)\left( {\dfrac{{(b - a)(n - 1)(n)}}{{2n}} + a(n - 1)} \right)$

After simplification we get,

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{{(b - a)}}{n}.\dfrac{{(b - a)({n^2} - n) + 2a{n^2} - 2an}}{{2n}}$

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{{(b - a)}}{n}.\dfrac{{(b + a){n^2} - (b + a)n}}{{2n}}$
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{(b + a)(b - a){n^2} - (b + a)(b - a)n}}{{2{n^2}}}$

Now, on computing we get,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{(b + a)(b - a)}}{2} - \dfrac{{(b + a)(b - a)}}{n}} \right)$

$= \dfrac{{(b + a)(b - a)}}{2}$

$= \dfrac{{{b^2} - {a^2}}}{2}$

2. Integrate the following $\int\limits_0^5 {(x + 1)dx}$

Ans: Given:

$\int\limits_0^5 {(x + 1)dx}$

Here, we know that $f\left( x \right)$ is continuous in $\left[ {a,b} \right]$ i.e., $\left[ {0,5} \right]$

Then we have,

$\int\limits_a^b {f(x)dx = \mathop {\lim }\limits_{n \to \infty } } h\sum\limits_{r = 0}^{n - 1} {f(a + rh),}$ where $h = \dfrac{{b - a}}{n}$

Substituting the value of $h$ in the above expression we get,

$\int\limits_0^5 {(x + 1)dx = \mathop {\lim }\limits_{n \to \infty } } \left( {\dfrac{5}{n}} \right)\sum\limits_{r = 0}^{n - 1} {f\left( {\dfrac{{5r}}{n}} \right)}$

Since $f\left( a \right) = a$,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{5}{n}} \right)\sum\limits_{r = 0}^{n - 1} {\left( {\dfrac{{5r}}{n}} \right) + 1}$

By expanding the summation we get,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{5}{n}} \right)\left( {\dfrac{{5(n - 1)(n)}}{{2n}} + (n - 1)} \right)$

Upon simplification we get,

$= \mathop {\lim \dfrac{5}{n}}\limits_{n \to \infty } .\dfrac{{5{n^2} - 5n + 2{n^2} - 2n}}{{2n}}$

$= \mathop {\lim \dfrac{5}{n}}\limits_{n \to \infty } .\dfrac{{7{n^2} - 7n}}{{2n}}$

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{{35{n^2} - 35n}}{{2{n^2}}}$

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{{35}}{2} - \left( {\dfrac{{35}}{{2n}}} \right)$

$= \dfrac{{35}}{2}$

3. Integrate the following $\int\limits_2^3 {{x^2}} dx$

Ans : Given :

$\int\limits_2^3 {{x^2}} dx$

As we know that $f\left( x \right)$ is continuous in $\left[ {a,b} \right]$ i.e., $\left[ {2,3} \right]$.

Then we have equation :

$\int\limits_a^b {f(x)dx = \mathop {\lim }\limits_{n \to \infty } h\sum\limits_{r = 0}^{n - 1} {f(a + rh),} }$where $h = \dfrac{{b - a}}{n}$

Now, substituting the value of $h$ in the above expression we get,

$\int\limits_2^3 {({x^2})dx = \mathop {\lim }\limits_{n \to \infty } } \left( {\dfrac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} {f\left( {2 + \left( {\dfrac{r}{n}} \right)} \right)}$

Since $f\left( a \right) = a$,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} {{{\left( {2 + \left( {\dfrac{r}{n}} \right)} \right)}^2}}$

By expanding the summation we get,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} {\left( {\dfrac{{{r^2}}}{{{n^2}}} + 4 + \dfrac{{4r}}{n}} \right)}$

After simplification we get,

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left( {\dfrac{{(n - 1)(n)(2n - 1)}}{{6{n^2}}} + 4n + \dfrac{{4(n - 1)(n)}}{{2n}}} \right)$

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left( {\dfrac{{({n^2} - n)(2n - 1)}}{{6{n^2}}} + 4n + \dfrac{{2({n^2} - n)}}{{2n}}} \right)$

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left( {\dfrac{{(2{n^3} - 2{n^2} - {n^2} + n)}}{{6{n^2}}} + 4n + \dfrac{{2({n^2} - n)}}{{2n}}} \right)$

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left( {\dfrac{{(2{n^3} - 3{n^2} + n) + (24{n^3}) + (12{n^3} - 12{n^2}}}{{6{n^2}}}} \right)$

$= \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left( {\dfrac{{38{n^3} - 15{n^2} + n}}{{6{n^2}}}} \right)$

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{38{n^3} - 15{n^2} + n}}{{6{n^3}}}} \right)$

Now, on computing we get,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{38}}{6}} \right) - \left( {\dfrac{{15}}{{6n}}} \right) + \left( {\dfrac{1}{{6{n^2}}}} \right)$

$= \left( {\dfrac{{38}}{6}} \right)$

$= \left( {\dfrac{{19}}{3}} \right)$

4. Integrate the following $\int\limits_1^4 {({x^2} - x)dx}$

Ans: Given:

$\int\limits_1^4 {({x^2} - x)dx}$

As we know that $f\left( x \right)$  is continuous in $\left[ {a,b} \right]$ i.e., $\left[ {1,4} \right]$ .

