# NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.8

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## Chapter 7 Ex.7.8 Question 1

Evaluate the definite integral as limit of sum: $$\int_a^b {x\;dx}$$

### Solution

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

Here, $$a = a,b = b\,\,{\text{and }}f\left( x \right) = x$$

\begin{align}\therefore \int_a^b {x\;dx}& = \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {a + \left( {a + h} \right) \ldots \left( {a + 2h} \right) \ldots a + \left( {n - 1} \right)h} \right]\\&= \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\left( {\underbrace {a + a + a + ...... + a}_{n\;times}} \right) + \left( {h + 2h + 3h + \ldots \left( {n - 1} \right)h} \right)} \right]\\&= \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {na + h\left( {1 + 2 + 3 + \ldots + \left( {n - 1} \right)} \right)} \right]\\&= \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {na + h\left\{ {\frac{{\left( {n - 1} \right)\left( n \right)}}{2}} \right\}} \right]\\&= \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {na + \frac{{n\left( {n - 1} \right)h}}{2}} \right] = \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } \frac{n}{n}\left[ {a + \frac{{\left( {n - 1} \right)h}}{2}} \right]\\&= \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } \left[ {a + \frac{{\left( {n - 1} \right)h}}{2}} \right] = \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } \left[ {a + \frac{{\left( {n - 1} \right)\left( {b - a} \right)}}{{2n}}} \right]\\&= \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } \left[ {a + \frac{{\left( {1 - \frac{1}{n}} \right)\left( {b - a} \right)}}{2}} \right] = \left( {b - a} \right)\left[ {a + \frac{{\left( {b - a} \right)}}{2}} \right]\\&= \left( {b - a} \right)\left[ {\frac{{2a + b - a}}{2}} \right]\\&= \frac{{\left( {b - a} \right)\left( {b + a} \right)}}{2}\\&= \frac{1}{2}\left( {{b^2} - {a^2}} \right)\end{align}

## Chapter 7 Ex.7.8 Question 2

Evaluate the definite integral as limit of sum: $$\int_0^b {\left( {x + 1} \right)dx}$$

### Solution

Let $$I = \int_0^b {\left( {x + 1} \right)dx}$$

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

Here, $$a = 0,b = 5\,\,{\text{and }}f\left( x \right) = \left( {x + 1} \right)$$

\begin{align}&\Rightarrow \; \; h = \frac{{5 - 0}}{n} = \frac{5}{n}\\ \therefore \int_0^5 {\left( {x + 1} \right)} dx &= \left( {5 - 0} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {f\left( 0 \right) + f\left( {\frac{5}{n}} \right) + \ldots + f\left( {\left( {n - 1} \right)\frac{5}{n}} \right)} \right]\\&= 5\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {1 + \left( {\frac{5}{n} + 1} \right) + \ldots \left\{ {1 + \left( {\frac{{5\left( {n - 1} \right)}}{n}} \right)} \right\}} \right]\\&= 5\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\left( {1 + \mathop 1\limits_{n\;times} + 1 \ldots 1} \right) + \left[ {\frac{5}{n} + 2.\frac{5}{n} + 3.\frac{5}{n} + \ldots + \left( {n - 1} \right)\frac{5}{n}} \right]} \right]\\&= 5\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {n + \frac{5}{n}\left\{ {1 + 2 + 3 \ldots \left( {n - 1} \right)} \right\}} \right]\\&= 5\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {n + \frac{5}{n}.\frac{{\left( {n - 1} \right)n}}{2}} \right] = 5\mathop {\lim }\limits_{n \to \infty } \left[ {1 + \frac{{5\left( {n - 1} \right)}}{{2n}}} \right]\\&= 5\mathop {\lim }\limits_{n \to \infty } \left[ {1 + \frac{5}{2}\left( {1 - \frac{1}{n}} \right)} \right] = 5\left[ {1 + \frac{5}{2}} \right]\\&= 5\left[ {\frac{7}{2}} \right]\\&= \frac{{35}}{2}\end{align}

