Question

Evaluate The Following Definite Integrals As Limit Of Sums:-$\purple {\boxed{ \red{ \boxed{\displaystyle \tt \green{ \leadsto\int_{0}^{2}} \blue{\left(e^{x}-x\right) d x}}}}}$Answer...

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm \: \displaystyle\int_0^2\rm ( {e}^{x} - x) \: dx$

Here,

Lower limit, a = 0

Upper limit, b = 2

$\rm \: f(x) = {e}^{x} - x$

Step:- 1

$\rm \: nh = b - a = 2 - 0 = 2 - - - - (1)$

where h is the width of interval and n is number of intervals of width h

Step :- 2

$\rm \: f(x) = {e}^{x} - x$

Put x = a + rh = 0 + rh = rh

So, above expression can be rewritten as

$\rm \: f(rh) = {e}^{rh} - rh$

Step :- 3

By definition of Limit as a Sum,

$\rm \: \displaystyle\int_a^b \: f(x) \: dx \: = \lim _{h \to \: 0}\bigg(h \sum_{r = 0}^{n - 1} f(a + rh)\bigg)$

So, on substituting the values, we get

$\rm \: \displaystyle\int_0^2 ({e}^{x} - x) \: dx \: = \lim _{h \to \: 0}\bigg(h \sum_{r = 0}^{n - 1} ({e}^{rh} - rh\bigg)$

$\rm \: = \lim _{h \to \: 0}h \sum_{r = 0}^{n - 1}{e}^{rh} - \lim _{h \to \: 0}h \sum_{r = 0}^{n - 1} rh$

$\rm \: = \lim _{h \to \: 0}h (1 + {e}^{h} + {e}^{2h} + - - + {e}^{(n - 1)h} ) - \lim _{h \to \: 0} {h}^{2} \sum_{r = 0}^{n - 1} r$

Now, first term forms a GP series, so using sum of n terms, we get

$\rm \: = \lim _{h \to \: 0}h \: \frac{1({e}^{hn} - 1)}{{e}^{h} - 1} - \lim _{h \to \: 0} {h}^{2} \dfrac{n(n - 1)}{2}$

$\rm \: = \: \frac{{e}^{2} - 1}{1} - \lim _{h \to \: 0} \dfrac{nh(nh - h)}{2}$

$\rm \: = {e}^{2} - 1 - \dfrac{2(2 - 0)}{2}$

$\rm \: = {e}^{2} - 1 - 2$

$\rm \: = {e}^{2} - 3$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \: \displaystyle\int_0^2\rm ( {e}^{x} - x) \: dx = {e}^{2} - 3}}$

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Formulae Used

1. Sum of n terms of GP series having first term a and common ratio r respectively is given by

$\boxed{\tt{ S_n \: = \: \frac{a( {r}^{n} - 1)}{r - 1} \: \: provided \: that \: r \: \ne \: 1 \: }}$

2. Sum of first n natural numbers is given by

$\boxed{\tt{ \sum_{r = 0}^{n - 1}r \: = \: \frac{n(n - 1)}{2} \: }}$

3. Limit result

$\boxed{\tt{ \: \lim _{x \to \: 0} \: \frac{{e}^{x} - 1}{x} = 1 \: }}$

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ADDITIONAL INFORMATION

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

Here,

Lower limit, a = 0

Upper limit, b = 2

\rm \: f(x) = {e}^{x} - xf(x)=e

x

−x

Step:- 1

\begin{gathered}\rm \: nh = b - a = 2 - 0 = 2 - - - - (1) \\ \end{gathered}

nh=b−a=2−0=2−−−−(1)

where h is the width of interval and n is number of intervals of width h

Step :- 2

\rm \: f(x) = {e}^{x} - xf(x)=e

x

−x

Put x = a + rh = 0 + rh = rh

So, above expression can be rewritten as

\begin{gathered}\rm \: f(rh) = {e}^{rh} - rh \\ \end{gathered}

f(rh)=e

rh

−rh

Step :- 3

By definition of Limit as a Sum,

\begin{gathered}\rm \: \displaystyle\int_a^b \: f(x) \: dx \: = \lim _{h \to \: 0}\bigg(h \sum_{r = 0}^{n - 1} f(a + rh)\bigg) \\ \end{gathered}

