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Tricks & Tips OF Area Under The Curve FOR NDA

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Topic Covered in This Lecture

In this lecture we will cover following topics :

1. Using transformation of graph

2. Using expansion of `|x-a|` in different intervals

3. General Graph

Using transformation of graph

Q 2166380275

Consider the curves `y = | x - 1 |` and `| x |= 2`

What is the area of the region bounded by the curves and
x-axis?
NDA Paper 1 2016
(A)

`3` sq units

(B)

`4` sq units

(C)

`5` sq units

(D)

`6` sq units

Solution:

Given curves

`y = | x - 1|` and `| x | = 2`

` => { tt ((x - 1, x >= 1 ),( -x (x-1) , x < 1 text(and) x = pm 2))`

`:.` Required area

` = int _(-2)^(1) - (x- 1) dx + int _(1)^(2) (x -1) dx`

` = int _(-2)^(1) -x) dx + int _(1)^(2) dx`

` = [x - x^( 2)/2]_(-2)^(1) + [ x^( 2)/2 - x]_(1)^(2)`

` = [(1- 1/2) -( -2- 4/2)] + [(4/2 -2 )-(1/2 -1 )]`

`= [1/2 - (-4) ] + [ 0 - (-1/2) ]`

` = (1/2 + 4) + 1/2 = 5` sq units
Correct Answer is `=>` (C) `5` sq units
Q 2751380224

What is the area bounded by the curves `|y| = 1-x^2 ?`
NDA Paper 1 2016
(A)

`4/3` square units

(B)

`8/3` square units

(C)

`4` square units

(D)

`16/3` square units

Solution:

`y = 1-x^2`

`|y| = 1-x^2`

`A = 4 int_(0)^(1) (1-x^2)dx`

` = 4 [ x - x^3/3]_(0)^(1) = 4 (1-1/3) = 8/3`
Correct Answer is `=>` (B) `8/3` square units

Using expansion of `|x-a|` in different intervals

Q 2116501479

Consider the function
`f(x) = | x - 1| + x^(2)` where `x in R`.

What is the area of the region bounded by X-axis, the
curve `y = f( x)` and the two ordinates `x =1/2` and `x = 1?`
NDA Paper 1 2016
(A)

`5/12` sq unit

(B)

`5/6` sq unit

(C)

`7/6` sq unit

(D)

`2` sq unit

Solution:

Given `f(x) = | x -1 | + x^2`

`:. f(x) = { tt (( x^2 + (x - 1) , text(for x > 1) ), ( x^2 - x + 1 ,text(for x < 1)) )`

`f(x) = | x - 1| + x^(2)`

When `1/2 < x < 1`

`f(x) = x^( 2) - x + 1`

:. Area of required region will be

` int _(1//2)^(1) f(x) dx = int _(1//2)^(1) (x^(2) - x -1)dx = [x^(3)/3 - x^(2)/2 + x]_(1//2)^(1)`

` = [ (1)^(3)/3 - (1)^(2)/2 + (1) ] -[(1/2)^(3)/3 - (1/2)^(2)/2 +1/2]`

` = ( 1/3 - 1/2 +1) -( 1/24 - 1/8 +1/2 )`

` = ((2-3+6)/6 ) - ((1-3 +12)/24)`

` 5/6 - 10/24`

`(20-10)/24 = 10/24 = 5/12` sq unit.
Correct Answer is `=>` (A) `5/12` sq unit
Q 2136601572

Consider the function
`f(x) = | x - 1| + x^(2)` where `x in R`.

What is the area of the region bounded by X-axis, the
curve `y = f(x)` and the two ordinates `x =1` and `x = 3/2?`

NDA Paper 1 2016
(A)

`5/12` sq unit

(B)

`7/12` sq unit

(C)

`2/3` sq unit

(D)

`11/12` sq unit

Solution:

Given `f(x) = | x -1 | + x^2`

`:. f(x) = { tt (( x^2 + (x - 1) , text(for x > 1) ), ( x^2 - x + 1 ,text(for x < 1)) )`

`f(x) =| x- 1| + x^(2)`

When `1 < x < 3/2`

`f(x) = x ^(2) + x- 1`

:. Area of required region will be

` int _(1)^(3//2) f(x) dx = int _(1)^(3//2) (x^(2) + x -1) dx`

` = [x^(3)/3 +x^(2)/2 - x]_(1)^(3//2)`

`= [(3/2)^(3)/3 + (3/2)^(2)/2 - 3/2 ] - [(1)^(3)/3 + (1)^(2)/2 -1]`

`= (27/24 +9/8 -3/2 )- (1/3 +1/2 -1)`

`= ((27+27-36)/24) - ((2+3-6)/6) `

` = 18/24 +1/6 = (18 +4)/24 = 22/24`

` = 11/12 `sq unit
Correct Answer is `=>` (D) `11/12` sq unit
Q 2136878772

Given curves are
`f(x) =x | x |- 1` and `g(x) = { tt (((3x)/2 ,x>0) , (2x, x <= 0))`