Then we have,

$= \int\limits_1^4 {{x^2}dx - \int\limits_1^4 {xdx} }$

Let I=I1-I2,

Where, ${I_1} = \int\limits_1^4 {{x^2}dx}$ and ${I_2} = \int\limits_1^4 {xdx}$                        …. (1)

$\int\limits_a^b {f(x)dx = (b - a)\mathop {\lim }\limits_{n \to \infty } } \mathop {\dfrac{1}{n}\left[ {f(a) + f(a + h) + f(a + (n - 1)h} \right]}\limits_{} ,$ where $h = \dfrac{{b - a}}{n}$

For  ${I_1} = \int\limits_1^4 {{x^2}dx}$, $a = 1$ , $b = 4$ , and $f(x) = {x^2}$

$h = \dfrac{{4 - 1}}{n} = \dfrac{3}{n}$

${I_1} = \int\limits_1^4 {{x^2}dx}$

$= (4 - 1)\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{n}\left[ {f(1) + f(1 + h) + ... + f(1 + (n - 1)h)} \right]$

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {{1^2} + {{\left( {1 + \dfrac{3}{n}} \right)}^2} + {{\left( {1 + 2.\dfrac{3}{n}} \right)}^2} + ...{{\left( {1 + \dfrac{{(n + 1)3}}{n}} \right)}^2}} \right]$( By substituting the values)

(on further calculation)

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {{1^2} + \left\{ {{1^2} + {{\left( {\dfrac{3}{n}} \right)}^2} + 2.\dfrac{3}{n}} \right\} + ...\left\{ {{1^2} + {{\left( {\dfrac{{(n - 1)3}}{n}} \right)}^2} + \dfrac{{2.(n - 1).3}}{n}} \right\}} \right]$

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {\left( {{1^2} + .... + {1^2}} \right) + {{\left( {\dfrac{3}{n}} \right)}^2}\left\{ {{1^2} + {2^2} + ... + {{(n - 1)}^2}} \right\} + 2.\dfrac{3}{n}\left\{ {1 + 2 + .. + (n - 1)} \right\}} \right]$

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {n + \dfrac{9}{{{n^2}}}\left\{ {\dfrac{{(n - 1)(n)(2n - 1)}}{6}} \right\} + \dfrac{6}{n}\left\{ {\dfrac{{(n - 1)(n)}}{2}} \right\}} \right]$( By further simplification)

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {n + \dfrac{{9n}}{6}\left( {1 - \dfrac{1}{n}} \right)\left( {2 - \dfrac{1}{n}} \right) + \dfrac{{6n - 6}}{2}} \right]$( separating the terms)

$= 3\mathop {\lim }\limits_{n \to \infty } \left[ {1 + \dfrac{9}{6}\left( {1 - \dfrac{1}{n}} \right)\left( {2 - \dfrac{1}{n}} \right) + 3 - \dfrac{3}{n}} \right]$

So we get,

$= 3[1 + 3 + 3]$

$= 3[7]$

${I_1} = 21$                                                                ….. (2)

For ${I_2} = \int\limits_1^4 {xdx}$ , $a = 1$ , $b = 4$ , and $f\left( x \right) = x$

$h = \dfrac{{4 - 1}}{n} = \dfrac{3}{n}$

${I_2} = (4 - 1)\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {f(1) + f(1 + h) + ...f(a + (n - 1)h)} \right]$

Substituting the values

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {1 + (1 + h) + ... + (1 + (n - 1)h)} \right]$

We can write it as

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {1 + \left( {1 + \dfrac{3}{n}} \right) + ... + \left\{ {1 + (n - 1)\dfrac{3}{n}} \right\}} \right]$

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {(1 + 1 + ... + 1) + \dfrac{3}{n}\left( {1 + 2 + ... + (n - 1)} \right)} \right]$

We get

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {n + \dfrac{3}{n}\left\{ {\dfrac{{(n - 1)n}}{2}} \right\}} \right]$

$= 3\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {1 + \dfrac{3}{2}\left( {1 - \dfrac{1}{n}} \right)} \right]$

By further calculation

$= 3\left[ {1 + \dfrac{3}{2}} \right]$

$= 3\left[ {\dfrac{5}{2}} \right]$

${I_2} = \dfrac{{15}}{2}$                                                        …. (3)

Now, by considering both the equations

$I = {I_1} - {I_2}$

$= 21 - \dfrac{{15}}{2} = \dfrac{{27}}{2}$

5. Integrate the following $\int\limits_{ - 1}^1 {{e^x}dx}$

Ans: Given:

$\int\limits_{ - 1}^1 {{e^x}dx}$

We know that $f\left( x \right)$ is continuous in $\left[ {a,b} \right]$ i.e., $\left[ { - 1,1} \right]$.

