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

# 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.

## Definite Integrals Excercise:19.5

Definite Integrals exercise 19 point 5 question 1

$\frac{33}{2}$
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

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

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

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

$\frac{35}{2}$
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}

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

-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

$\frac{14}{3}$
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

$\frac{7}{3}$
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

$\frac{43}{3}$

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

$\frac{4}{3}$
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

$\frac{32}{3}$
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

$\frac{27}{2}$
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

$\frac{7}{2}$
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

$e^{2}-1$
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

$\frac{33}{2}$
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.

All the answers in the Class 12 RD Sharma Chapter 19 Exercise 19.5 Solution book are given by the faculty members who have expertise in the respective domains. Solving the sums in Definite Integrals at the starting stage would be challenging, but when the students follow the shortcut methods and understand the concept given in the RD Sharma books, they need not worry.

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## RD Sharma Chapter-wise Solutions

1. Which solution book can I use to clarify my doubts on the mathematics chapter 19 portion?

The RD Sharma Class 12th Exercise 19.5 solution book can be used to clarify your doubts regarding the respective chapter

2. How many exercises does the Class 12 mathematics chapter 19 contain?

There are five exercises, ex 19.1 to 19.5 in mathematics, chapter 19. Refer to the RD Sharma books for the solutions to these exercises.

3. How can I find a particular portion on the Career360 website?

Visit the career 360 website and search for the name of the solution book that you need. For instance, the RD Sharma class 12th Exercise 19.5 book will give you the answers to all the questions given in the particular exercise.

4. How can I access the RD Sharma solution books in offline mode?

Once you download the PDFs of the RD Sharma solution books to your device, you can refer to the solution offline whenever needed.

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The solutions given in the RD Sharma books are provided by proficient experts in the respective field. And the answer keys are verified to check the accuracy too. Therefore, everyone can drop their hesitation in using it.

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