What is the area bounded by the curves ?
NDA Paper 1 2016
(A)

`(17)/6` sq units

(B)

`8/3` sq units

(C)

`2` sq units

(D)

`1/3` sq units

Solution:

When `x > 0, f(x) =x^(2) -1, g(x) = (3x)/2`

and when `x < 0, f(x) = - x^(2) - 1, g(.x) = 2x`

` = | int_(0)^(2) [(3x)/2 - (x^(2) -1)] dx | + | int_(-1)^(0) [(-x^(2) -1)-2x]| dx`

` = | int_(0)^(2) ((3x)/2 - x^(2) + 1) dx | + | int_(-1)^(0) (-1) (x^(2) + 2x - 1) | dx`

` = [3/2 .x^(2)/2 - x^(3)/3 +x ]_(0)^(2) + [x^(3)/3 - (2x^2)/2 + x]_(-1)^(0)`

` =(3/2 xx 4/2 - 8/3 +2 ) + [ 0 - (-1/3 +1 -1)]`

` = (3 - 8/3 +2 ) +1/3 `

`= 5 -7/3 = 8/3` sq units
Correct Answer is `=>` (B) `8/3` sq units

General Graph

Q 1649801713

Consider the line `x = sqrt(3)y`
and the circle `x^2 + y^ 2 = 4`.

What is the area of the region in the first quadrant
enclosed by the `X`-axis, the line `x = sqrt(3)` and the
circle?
NDA Paper 1 2015
(A)

`pi/3 - sqrt(3)/2`

(B)

`pi/2 - sqrt(3)/2`

(C)

`pi/3 - 1/2`

(D)

None of the above

Solution:

Required area is shaded area represented in the
following figure

:. Required area `= int _sqrt(3)^2 sqrt(4 - x^2 ) dx`

` = 1/2 [ x sqrt(4 - x^2 ) + 4 sin^(-1)(x/2)]_sqrt(3)^2`

` = 1/2 [ 4 sin^(-1) (1) - sqrt(3) sqrt(1) - 4 sin^(-1) ( sqrt(3)/2)]`

` = 1/2 [ 4 . pi/2 - sqrt(3) -4 xx pi/3]`

` = pi - (2pi)/3 - sqrt(3)/2 = pi/3 - sqrt(3)/2`
Correct Answer is `=>` (A) `pi/3 - sqrt(3)/2`
Q 1609801718

Consider the curves `y = sin x` and `y = cos x`.

What is the area of the region bounded by the
above two curves and the lines `x = 0` and `x = pi/4 `?
NDA Paper 1 2015
(A)

`sqrt(2) - 1`

(B)

`sqrt(2) + 1`

(C)

`sqrt(2)`

(D)

`2`

Solution:

Given equation of curves are

`y = sinx ... (i)`

and `y = cosx ... (ii)`

The graph of above curves between `0` to `pi/2` is


Required area = Area of region `OABO`

`= int _0^(pi/4) (cosx-sinx) dx`

`=[sin x + cos x]_0^(pi/4)`

`= 1/sqrt(2) + 1/sqrt(2) - 0 - 1 = sqrt(2) - 1`
Correct Answer is `=>` (A) `sqrt(2) - 1`
Q 1629001811

Consider the curves `y = sin x` and `y = cos x`.


What is the area of the region bounded by the
above two curves and the lines `x = pi/4` and `x = pi/2`?
NDA Paper 1 2015
(A)

`sqrt(2) -1`

(B)

`sqrt(2) +1`

(C)

`2sqrt(2)`

(D)

`2`

Solution:

Given equation of curves are

`y = sinx ... (i)`

and `y = cosx ... (ii)`

The graph of above curves between `0` to `pi/2` is

Required area = Area of region `ACDA`

` = int_(pi/4)^(pi/4) ( sinx - cosx )dx`

`= [- cosx- sinx]_(pi/4)^(pi/2)`

`=- [cosx + sinx]_(pi/4)^(pi/2)`

`= - [0+1-(1/sqrt(3) + 1/sqrt(3) )]= sqrt(2) -1`
Correct Answer is `=>` (A) `sqrt(2) -1`
Q 2241480323

The area bounded by the coordinate axes and the curve
`sqrt(x) + sqrt(y) = 1` is
NDA Paper 1 2015
(A)

`1` sq unit

(B)

