RD Sharma Class 12 Exercise 19.5 Definite Integrals Solutions Maths

RD Sharma Class 12 Exercise 19.5 Definite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 02:22 PM IST

The class 12 students need not possess the best set of solution books to understand the concepts profoundly and score well in the exams. However, the RD Sharma solution books are a great source of trust for the students. Especially, to solve sums in the RD Sharma solutions chapter like Definite Integrals, many students use the RD Sharma Class 12th Exercise 19.5 solution books. They use it while doing homework, assignments, and even while preparing for their exams.

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RD Sharma Class 12 Solutions Chapter19 Definite Integrals - Other Exercise
Definite Integrals Excercise:19.5
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter19 Definite Integrals - Other Exercise

Definite Integrals Excercise:19.5

Definite Integrals exercise 19 point 5 question 1

Answer:

$\frac{33}{2}$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\begin{aligned} &\int_{0}^{3}(x+4) d x \\ \end{aligned}$
Solution:
We have,
$\begin{aligned} &\int_{0}^{3}(x+4) d x \\ \end{aligned}$
$\begin{aligned}&\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)] \end{aligned}$
Where, $h=\frac{b-a}{n}$
Here,
$\begin{aligned} &a=0, b=3, f(x)=(x+4) \\ &h=\frac{3}{n} \Rightarrow n h=3 \end{aligned}$
Thus, we have
$\begin{aligned} I &=\int_{0}^{3}(x+4) d x \\ \end{aligned}$
$\begin{aligned}& I =\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f(n-1) h] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h[4+(h+4)+(2 h+4)+\ldots((n-1) h+4)] \\ \end{aligned}$
$\begin{aligned}&=\lim _{h \rightarrow 0} h[4 n+h(1+2+3+\ldots(n-1))] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[4 n+h\left(\frac{n(n-1)}{2}\right)\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{3}{n}\left[4 n+\frac{3}{n} \frac{\left(n^{2}-1\right)}{2}\right]\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{3}{n} \\ n \rightarrow \infty \end{array}\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} 12+\frac{9}{2}\left(1-\frac{1}{n}\right) \\ &=12+\frac{9}{2}=\frac{33}{2} \end{aligned}$

Definite Integrals exercise 19.5 question 2

Answer:

8
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{0}^{2}(x+3) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Where, $h=\frac{b-a}{n}$
Here,
$\begin{aligned} &a=0, b=2, f(x)=(x+3) \\ &h \rightarrow \frac{2}{n} \end{aligned}$
Thus, we have
$\begin{aligned} I &=\int_{0}^{2}(x+3) d x \\ I &=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f(n-1) h] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h[3+(h+3)+(2 h+3)+\ldots((n-1) h+3)] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h[3 n+h(1+2+3+\ldots(n-1))] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[3 n+h\left(\frac{n(n-1)}{2}\right)\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[3 n+\frac{2}{n} \frac{n(n-1)}{2}\right] & &&&&&\quad\left[\begin{array}{l} \left.\mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n}\right] \\ n \rightarrow \infty \end{array}\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty}\left[6+\frac{2}{n} \cdot n^{2}\left(1-\frac{1}{n}\right)\right] \\ &=6+2=8 \end{aligned}$

Definite Integrals exercise 19 point 5 question 3

Answer:

8
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{1}^{3}(3x-2) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Where, $h=\frac{b-a}{n}$
Here,
$\begin{aligned} &a=1, b=3, f(x)=3x-1 \\ &h \rightarrow \frac{2}{n} \end{aligned}$
Thus, we have
$\begin{aligned} &I=\int_{1}^{3}(3 x-2) d x \\ &I=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h[1+(3(1+h)-2)+\{3(1+2 h)-2\}+\ldots\{3(1+(n-1) h)-2\}] \\ &=\lim _{h \rightarrow 0} h[n+3 h(1+2+3+\ldots(n-1))] \end{aligned}$ $=\lim _{h \rightarrow 0} h\left[n+3 h\left(\frac{n(n-1)}{2}\right)\right]$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[ n+\frac{6}{n} \frac{n(n-1)}{2}\right] & &&&&&\quad\left[\begin{array}{l} \left.\mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n}\right] \\ n \rightarrow \infty \end{array}\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty}\left[2+\frac{6}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)\right] \\ &=6+2=8 \end{aligned}$

Definite Integrals exercise 19.5 question 4

Answer:

