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1

### AIEEE 2005

The area enclosed between the curve $$y = {\log _e}\left( {x + e} \right)$$ and the coordinate axes is
A
$$1$$
B
$$2$$
C
$$3$$
D
$$4$$

## Explanation

The graph of the curve $$y = {\log _e}\left( {x + e} \right)$$ is as shown in the fig.

Required area

$$A = \int\limits_{1 - e}^0 {ydx} = \int\limits_{1 - e}^0 {{{\log }_e}} \left( {x + e} \right)dx$$

put $$x + e = t \Rightarrow dx = dt$$

also At $$x = 1 - e,t = 1$$

At $$x = 0,\,\,t = e$$

$$\therefore$$ $$A = \int\limits_1^e {{{\log }_e}} \,tdt = \left[ {t\,{{\log }_e}t - t_1^e} \right]$$

$$e - e - 0 + 1 = 1$$

Hence the required area is $$1$$ square unit.
2

### AIEEE 2005

If $${I_1} = \int\limits_0^1 {{2^{{x^2}}}dx,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx,\,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} } }$$ and $${I_4} = \int\limits_1^2 {{2^{{x^3}}}dx}$$ then
A
$${I_2} > {I_1}$$
B
$${I_1} > {I_2}$$
C
$${I_3} = {I_4}$$
D
$${I_3} > {I_4}$$

## Explanation

$${I_1} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_2} = \int\limits_0^1 {{2^{{x^3}}}} dx,$$

$$= {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,$$

$${I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\,$$

$$\forall 0 < x < 1,\,{x^2} > {x^3}$$

$$\Rightarrow \int\limits_0^1 {{2^{{x^2}}}} \,dx > \int\limits_0^1 {{2^{{x^3}}}} dx$$

and $$\int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}dx}$$

$$\Rightarrow {I_1} > {I_2}$$ and $${I_4} > {I_3}$$
3

### AIEEE 2004

The area of the region bounded by the curves
$$y = \left| {x - 2} \right|,x = 1,x = 3$$ and the $$x$$-axis is
A
$$4$$
B
$$2$$
C
$$3$$
D
$$1$$

## Explanation

The required area is shown by shaded region

$$A = \int\limits_1^3 {\left| {x - 2} \right|dx = 2\int\limits_2^3 {\left( {x - 2} \right)} } dx$$

$$= 2\left[ {{{{x^2}} \over 2} - 2x} \right]_2^3 = 1$$
4

### AIEEE 2004

If $$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx}$$
and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,$$ then the value of $${{{I_2}} \over {{I_1}}}$$ is
A
$$1$$
B
$$-3$$
C
$$-1$$
D
$$2$$

## Explanation

$$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$$

$$\Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}$$

$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$

Now $${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$$

$$= \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$$

$$\left[ {} \right.$$ using $$\int\limits_a^b {f\left( x \right)} dx\,a$$

$$= \int\limits_a^b {f\left( {a + b - x} \right)dx}$$ $$\left. \, \right]$$

$$= {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$$

$$\therefore$$ $${{{I_2}} \over {{I_1}}} = 2$$

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