∫

a

b

f(x)dx=

h→0

lim

(h

r=0

∑

n−1

f(a+rh))

So, on substituting the values, we get

\begin{gathered}\rm \: \displaystyle\int_0^2 ({e}^{x} - x) \: dx \: = \lim _{h \to \: 0}\bigg(h \sum_{r = 0}^{n - 1} ({e}^{rh} - rh\bigg) \\ \end{gathered}

∫

0

2

(e

x

−x)dx=

h→0

lim

(h

r=0

∑

n−1

(e

rh

−rh)

\begin{gathered}\rm \: = \lim _{h \to \: 0}h \sum_{r = 0}^{n - 1}{e}^{rh} - \lim _{h \to \: 0}h \sum_{r = 0}^{n - 1} rh \\ \end{gathered}

=

h→0

lim

h

r=0

∑

n−1

e

rh

−

h→0

lim

h

r=0

∑

n−1

rh

\begin{gathered}\rm \: = \lim _{h \to \: 0}h (1 + {e}^{h} + {e}^{2h} + - - + {e}^{(n - 1)h} ) - \lim _{h \to \: 0} {h}^{2} \sum_{r = 0}^{n - 1} r \\ \end{gathered}

=

h→0

lim

h(1+e

h

+e

2h

+−−+e

(n−1)h

)−

h→0

lim

h

2

r=0

∑

n−1

r

Now, first term forms a GP series, so using sum of n terms, we get

\begin{gathered}\rm \: = \lim _{h \to \: 0}h \: \frac{1({e}^{hn} - 1)}{{e}^{h} - 1} - \lim _{h \to \: 0} {h}^{2} \dfrac{n(n - 1)}{2} \\ \end{gathered}

=

h→0

lim

h

e

h

−1

1(e

hn

−1)

−

h→0

lim

h

2

2

n(n−1)

\begin{gathered}\rm \: = \: \frac{{e}^{2} - 1}{1} - \lim _{h \to \: 0} \dfrac{nh(nh - h)}{2} \\ \end{gathered}

=

1

e

2

−1

−

h→0

lim

2

nh(nh−h)

\begin{gathered}\rm \: = {e}^{2} - 1 - \dfrac{2(2 - 0)}{2} \\ \end{gathered}

=e

2

−1−

2

2(2−0)

\begin{gathered}\rm \: = {e}^{2} - 1 - 2 \\ \end{gathered}

=e

2

−1−2

\begin{gathered}\rm \: = {e}^{2} - 3 \\ \end{gathered}

=e

2

−3

Hence,

\begin{gathered}\rm\implies \:\boxed{\tt{ \rm \: \displaystyle\int_0^2\rm ( {e}^{x} - x) \: dx = {e}^{2} - 3}}\\ \end{gathered}

⟹

∫

0

2

(e

x

−x)dx=e

2

−3

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Formulae Used

1. Sum of n terms of GP series having first term a and common ratio r respectively is given by

\begin{gathered}\boxed{\tt{ S_n \: = \: \frac{a( {r}^{n} - 1)}{r - 1} \: \: provided \: that \: r \: \ne \: 1 \: }} \\ \end{gathered}

S

n

=

r−1

a(r

n

−1)

providedthatr



=1

2. Sum of first n natural numbers is given by

\begin{gathered}\boxed{\tt{ \sum_{r = 0}^{n - 1}r \: = \: \frac{n(n - 1)}{2} \: }} \\ \end{gathered}

r=0

∑

n−1

r=

2

n(n−1)

3. Limit result

\begin{gathered}\boxed{\tt{ \: \lim _{x \to \: 0} \: \frac{{e}^{x} - 1}{x} = 1 \: }} \\ \end{gathered}

x→0

lim

x

e

x

−1

=1

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}

f(x)

k

sinx

cosx

sec

2

x

cosec

2

x

secxtanx

cosecxcotx

tanx

x

1

e

x

∫f(x)dx

kx+c

−cosx+c

sinx+c

tanx+c

−cotx+c

secx+c

−cosecx+c

logsecx+c

logx+c

e

x

+c