6
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{-1}^{1}(x+3) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Here,
$\begin{aligned} &a=-1, b=1, f(x)=x+3 \\ &h=\frac{b-a}{n}=\frac{1-(-1)}{n}=\frac{2}{n} \end{aligned}$
Thus, we have
$\begin{aligned} I &=\int_{-1}^{1}(x+3) d x \\ I &=\lim _{h \rightarrow 0} h[f(-1)+f(-1+h)+f(-1+2 h)+\ldots f(-1+(n-1)) h] \\ \end{aligned}$ $\begin{aligned} &=\lim _{h \rightarrow 0} h[2+(2+h)+(2+2 h)+\ldots((n-1) h+2)] \\ &=\lim _{h \rightarrow 0} h[2 n+h(1+2+3+\ldots(n-1))] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[2 n+\frac{2}{n} \frac{n(n-1)}{2}\right] \quad\left[\begin{array}{l} \left.\mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n}\right] \\ n \rightarrow \infty \end{array}\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} 2+\frac{2}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow 0} 4+2\left(1-\frac{1}{\infty}\right) \\ &=4+2=6 \end{aligned}$

Definite Integrals exercise 19.5 question 5

Answer:

$\frac{35}{2}$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{0}^{5}(x+1) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Where, $h=\frac{b-a}{n}$
Here,
$\begin{aligned} &a=0, b=5, f(x)=x+1 \\ &h=\frac{5-0}{n}=\frac{5}{n} \end{aligned}$
Thus, we have
$\begin{aligned} &=\int_{0}^{5}(x+1) d x \\ &=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f(n-1) h] \\ \end{aligned}$
$\begin{aligned}&=\lim _{h \rightarrow 0} h[1+(h+1)+(2 h+1)+\ldots((n-1) h+1)] \\ \end{aligned}$
$\begin{aligned}&=\lim _{h \rightarrow 0} h[n+h(1+2+3+\ldots(n-1))] \\ \end{aligned}$
$\begin{aligned} &\left.=\lim _{n \rightarrow \infty} \frac{5}{n}\left[n+\frac{5}{n} \frac{n(n-1)}{2}\right] \quad \left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \\ n \rightarrow \infty \end{array}\right] h=\frac{5}{n}\right] \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} 5+\frac{25}{2 n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ &=\lim _{n \rightarrow 0} 5+\frac{25}{2}\left(1-\frac{1}{\infty}\right) \\ &=5+\frac{25}{2}=\frac{35}{2} \end{aligned}$

Definite Integrals exercise 19.5 question 6
Answer:

14
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{1}^{3}(2 x+3) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Where, $h=\frac{b-a}{n}$
Here,
$\begin{aligned} &a=1, b=3, f(x)=2 x+3 \\ &h=\frac{3-1}{n}=\frac{2}{n} \end{aligned}$
Thus, we have
$\begin{aligned} I &=\int_{1}^{3}(2 x+3) d x \\ I &=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h[2+3+\{2(1+h)+3\}+\ldots 2(1+(n-1)+h)] \\ &=\lim _{h \rightarrow 0} h[5+(5+2 h)+(5+4 h)+\ldots 5+2(n-1) h] \\ &=\lim _{h \rightarrow 0} h[5 n+2 h(1+2+3+\ldots(n-1))] \end{aligned}$
$=\lim _{n \rightarrow \infty} \frac{2}{n}\left[5 n+\frac{4}{n} \frac{n(n-1)}{2}\right] \left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n} \\ n \rightarrow \infty \end{array}\right]$
$\begin{aligned} &=\lim _{n \rightarrow \infty}\left[10+\frac{4}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)\right] \\ &=\lim _{n \rightarrow \infty} 10+4\left(1-\frac{1}{n}\right) \\ &=10+4=14 \end{aligned}$

Definite Integrals exercise 19.5 question 7

Answer:

-4
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{3}^{5}(2-x) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Here,
$\begin{aligned} &a=3, b=5, f(x)=2-x \\ &h=\frac{2}{n} \end{aligned}$
Thus, we have
$\begin{aligned} &I=\int_{3}^{5}(2-x) d x \\ &I=\lim _{h \rightarrow 0} h[f(3)+f(3+h)+f(3+2 h)+\ldots f(3+(n-1)) h] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h[(2-3)+\{2-(3+h)\}+(2-(3+2 h)) \ldots 2-(3+(n-1) h)\rfloor \\ &=\lim _{h \rightarrow 0} h[-n-h(1+2+3+\ldots(n-1) h)] \end{aligned}$
$=\lim _{n \rightarrow \infty} \frac{2}{n}\left[-n-\frac{2}{n} \cdot \frac{n(n-1)}{2}\right] \quad\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n} \\ n \rightarrow \infty \end{array}\right]$
$\begin{aligned} &=\lim _{n \rightarrow \infty}\left[-2-\frac{2}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)\right] \\ &=\lim _{n \rightarrow \infty}-2-2\left(1-\frac{1}{n}\right) \\ &=-2-2=-4 \end{aligned}$

