Area Under The Curves

The area enclosed between the curves

${y^2} = x$ and $y = \left| x \right|$

From the figure, area lies between

${y^2} = x$ and $y = x$



$\therefore$ Required area $ = \int_0^1 {\left( {{y_2} - {y_1}} \right)} dx$

$ = \int_0^1 {\left( {\sqrt x - x} \right)dx = \left[ {{{{x^{3/2}}} \over {3/2}} - {{{x^2}} \over 2}} \right]} _0^1$

$\therefore$ Required area $ = {2 \over 3}\left[ {{x^{3/2}}} \right]_0^1 - {1 \over 2}\left[ {{x^2}} \right]_0^1$

$ = {2 \over 3} - {1 \over 2} = {1 \over 6}$

Intersection points of ${x^2} = 4y$ and ${y^2} = 4x$ are $\left( {0,0} \right)$ and $\left( {4,4} \right).$ The graph is as shown in the figure.



By symmetry, we observe

${S_1} = {S_3} = \int\limits_0^4 {ydx = \int\limits_0^4 {{{{x^2}} \over 4}dx} } $

$ = \left[ {{{{x^3}} \over {12}}} \right]_0^4 = {{16} \over 3}$ sq. unit

Also ${S_2} = \int\limits_0^4 {\left( {2\sqrt x - {{{x^2}} \over 4}} \right)} dx$

$\,\,\,\,\,\,\,\,\,\,\,\,$ $ = \left[ {{{2{x^{{3 \over 2}}}} \over {{3 \over 2}}} - {{{x^3}} \over {12}}} \right]_0^4$

$\,\,\,\,\,\,\,\,\,\,\,\,$ $ = {4 \over 3} \times 8 - {{16} \over 3} = {{16} \over 3}$

$\therefore$ ${S_1}:{S_2}:{S_3} = 1:1:1$

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.

Given that

$\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta $

Differentiating $w.r.t$ $\beta $

$f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2 $

$f\left( {{\pi \over 2}} \right) = \left( {1 - {\pi \over 4}} \right)\sin {\pi \over 2} + \sqrt 2 $

$ = 1 - {\pi \over 4} + \sqrt 2 $

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$

$A = \int\limits_{ - 1}^0 {\left\{ {\left( {3 + x} \right) - \left( { - x + 1} \right)} \right\}dx + } $

$\,\,\,\,\,\,\,\,\,\,\,\,$ $\int\limits_0^1 {\left\{ {\left( {3 - x} \right) - \left( { - x + 1} \right)} \right\}dx + } $

$\,\,\,\,\,\,\,\,\,\,\,\,$ $\int\limits_1^2 {\left\{ {\left( {3 - x} \right) - \left( {x - 1} \right)} \right\}dx} $

$ = \int\limits_{ - 1}^0 {\left( {2 + 2x} \right)dx + \int\limits_0^1 {2dx + \int\limits_1^2 {\left( {4 - 2x} \right)dx} } } $

$ = \left[ {2x - {x^2}} \right]_{ - 1}^0 + \left[ {2x} \right]_0^1 + \left[ {4x - {x^2}} \right]_1^2$

$ = 0 - \left( { - 2 + 1} \right) + \left( {2 - 0} \right) + \left( {8 - 4} \right) - \left( {4 - 1} \right)$

$ = 1 + 2 + 4 - 3 = 4$ sq. units

First we draw each curve as separate graph



NOTE : Graph of $y = \left| {f\left( x \right)} \right|$ can be obtained from the graph of the curve $y = f\left( x \right)$ by drawing the mirror image of the portion of the graph below $x$-axis, with respect to $x$-axis.

Clearly the bounded area is as shown in the following figure.



Required area $ = 4\int\limits_0^1 {\left( { - \ln x} \right)} dx$

$ = - 4\left[ {x\,\ln x + - x_0^1} \right] = 4\,\,$ sq. units