Applications Of Integrals - Class 12 Commerce Applied Mathematics - Extra Questions

$$\int _{ 0 }^{  3/2 }{ (1+4x-x^2)dx}=2\int _{ 0 }^{ 3/2 }{  mxdx}  $$
$$ \Rightarrow { |x+2x^2-x^3/3| }_{ 0 }^{ 3/2 }={ |mx^2| }_{ 0 }^{ 3/2 }$$
$$ \Rightarrow 6m=13$$

The given equation of the ellipse can be represented as

The given equation of the ellipse, 

The area bounded by the parabola, $$x^2=y$$, and the line, $$y=|x|$$, can be represented as
The given area is symmetrical about $$y$$-axis
$$\therefore$$ Area $$OACO$$ $$=$$ Area $$ODBO$$
The point of intersection of parabola, $$x^2=y$$, and line, $$y=x$$, is $$A(1,1)$$.
Area of $$OACO$$ $$=$$ Area $$\Delta OAB - \text{Area} \ OBACO$$
$$\therefore \text{Area of}   \Delta OAB=\dfrac {1}{2}\times OB\times AB=\dfrac {1}{2}\times 1\times 1=\dfrac {1}{2}$$
Area of $$OBACO$$ $$=\displaystyle \int_0^1 y dx=\int_0^1 x^2 dx=\left [\frac {x^3}{3}\right ]_0^1=\frac {1}{3}$$
$$\Rightarrow$$ Area of $$OACO$$ $$=$$ Area of $$\Delta OAB -$$ Area of $$OBACO$$
$$=\dfrac {1}{2}-\dfrac {1}{3}=\dfrac {1}{6}$$
Therefore, required area $$=2\left [\dfrac {1}{6}\right ]=\dfrac {1}{3}$$ sq.units.

The area bounded by the curve, $$x^2=4y$$, and line, $$x=4y-2$$, is represented by the shaded area $$OBAO$$.
Let $$A$$ and $$B$$ be the points of intersection of the line and parabola.
Coordinates of point $$A\ \text{are} \left (-1, \dfrac {1}{4}\right )$$
Coordinates of point $$B$$ are $$(2, 1)$$.
We draw $$AL$$ and $$BM$$ perpendicular to x-axis.

The area enclosed between the parabola, $$y^2=4ax$$ and the line, $$y=mx$$ is represented by the shaded area $$OABO$$ as
The points of intersection of both the curves are $$(0, 0)$$ and $$\left (\dfrac {4a}{m^2}, \dfrac {4a}{m}\right )$$.
We draw $$AC$$ perpendicular to $$x$$-axis.

The area bounded by the curves, $$y=x^2+2, y=x, x=0$$, and $$x=3$$, is represented by the shaded area $$OCBAO$$ as shown in the diagram.

The area of the smaller region bounded by the ellipse $$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1$$ and the line $$\dfrac {x}{a}+\dfrac {y}{b}=1$$ is represented by the shaded region $$BCAB$$
$$\therefore Area\ BACB = Area\ (OBCAO) - Area (OBAO)$$
$$=\displaystyle \int_{0}^{a}b\sqrt {1-\dfrac {x^2}{a^2}}dx-\int_{0}^{a}b\left (1-\dfrac {x}{a}\right )dx$$
$$=\dfrac {b}{a}\displaystyle \int_{0}^{a}\sqrt {a^2-x^2}dx-\dfrac {b}{a}\int_{0}^{a}(a-x)dx$$
$$=\dfrac {b}{a}\left [\left \{\dfrac {x}{2}\sqrt {a^2-x^2}+\dfrac {a^2}{2}\sin^{-1}\dfrac {x}{a}\right \}_{0}^{a}-\left \{ax-\dfrac {x^2}{2}\right \}_{0}^{a}\right ]$$
$$=\dfrac {b}{a}\left [\left \{\dfrac {a^2}{2}\left (\dfrac {\pi}{2}\right )\right \}-\left \{a^2-\dfrac {a^2}{2}\right \}\right ]$$
$$=\dfrac {b}{a}\left [\dfrac {a^2\pi}{4}-\dfrac {a^2}{2}\right ]$$
$$=\dfrac {ba^2}{2a}\left [\dfrac {\pi}{2}-1\right ]$$
$$=\dfrac {ab}{2}\left [\dfrac {\pi}{2}-1\right ]$$
$$=\dfrac {ab}{4}(\pi-2)$$

The area enclosed between the parabola, $$4y=3x^2$$ and the line, $$2y=3x+12$$ is represented by the shaded are $$OBAO$$
The points of intersection of the given curves are $$A (-2, 3)$$ and $$(4, 12)$$.
We draw $$AC$$ and $$BD$$ perpendicular to $$x$$-axis.