Definite Integrals exercise 19 point 5 question 8

Answer:

$\frac{14}{3}$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{0}^{2}\left(x^{2}+1\right) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Where, $h=\frac{b-a}{n}$
Here,
$\begin{aligned} &a=0, b=2, f(x)=x^{2}+1 \\ &h=\frac{2}{n} \end{aligned}$
Thus, we have
$\begin{aligned} &I=\int_{0}^{2}\left(x^{2}+1\right) d x \\ &I=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f((n-1)) h] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[1+\left(h^{2}+1\right)+\left((2 h)^{2}+1\right)+\ldots\left(((n-1) h)^{2}+1\right)\right] \\ &=\lim _{h \rightarrow 0} h\left[n+h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots(n-1)^{2}\right)\right] \end{aligned}$
$=\lim _{n \rightarrow \infty} \frac{2}{n}\left[n+\frac{4}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \quad\left[\begin{array}{l} \mathrm{Q} h=-\mathrm{if} \\ n \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right]$
$\begin{aligned} &=\lim _{n \rightarrow \infty} 2+\frac{4}{3 n^{2}} \cdot n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ &=2+\frac{4}{3} \times 2 \times 1 \\ &=\frac{14}{3} \end{aligned}$

Definite Integrals exercise 19.5 question 9

Answer:

$\frac{7}{3}$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{1}^{2}\left(x^{2}\right) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Where, $h=\frac{b-a}{n}$
Here,
$\begin{aligned} &a=1, b=2, f(x)=x^{2} \\ &h=\frac{2-1}{n}=\frac{1}{n} \end{aligned}$
Thus, we have
$\begin{aligned} &I=\int_{1}^{2}\left(x^{2}\right) d x \\ &I=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[1+(1+h)^{2}+(1+2 h)^{2}+\ldots(1+(n-1) h)^{2}\right] \\ &=\lim _{h \rightarrow 0} h\left[n+2 h(1+2+3+\ldots(n-1))+h^{2}\left(1^{2}+2^{2}+3^{2}+(n-1)^{2}\right)\right] \end{aligned}$
$=\lim _{n \rightarrow \infty} \frac{2}{n}\left[n+\frac{2}{n}\left(\frac{n(n-1)}{2}\right)+\frac{1}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \quad\left[\begin{array}{l} \mathbf{Q} h=\frac{1}{n} \text { if } \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right]$
$=\lim _{n \rightarrow \infty} 1+\frac{n^{2}}{n^{2}}\left(1-\frac{1}{n}\right)+\frac{1}{6 n^{3}} n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)$
$\begin{aligned} &=1+1+\frac{2}{6} \\ &=\frac{6+6+2}{6}=\frac{14}{6} \\ &=\frac{7}{3} \end{aligned}$

Definite Integrals exercise 19.5 question 10

Answer:

$\frac{43}{3}$

Hint:

To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{2}^{3}\left(2 x^{2}+1\right) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Where, $h=\frac{b-a}{n}$
Here,
$\begin{aligned} &a=2, b=3, f(x)=2 x^{2}+1 \\ &h=\frac{3-2}{n}=\frac{1}{n} \end{aligned}$
Thus, we have
$\begin{aligned} I &=\int_{2}^{3}\left(2 x^{2}+1\right) d x \\ I &=\lim _{h \rightarrow 0} h[f(2)+f(2+h)+f(2+2 h)+\ldots f(2+(n-1)) h] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[\left(2 \times 2^{2}+1\right)\left(2(2+h)^{2}+1\right)+\left(2(2+2 h)^{2}+1\right)+\ldots+\left(2(2+(n-1) h)^{2}+1\right)\right] \\ &=\lim _{h \rightarrow 0} h\left[9 n+8 h(1+2+3+\ldots)+2 h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots\right)\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{n}\left[9 n+\frac{8}{n}\left(\frac{n(n-1)}{2}\right)+\frac{2}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} 9+\frac{4}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)+\frac{1}{3 n^{3}} \cdot n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ \end{aligned}$
$\begin{aligned} &=9+4-0+\frac{2}{3} \\ &=\frac{29+12+2}{3}=\frac{43}{3} \end{aligned}$