`1/2` sq unit

(C)

`1/3` sq unit

(D)

`1/6` sq unit

Solution:

`:.` Required area `= int _0^1 ( 1- sqrt(x))^2 dx`

` = int _0^1 (1+ x-2 sqrt(x)) dx`

`= [ x +x^2/2 -2 . x^(3//2)/(3//2)]_0^1`

` = 1 + 1/2 - 2 xx 2/3 (1)`

` = 1 + 1/2 - 4/3 = 1/6 `sq unit
Correct Answer is `=>` (D) `1/6` sq unit
Q 1659267114

The line `2y = 3x + 12` cuts the
parabola `4 y = 3x^2`

What is the area enclosed by the parabola and the line?
NDA Paper 1 2014
(A)

27 sq units

(B)

36 sq units

(C)

48 sq units

(D)

S4 sq units

Solution:

Area enclosed by the parabola and the line

` = int _(-2)^4 [((3x + 12)/2 - (3x^2)/4 )] dx`

` = 1/2 [ (3x^2)/2 + 12x ]_(-2)^(4) - 3/4 [(x^3)/3] _(-2)^(4)`

` = 1/2 [ {(3(4)^2)/2 + 12(4) } - { (3(-2)^2)/2 + 12(-2) }] - 3/4 [ 4^3/3 - (-2)^3/3]`

` = 1/2 [24 + 48- 6 + 24] - 3/4 [(64 + 8)/3]`

` = 1/2 (90) - 18`

` = 45 - 18 = 27` sq units
Correct Answer is `=>` (A) 27 sq units
Q 1619767610

What is the area of the parabola `y^2 = 4bx` bounded by its
latus rectum?
NDA Paper 1 2014
(A)

`(2b^2) /3` sq units

(B)

`(4b^2) /3` sq units

(C)

` b^2 ` sq units

(D)

`(8b^2) /3` sq units

Solution:

Equation of parabola is `y^2 = 4bx`.

Area of parabola bounded by its latus rectum

` 2. int _0^b sqrt(4bx) dx`

` = 4 sqrt(b) x 2/3 [ x^(3//2)] _0^b = (8 sqrt(b))/3 [ b^(3//2)-0]`

` = (8b^2)/3 ` sq units
Correct Answer is `=>` (D) `(8b^2) /3` sq units
Q 1753223144

Consider an ellipse, `x^2/a^2 + y^2/b^2 = 1`

What is the area included between the ellipse and the
greatest rectangle inscribed in the ellipse?
NDA Paper 1 2014
(A)

`ab (pi - 1)`

(B)

`2ab (pi - 1)`

(C)

`ab (pi - 2)`

(D)

None of these

Solution:

Given ellipse, `x^2/a^2 + y^2/b^2 = 1`

We know that,

Area of the ellipse `x^2/a^2 + y^2/b^2 = 1` is, `Delta ' = pi ab`

`:.` Required area =Area of shaded region

=Area of an ellipse

-Area of greatest rectangle

`= pi ab - 2ab`

` = ab (pi -2)`
Correct Answer is `=>` (C) `ab (pi - 2)`
Q 2379667516

What is the area bounded by the lines `x = 0, y = 0`
and `x + y + 2 = 0?`
NDA Paper 1 2013
(A)

`1/2` sq unit

(B)

`1` sq unit

(C)

`2` sq units

(D)

`4` sq units

Solution:

Given equation of lines `x= 0, y = 0` and `x + y + 2 = 0`


`therefore` Required area = Area of `Delta OAB`


` = 1/2 xx OA xx OB = 1/2xx 2 xx 2 = 2 qs ` units


Alternate Method

`therefore` Required area = `| underset(-2) overset(0) inty dx| = |underset(-2) overset(0) int(-2-x)dx|`

` = |[-2x-x^2/2]_(-2)^(0)| = |[0+2(-2)+(-2)^2/2]|`


` = |-4+2| = 2 qs ` units
Correct Answer is `=>` (C) `2` sq units
Q 2349767613

What is the area of the parabola `x^2 = y` bounded
by the line `y = 1?`
NDA Paper 1 2013
(A)

`1/3` qs units

(B)

`2/3` qs units

(C)

`4/3` qs units

(D)

`2` qs units

Solution:

Given equation of parabola and line are


`x^2 = y` ................(i)


and `y = 1` ................(ii)

