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.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 7 Exercise 7.8 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 7 Exercise 7.8, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 7 Exercise 7.8 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.