## Chapter 7 Ex.7.8 Question 3

Evaluate the definite integral as limit of sum: $$\int_2^3 {{x^2}dx}$$

### Solution

Since,

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

Here, $$a = 2,b = 3\,\,{\text{and }}f\left( x \right) = {x^2}$$

\begin{align}&\Rightarrow \; \; h = \frac{{3 - 2}}{n} = \frac{1}{n}\\ \therefore \int_2^3 {{x^2}} dx &= \left( {3 - 2} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {f\left( 2 \right) + f\left( {2 + \frac{1}{n}} \right) + f\left( {2 + \frac{2}{n}} \right) \ldots f\left\{ {2 + \left( {n - 1} \right)\frac{1}{n}} \right\}} \right]\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{{\left( 2 \right)}^2} + {{\left( {2 + \frac{1}{n}} \right)}^2} + {{\left( {2 + \frac{2}{n}} \right)}^2} + \ldots \left( {2 + \frac{{{{\left( {n - 1} \right)}^2}}}{n}} \right)} \right]\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{2^2} + \left\{ {{2^2} + {{\left( {\frac{1}{n}} \right)}^2} + 2.2\frac{1}{n}} \right\} + \ldots + \left\{ {{{\left( 2 \right)}^2} + \frac{{{{\left( {n - 1} \right)}^2}}}{{{n^2}}} + 2.2.\frac{{\left( {n - 1} \right)}}{n}} \right\}} \right]\end{align}

\begin{align}&\Rightarrow \; \; h = \frac{{3 - 2}}{n} = \frac{1}{n}\\&\therefore \int_2^3 {{x^2}} dx = \left( {3 - 2} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {f\left( 2 \right) + f\left( {2 + \frac{1}{n}} \right) + f\left( {2 + \frac{2}{n}} \right) \ldots f\left\{ {2 + \left( {n - 1} \right)\frac{1}{n}} \right\}} \right]\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{{\left( 2 \right)}^2} + {{\left( {2 + \frac{1}{n}} \right)}^2} + {{\left( {2 + \frac{2}{n}} \right)}^2} + \ldots \left( {2 + \frac{{{{\left( {n - 1} \right)}^2}}}{n}} \right)} \right]\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{2^2} + \left\{ {{2^2} + {{\left( {\frac{1}{n}} \right)}^2} + 2.2\frac{1}{n}} \right\} + \ldots + \left\{ {{{\left( 2 \right)}^2} + \frac{{{{\left( {n - 1} \right)}^2}}}{{{n^2}}} + 2.2.\frac{{\left( {n - 1} \right)}}{n}} \right\}} \right]\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\left( {{2^2} + \mathop \ldots \limits_{n\;times} + {2^2}} \right) + \left\{ {{{\left( {\frac{1}{n}} \right)}^2} + {{\left( {\frac{2}{n}} \right)}^2} + \ldots + {{\left( {\frac{{n - 1}}{n}} \right)}^2}} \right\} + 2.2.\left\{ {\frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \ldots + \frac{{\left( {n - 1} \right)}}{n}} \right\}} \right]\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {4n + \frac{1}{{{n^2}}}\left\{ {{1^2} + {2^2} + {3^2} \ldots + {{\left( {n - 1} \right)}^2}} \right\} + \frac{4}{n}\left\{ {1 + 2 + \ldots + \left( {n - 1} \right)} \right\}} \right]\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {4n + \frac{1}{{{n^2}}}\left\{ {\frac{{n\left( {n - 1} \right)\left( {2n - 1} \right)}}{6}} \right\} + \frac{4}{n}\left\{ {\frac{{n\left( {n - 1} \right)}}{2}} \right\}} \right]\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {4n + \frac{{n\left( {1 - \frac{1}{n}} \right)\left( {2 - \frac{1}{n}} \right)}}{6} + \frac{{4n - 4}}{2}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {4 + \frac{1}{6}\left( {1 - \frac{1}{n}} \right)\left( {2 - \frac{1}{n}} \right) + 2 - \frac{2}{n}} \right]\\&= 4 + \frac{2}{6} + 2\\&= \frac{{19}}{3}\end{align}

## Chapter 7 Ex.7.8 Question 4

Evaluate the definite integral as limit of sum: $$\int_1^4 {\left( {{x^2} - x} \right)dx}$$

### Solution

Let $$I = \int_1^4 {\left( {{x^2} - x} \right)dx}$$

$$\;\qquad = \int_1^4 {{x^2}dx} - \int_1^4 {x\;dx}$$

Let $$I = {I_1} - {I_2},$$ where $${I_1} = \int_1^4 {{x^2}dx} \,\,{\text{and }}{I_2} = \int_1^4 {x\;dx} \,\,\,\,\, \ldots \left( 1 \right)$$