On solving Eqs. (i) and (ii), we get


`x^2 = 1 => x = pm1`


`therefore` Required area = `2 xx` Area of OPBO


` = 2 int_0^1 sqrty dy`


`2[(2y^(3/2))/3]_0^1 = 4/3(1-0) = 4/3` sq units

Area along x axis

`= int_(-1)^1 (1 - y) dx = int_(-1)^1 [1 - x^2] dx `

`= 2 int_0^1 [1 - x^2] dx `

`= 2 ([x - x^3/3]_0)^1 = 2 [1-1/3]`

`4/3`
Correct Answer is `=>` (C) `4/3` qs units
Q 2369867715

What is the area bounded by `y = tanx, y = 0` and `x = pi/4 ?`
NDA Paper 1 2013
(A)

`log 2` sq units

(B)

`(log2)/2` sq units

(C)

`2 (log 2)` sq units

(D)

None of these

Solution:

Given equation of curves
`y = tanx` ................(i)

and `y= 0` and `x = pi/4` .........(ii)


`therefore` Requrred area = `int_0^(pi/4) ydx = int_0^(pi/4) tanxdx`


`[log|secx|_0^(pi/4) = log|sec(pi)|-log|sec0|`


`log|sqrt2|-log|1| = log sqrt2-0 = 1/2 log2` sq units
Correct Answer is `=>` (B) `(log2)/2` sq units
Q 2319067810

What is the area of the region enclosed by
`y = 2 |x|` and `y = 4?`
NDA Paper 1 2013
(A)

2 sq units

(B)

4 sq units

(C)

8 sq units

(D)

16 sq units

Solution:

Given curves



`therefore` Required area = area of `Delta OAB = 1/2xxOP xx AB`

` = 1/2xx4xx4 = 8` sq units
Correct Answer is `=>` (C) 8 sq units
Q 2379067816

What is the area of the parabola `y^2 = x` bounded
by its latusrectum?
NDA Paper 1 2013
(A)

`1/12` sq units

(B)

`1/6` sq units

(C)

`1/3` sq units

(D)

None of these

Solution:

Given curve, `y^2= x`

On compare with `y^2 = 4ax`, we get



`4a = 1`


`a = 1/4`

`therefore` Focus of the parabola, `S = (a , 0) = (1/4 , 0)`


`therefore` Required area =`2 * int_0^(1/4) sqrtxdx =2[2/3x^(3/2)]_0^(1/4)`


` = 4/3[(1/2)^3-0]`


` = 4/3 * (1/2)^3 = 4/3xx1/8 = 1/6 ` sq units
Correct Answer is `=>` (B) `1/6` sq units
Q 2319167910

What is the area of the portion of the curve
`y = sin x`, lying between `x = 0, y = 0` and `x = 2pi?`
NDA Paper 1 2012
(A)

1 sq unit

(B)

2 sq units

(C)

4 sq units

(D)

8 sq units

Solution:

`therefore` Required area (OBAB'C)

` = int_0^pisinxdx+int_pi^(2pi)-sinxdx`


` = [-cosx]_0^(pi)+[cosx]_0^(2pi)`


` = (cospi-cos0)+(cos2pi-cospi)`

` = -(-1-1)+(1+1) = 4` sq units
Correct Answer is `=>` (C) 4 sq units
Q 2319178919

The area bounded by the curve `x = f(y)`, the
Y-axis and the two lines `y = a` and `y = b` is equa to
NDA Paper 1 2012
(A)

`int_a^bydx`

(B)

`int_a^by^2dx`

(C)

`int_a^bxdy`

(D)

None of these

Solution:

Requires area ` = int_(y = a)^(y = b)xdy`
Correct Answer is `=>` (C) `int_a^bxdy`
Q 2359180014

What is the i area bounded by the curve
`sqrtx + sqrty = sqrta` where, `(x,y ge 0)` and the coordinate
axes?
NDA Paper 1 2011
(A)

`(5a^2)/6`

(B)

`a^2/3`

(C)

`a^2/2`

(D)

`a^2/6`

Solution:

The given equation of curve


`sqrtx+sqrty = sqrta` (where,`x,y ge 0`)


`=> sqrty = sqrta-sqrtx => (sqrty)^2 = (sqrta-sqrtx)^2`


`y = ( sqrta-sqrtx)^2`


At `x = 0 , sqrty = sqrta => y = a`

At `y = 0 , sqrtx = sqrta => x = a`


So, curve cuts the axes at (a, 0) and (0, a), respectively


`therefore` Required area `= int_(x = 0)^(x = a) ydx = int_0^a (sqrta-sqrtx)^2dx`


` = int_0^a (a+x-2sqrtasqrtx)dx = [ax+x^2/2-4/3 sqrt(ax^(3/2))dx]_0^a`

` = a^2+a^2/2-4/3 sqrta * a^(3/2) = (3a^2)/2-4/3a^2 = (9-8)/6a^2 = a^2/6`
Correct Answer is `=>` (D) `a^2/6`
 
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