Then we have,

$\int\limits_a^b {f(x)dx = \mathop {\lim }\limits_{n \to \infty } } h\sum\limits_{r = 0}^{n - 1} {f(a + rh),}$ where $h = \dfrac{{b - a}}{n}$

Substituting the value of $h$ in the above expression we get,

$\int\limits_0^2 {({e^x}} )dx = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{2}{n}} \right)\sum\limits_{r = 0}^{n - 1} {f\left( { - 1 + \dfrac{{2r}}{n}} \right)}$

Since $f\left( a \right) = a$

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{2}{n}} \right)\sum\limits_{r = 0}^{n - 1} {{e^{\dfrac{{2r}}{n} - 1}}}$

By expanding the summation we get,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{2}{{ne}}} \right)({e^0} + {e^h} + {e^{2h}} + ..... + {e^{nh}})$

Sum of ${e^0} + {e^h} + {e^{2h}} + ..... + {e^{nh}}$ whose GP has common ratio with ${e^{\dfrac{1}{n}}}$

Whose sum is :

$= \dfrac{{{e^h}(1 - {e^{nh}})}}{{1 - {e^h}}}$

After simplification we get,

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{2}{{ne}}} \right)\left( {\dfrac{{{e^h}(1 - {e^{nh}})}}{{1 - {e^h}}}} \right)$

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{2}{{ne}}} \right).\dfrac{{{e^h}(1 - {e^{nh}})}}{{\dfrac{{1 - {e^h}.h}}{h}}}$

$\mathop {\lim }\limits_{h \to 0} \dfrac{{1 - {e^h}}}{h}$

= -1

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{2}{{ne}}} \right)\left( {\dfrac{{{e^h}(1 - {e^{nh}})}}{{ - h}}} \right)$

$= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{2}{{ne}}} \right)\left( {\dfrac{{{e^{\left( {\dfrac{2}{n}} \right)}}(1 - {e^{n\left( {\dfrac{2}{n}} \right)}})}}{{ - \dfrac{2}{n}}}} \right)$[ Since, $h = \dfrac{2}{n}$]

$= \dfrac{{{e^2} - 1}}{e}$

$= e - {e^{ - 1}}$

6. Integrate the following $\int\limits_0^4 {(x + {e^{2x}})dx}$

Ans: Given:

$\int\limits_0^4 {(x + {e^{2x}})dx}$

As we know that

$\int\limits_a^b {f(x)dx = (b - a)\mathop {\lim }\limits_{n \to \infty } } \mathop {\dfrac{1}{n}\left[ {f(a) + f(a + h) + f(a + (n - 1)h} \right]}\limits_{} ,$ where $h = \dfrac{{b - a}}{n}$.

$a = 0$ , $b = 4$ and $f(x) = x + {e^{2x}}$

$h = \dfrac{{4 - 0}}{n} = \dfrac{4}{n}$

It can be written as

$\int\limits_0^4 {(x + {e^{2x}})dx = (4 - 0)\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {f(0) + f(h) + f(2h) + ... + f((n - 1)h)} \right]}$

$= 4\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {\left( {0 + {e^0}} \right) + (h + {e^{2h}}) + (2h + {e^{2.2h}}) + ... + \left\{ {(n - 1)h + {e^{2(n - 1)}}} \right\}} \right]$

$= 4\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {1 + \left( {h + {e^{2h}}} \right) + (2h + {e^{4h}}) + ... + \left\{ {(n - 1)h + {e^{2(n - 1)h}}} \right\}} \right]$

$= 4\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {\left\{ {h + 2h + 3h + ... + (n - 1)h} \right\} + \left( {1 + {e^{2h}} + {e^{4h}} + ... + {e^{2(n - 1)h}}} \right)} \right]$

We can write it as

$= 4\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {h\left\{ {1 + 2 + ...(n - 1)} \right\} + \left( {\dfrac{{{e^{2hn}} - 1}}{{{e^{2h}} - 1}}} \right)} \right]$

$= 4\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {\dfrac{{(h(n - 1)n)}}{2} + \left( {\dfrac{{{e^{2hn}} - 1}}{{{e^{2h}} - 1}}} \right)} \right]$

$= 4\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {\dfrac{4}{n}.\dfrac{{(n - 1)n}}{2} + \left( {\dfrac{{{e^8} - 1}}{{{e^{\dfrac{8}{n}}} - 1}}} \right)} \right]$

By further simplification

$= 4(2) + 4\mathop {\lim }\limits_{n \to \infty } \dfrac{{({e^8} - 1)}}{{\left( {\dfrac{{{e^{\dfrac{8}{n}}} - 1}}{{\dfrac{8}{n}}}} \right)8}}$

$= 8 + \dfrac{{4.({e^8} - 1)}}{8}$$\left( {\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x} = 1} \right)$

So we get,

$= 8 + \dfrac{{{e^8} - 1}}{2}$

$= \dfrac{{15 + {e^8}}}{2}$

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

Opting for the NCERT solutions for Ex 7.8 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.8 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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