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

For, $${I_1} = \int_1^4 {{x^2}dx} ,$$

\begin{align}&a = 1,b = 4\,\,{\text{and }}f\left( x \right) = {x^2}\\ \therefore h &= \frac{{4 - 1}}{n} = \frac{3}{n}\\{I_1}& = \int_1^4 {{x^2}dx} = \left( {4 - 1} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {f\left( 1 \right) + f\left( {1 + h} \right) + \ldots + f\left( {1 + \left( {n - 1} \right)h} \right)} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{1^2} + {{\left( {1 + \frac{3}{n}} \right)}^2} + {{\left( {1 + 2.\frac{3}{n}} \right)}^2} + \ldots {{\left( {1 + \frac{{\left( {n - 1} \right)3}}{n}} \right)}^2}} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{1^2} + \left\{ {{1^2} + {{\left( {\frac{3}{n}} \right)}^2} + 2.\frac{3}{n}} \right\} + \ldots + \left\{ {{1^2} + {{\left( {\frac{{\left( {n - 1} \right)3}}{n}} \right)}^2} + \frac{{2.\left( {n - 1} \right).3}}{2}} \right\}} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\left( {{1^2} + \mathop \ldots \limits_{n\;times} + {1^2}} \right) + {{\left( {\frac{3}{n}} \right)}^2}\left\{ {{1^2} + {2^2} + \ldots + {{\left( {n - 1} \right)}^2}} \right\} + 2.\frac{3}{n}\left\{ {1 + 2 + \ldots + \left( {n - 1} \right)} \right\}} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {n + \frac{9}{{{n^2}}}\left\{ {\frac{{\left( {n - 1} \right)\left( n \right)\left( {2n - 1} \right)}}{6}} \right\} + \frac{6}{n}\left\{ {\frac{{\left( {n - 1} \right)\left( n \right)}}{2}} \right\}} \right]\\&= \mathop {3\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {n + \frac{{9n}}{6}\left( {1 - \frac{1}{n}} \right)\left( {2 - \frac{1}{n}} \right) + \frac{{6n - 6}}{2}} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \left[ {1 + \frac{9}{6}\left( {1 - \frac{1}{n}} \right)\left( {2 - \frac{1}{n}} \right) + 3 - \frac{3}{n}} \right]\\&= 3\left[ {1 + 3 + 3} \right]\\&= 3\left[ 7 \right]\\{I_1}& = 21\,\,\,\,\, \ldots \left( 2 \right)\end{align}

For $${I_2} = \int_1^4 {x\;dx}$$

\begin{align}&a = 1,b = 4\,\,{\text{and }}f\left( x \right) = x\\&\Rightarrow \; \; h = \frac{{4 - 1}}{n} = \frac{3}{n}\\&\therefore {I_2} = \left( {4 - 1} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {f\left( 1 \right) + f\left( {1 + h} \right) + \ldots + f\left( {a + \left( {n - 1} \right)h} \right)} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {1 + \left( {1 + h} \right) + \ldots + \left( {1 + \left( {n - 1} \right)h} \right)} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {1 + \left( {1 + \frac{3}{n}} \right) + \ldots + \left\{ {1 + \left( {n - 1} \right)\frac{3}{n}} \right\}} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\left( {1 + 1 + \mathop \ldots \limits_{n\;times} + 1} \right) + \frac{3}{n}\left( {1 + 2 + \ldots + \left( {n - 1} \right)} \right)} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {n + \frac{3}{n}\left\{ {\frac{{\left( {n - 1} \right)n}}{2}} \right\}} \right]\\&= 3\mathop {\lim }\limits_{n \to \infty } \left[ {1 + \frac{3}{2}\left( {1 - \frac{1}{n}} \right)} \right]\\&= 3\left[ {1 + \frac{3}{2}} \right] = 3\left[ {\frac{5}{2}} \right]\\{I_2} &= \frac{{15}}{2}\,\,\,\,\, \ldots \left( 3 \right)\end{align}