Definite Integrals exercise 19.5 question 11

Answer:

$\frac{4}{3}$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{1}^{2}\left ( x^{2}-1 \right )dx$
Solution:
We have,
$\int_{a}^{b}\left ( fx \right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$
Where, $h=\frac{b-a}{n}$
Here,
$a=1,b=2,f(x)=x^{2}-1\\ h=\frac{2-1}{n}=\frac{1}{n}$
Thus, we have
$\begin{aligned} I &=\int_{1}^{2}\left(x^{2}-1\right) d x \\ I &=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[\left(1^{2}-1\right)\left((1+h)^{2}-1\right)+\left((1+2 h)^{2}-1\right)+\ldots+\left(1+((n-1) h)^{2}-1\right)\right] \\ &=\lim _{h \rightarrow 0} h\left[0+2 h(1+2+3+\ldots(n-1))+h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots\right)\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{2}{n}\left(\frac{n(n-1)}{2}\right)+\frac{1}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \quad\left[\begin{array}{l} \mathrm{Q} h=\frac{1}{n} \text { if } \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)+\frac{1}{6 n^{3}} \cdot n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ \end{aligned}$
$\begin{aligned} &=1+\frac{2}{6} \\ &=\frac{6+2}{6}=\frac{8}{6} \\ &=\frac{4}{3} \end{aligned}$

Definite Integrals exercise 19.5 question 12

Answer:

$\frac{32}{3}$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{0}^{2}\left ( x^{2}+4 \right )dx$
Solution:
We have,
$\int_{a}^{b}\left ( fx \right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$
Where, $h=\frac{b-a}{n}$
Here,
$a=0,b=2,f(x)=x^{2}-x\\ h=\frac{2-0}{n}=\frac{2}{n}$
Thus, we have
$\int_{0}^{2}\left ( x^{2}+4 \right )dx$
$\begin{aligned} I &=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f(0+(n-1)) h] \\ &=\lim _{h \rightarrow 0} h\left[4\left(h^{2}+4\right)+\left((2 h)^{2}+4\right)+\ldots+\left((n-1)^{2} h+4\right)\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[4 n+\frac{4}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \quad\left[\begin{array}{l} \text { Q } h=\frac{2}{n} \text { if } \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} 8+\frac{4}{3 n^{3}} . n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ \end{aligned}$
$\begin{aligned} &=8+\frac{4}{3}(1-0)(2-0) \\ &=8+\frac{4}{3} \times 1 \times 2 \\ &=8+\frac{8}{3} \\ &=\frac{24+8}{3} \\ &=\frac{32}{3} \end{aligned}$

Definite Integrals exercise 19.5, question 13

Answer:

$\frac{27}{2}$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{4}^{1}\left ( x^{2}-x \right )dx$
Solution:
$\int_{a}^{b}\left ( fx \right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$
Where, $h=\frac{b-a}{n}$
Here,
$a=1,b=4,f(x)=x^{2}-x\\ h=\frac{4-1}{n}=\frac{3}{n}$
Thus, we have
$\begin{aligned} &I=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \\ &=\lim _{h \rightarrow 0} h\left[\left(1^{2}-1\right)+\left((1+h)^{2}-(1+h)\right)+\left((1+2 h)^{2}-(1+h)\right)\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[0+\left(h+h^{2}\right)+\left(2 h+(2 h)^{2}\right)+\ldots\right] \\ &=\lim _{h \rightarrow 0} h\left[h+\left(1+2+3+\ldots(n-1)+h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots(n-1)^{2}\right)\right)\right] \\ \end{aligned}$
$\begin{aligned}&=\lim _{n \rightarrow \infty} \frac{3}{n}\left[\frac{3}{n}\left(\frac{n(n-1)}{2}\right)+\frac{9}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right]\left[\begin{array}{c} \mathrm{Q} h=\frac{3}{n} \text { if } \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{9}{2 n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)+\frac{9}{2 n^{3}} \cdot n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ &=\lim _{n \rightarrow \infty} \frac{9}{2}\left(1-\frac{1}{n}\right)+\frac{9}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \end{aligned}$
$=\frac{9}{2}\left ( 1-0 \right )+\frac{9}{2}\left ( 1-0 \right )(2-0)\\ =\frac{9}{2}+9\\ =\frac{9+18}{2}\\ =\frac{27}{2}$

Definite Integrals exercise 19.5, question 14

Answer:

$\frac{7}{2}$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{1}^{0}\left ( 3x^{2}-5x \right )dx$
Solution:
$\int_{a}^{b}\left ( fx \right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$
Where, $h=\frac{b-a}{n}$
Here,
$a=0,b=1,f(x)=3x^{2}+5x\\ h=\frac{1-0}{n}=\frac{1}{n}$
Thus, we have
$\begin{aligned} I &=\int_{0}^{1}\left(3 x^{2}+5 x\right) d x \\ I &=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f(0+(n-1)) h] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[0+\left(3 h^{2}+5 h\right)+\left(3(2 h)^{2}+5(2 h)\right)+\ldots\right] \\ &=\lim _{h \rightarrow 0} h\left[3 h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots(n-1)^{2}\right)+5 h(1+2+3+\ldots(n-1))\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{3}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)+\frac{5}{n}\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{3}{n^{3}} \cdot \frac{n^{3}}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{5}{2 n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{5}{2}\left(1-\frac{1}{n}\right) \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{2}(1-0)(2-0)+\frac{5}{2}(1-0) \\ &=1+\frac{5}{2} \\ &=\frac{2+5}{2} \\ &=\frac{7}{2} \end{aligned}$

Definite Integrals exercise 19.5, question 15

Answer:

$e^{2}-1$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{2}^{0}(e^{x})dx$
Solution:
We have,
$\int_{a}^{b}\left ( fx \right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$
Where, $h=\frac{b-a}{n}$
Here,
$a=0,b=2,f(x)=e^{x}\\ h=\frac{2-0}{n}=\frac{2}{n},nh=2$
Thus, we have
$\begin{aligned} I &=\int_{0}^{2}\left(e^{x}\right) d x \\ I &=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f((n-1)) h] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[1+e^{h}+e^{2 h}+\ldots+e^{(n-1) h}\right] \\ &=\lim _{h \rightarrow 0} h\left[\frac{\left(e^{h}\right)^{n}-1}{e^{h}-1}\right] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow 0} h\left[\frac{e^{n h}-1}{e^{h}-1}\right] \\ &=\lim _{n \rightarrow 0} h\left[\frac{e^{2}-1}{e^{h}-1}\right] \quad[n h=2] \\ \end{aligned}$
$\begin{aligned} &=\lim _{n \rightarrow 0}\left[\frac{e^{2}-1}{e^{h}-1}{h}\right] \\ &=e^{2}-1 &\left[\mathrm{Q} \frac{e^{a}-1}{a} \Rightarrow 1\right] \end{aligned}$

Definite Integrals exercise 19.5 question 1

Answer:

$\frac{33}{2}$
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{0}^{3}\left ( x+4 \right )dx$
Solution:
We have,
$\int_{0}^{3}\left ( x+4 \right )dx$
$\int_{a}^{b}\left ( fx\right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$
Where, $h=\frac{b-a}{n}$
Here,
$a=0,b=3,f(x)=(x+4)\\ h=\frac{3}{n}\Rightarrow nh=3$
Thus, we have
$\begin{aligned} I &=\int_{0}^{3}(x+4) d x \\ I &=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f(n-1) h] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h[4+(h+4)+(2 h+4)+\ldots((n-1) h+4)] \\ &=\lim _{h \rightarrow 0} h[4 n+h(1+2+3+\ldots(n-1))] \\ \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} h\left[4 n+h\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{3}{n}\left[4 n+\frac{3}{n} \frac{\left(n^{2}-1\right)}{2}\right] \\ \end{aligned}$ $\begin{aligned} &=\lim _{n \rightarrow \infty} 12+\frac{9}{2}\left(1-\frac{1}{n}\right) \\ &=12+\frac{9}{2}=\frac{33}{2} \end{aligned}$

Chapter 19th in Mathematics of class 12 is a bit hard for the students to solve independently. This chapter consists of five exercises, ex 19.1 to ex 19.5. in the last exercise, ex 19.5, there are around 34 questions given in the textbook to be solved. The concept in this exercise is to elaborate on the integrals as a limit of sums. These answers are given in shortcuts and elaborated methods in the RD Sharma Class 12 Chapter 19 Exercise 19.5 book. The students can themselves select the method that they feel comfortable working with.

As ex 19.5 is the last exercise, it might be even more challenging for the students to solve. This book follows the NCERT pattern giving a valid reason for the CBSE students to use it as their reference material. As there are plenty of practice questions, the students will be able to prepare for their exams in a better way. The practice question in the RD Sharma Class 12th Exercise 19.5 book helps them face the exams with confidence.

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