From equations (2) and (3), we get

$I = {I_1} - {I_2} = 21 - \frac{{15}}{2} = \frac{{27}}{2}$

## Chapter 7 Ex.7.8 Question 5

Evaluate the definite integral as limit of sum: $$\int_{ - 1}^1 {{e^x}dx}$$

### Solution

Let $$I = \int_{ - 1}^1 {{e^x}dx} \,\,\,\,\, \ldots \left( 1 \right)$$

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

Here, $$a = - 1,\;b = 1\,\,{\text{and }}f\left( x \right) = {e^x}$$

\begin{align}\therefore h &= \frac{{1 + 1}}{n} = \frac{2}{n}\\\therefore I &= \left( {1 + 1} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {f\left( { - 1} \right) + f\left( { - 1 + \frac{2}{n}} \right) + f\left( { - 1 + 2.\frac{2}{n}} \right) + \ldots + f\left( { - 1 + \frac{{\left( {n - 1} \right)2}}{n}} \right)} \right]\\&= 2\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{e^{ - 1}} + {e^{\left( { - 1 + \frac{2}{n}} \right)}} + {e^{\left( { - 1 + 2\frac{2}{n}} \right)}} + {e^{\left( { - 1 + \left( {n - 1} \right)\frac{2}{n}} \right)}}} \right]\\&= 2\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{e^{ - 1}}\left\{ {1 + {e^{\frac{2}{n}}} + {e^{\frac{4}{n}}} + {e^{\frac{6}{n}}} + {e^{\left( {n - 1} \right)\frac{2}{n}}}} \right\}} \right]\\&= 2\mathop {\lim }\limits_{n \to \infty } \frac{{{e^{ - 1}}}}{n}\left[ {\frac{{{e^{\frac{{2n}}{n}}} - 1}}{{{e^{\frac{2}{n}}} - 1}}} \right] = {e^{ - 1}} \times 2\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\frac{{{e^2} - 1}}{{{e^{\frac{2}{n}}} - 1}}} \right]\\&= \frac{{{e^{ - 1}} \times 2\left( {{e^2} - 1} \right)}}{{\mathop {\lim }\limits_{\frac{2}{n} \to 0} \left( {\frac{{{e^{\frac{2}{n}}} - 1}}{{\frac{2}{n}}}} \right) \times 2}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{{e^h} - 1}}{h}} \right) = 1} \right]\\&= \frac{{{e^2} - 1}}{e}\\&= \left( {e - \frac{1}{e}} \right)\end{align}

## Chapter 7 Ex.7.8 Question 6

Evaluate the definite integral as limit of sum: $$\int_0^4 {\left( {x + {e^{2x}}} \right)} dx$$

### Solution

Since,

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

Here, $$a = 0,\;b = 4\,\,{\text{and }}f\left( x \right) = x + {e^{2x}}$$

\begin{align}\therefore h& = \frac{{4 - 0}}{n} = \frac{4}{n}\\[10pt]&\Rightarrow \; \; \int_0^4 {\left( {x + {e^{2x}}} \right)dx} = \left( {4 - 0} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {f\left( 0 \right) + f\left( h \right) + f\left( {2h} \right) + \ldots + f\left( {\left( {n - 1} \right)h} \right)} \right]\\[10pt] &= 4\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\left( {0 + {e^0}} \right) + \left( {h + {e^{2h}}} \right) + \left( {2h + {e^{2.2h}}} \right) + \ldots + \left\{ {\left( {n - 1} \right)h + {e^{2\left( {n - 1} \right)h}}} \right\}} \right]\\[10pt]&= 4\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {1 + \left( {h + {e^{2h}}} \right) + \left( {2h + {e^{4h}}} \right) + \ldots + \left\{ {\left( {n - 1} \right)h + {e^{2\left( {n - 1} \right)h}}} \right\}} \right]\\[10pt]&= 4\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\left\{ {h + 2h + 3h + ...... + \left( {n - 1} \right)h} \right\} + \left( {1 + {e^{2h}} + {e^{4h}} + ..... + {e^{2\left( {n - 1} \right)h}}} \right)} \right]\\[10pt]&= 4\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {h\left\{ {1 + 2 + \ldots \left( {n - 1} \right)} \right\} + \left( {\frac{{{e^{2hn}} - 1}}{{{e^{2h}} - 1}}} \right)} \right] = 4\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\frac{{\left[ {h\left( {n - 1} \right)n} \right]}}{2} + \left( {\frac{{{e^{2hn}} - 1}}{{{e^{2h}} - 1}}} \right)} \right]\\[10pt]&= 4\mathop {\lim }\limits_{x \to \infty } \frac{1}{n}\left[ {\frac{4}{n}\frac{{\left( {n - 1} \right)n}}{2} + \left( {\frac{{{e^8} - 1}}{{{e^{\frac{8}{n}}} - 1}}} \right)} \right] = 4\left( 2 \right) + 4\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {{e^8} - 1} \right)}}{{\left( {\frac{{{e^{\frac{8}{n}}} - 1}}{{\frac{8}{n}}}} \right)8}}\\[10pt]&= 8 + \frac{{4\left( {{e^8} - 1} \right)}}{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = 1} \right)\\[10pt]&= 8 + \frac{{{e^8} - 1}}{2} = \frac{{15 + {e^8}}}{2}\end{align}

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