The region enclosed by the parabola $$x^2=y$$, the line $$y=x+2$$, and $$x$$-axis is represented by the shaded region $$OABCO$$ as
The point of intersection of the parabola $$x^2=y$$ and the line $$y=x+2$$ is $$A (-1, 1)$$.
$$\therefore Area\ OABCO = Area\ (BCA) + Area\ COAC$$
$$=\displaystyle \int_{-2}^{-1}(x+2)dx+\int_{-1}^{0}x^2dx$$
$$=\left [\dfrac {x^2}{2}+2x\right ]_{-2}^{-1}+\left [\dfrac {x^3}{3}\right ]_{-1}^{0}$$
$$=\left [\dfrac {(-1)^2}{2}+2(-1)-\dfrac {(-2)^2}{2}-2(-2)\right ]+\left [0-\dfrac {(-1)^3}{3}\right ]$$
$$=\left [\dfrac {1}{2}-2-2+4+\dfrac {1}{3}\right ]$$
$$=\dfrac {5}{6}$$ sq. units

The area bounded by curves $$\left \{(x, y): y\geq x^2 \ and \ y=|x|\right \}$$, is represented by the shaded region as
It can be observed that the required area is symmetrical about $$y$$-axis.
Required area $$=2\left [Area (OCAO)- Area (OCADO)\right ]$$
$$=2\left[\displaystyle \int_{0}^{1}x dx-\int_{0}^{1}x^2dx\right ]$$
$$=2\left [\left [\dfrac {x^2}{2}\right ]_{0}^{1}-\left [\dfrac {x^3}{3}\right ]_{0}^{1}\right ]$$
$$=2\left [\dfrac {1}{2}-\dfrac {1}{3}\right ]$$
$$=2\left [\dfrac {1}{6}\right ]=\dfrac {1}{3}$$ sq. units

The area of the smaller region by the ellipse $$\dfrac {x^2}{9}+\dfrac {y^2}{4}=1$$ and the line $$\dfrac {x}{3}+\dfrac {y}{2}=1$$ represented by the shaded region $$BCAB$$
$$\therefore Area\ BCAB = Area\ (OBCAO) - Area\ (OBAO)$$
$$=\displaystyle \int_{0}^{3}2\sqrt {1-\dfrac {x^2}{9}}dx-\int_{0}^{3}2\left (1-\dfrac {x}{3}\right )dx$$
$$=\dfrac {2}{3}\left [\displaystyle \int_{0}^{3}\sqrt {9-x^2}dx\right ]-\dfrac {2}{3}\displaystyle \int_{0}^{3}(3-x)dx$$
$$=\dfrac {2}{3}\left [\dfrac {x}{2}\sqrt {9-x^2}+\dfrac {9}{2}\sin^{-1}\dfrac {x}{3}\right ]_{0}^{3}-\dfrac {2}{3}\left [3x-\dfrac {x^2}{2}\right ]_{0}^{3}$$
$$=\dfrac {2}{3}\left [\dfrac {9}{2}\left (\dfrac {\pi}{2}\right )\right ]-\dfrac {2}{3}\left [9-\dfrac {9}{2}\right ]$$
$$=\dfrac {2}{3}\left [\dfrac {9\pi}{4}-\dfrac {9}{2}\right ]$$
$$=\dfrac {2}{3}\times \dfrac {9}{4}(\pi-2)$$
$$=\dfrac {3}{2}(\pi-2)$$ sq. units

The required region is bounded by lines $$x+y=1, x-y=1, -x+y=1$$ and $$-x-y=11$$
The area bounded by the curve $$|x|+|y|=1$$ is represented by the shaded region $$ADCB$$
The curve intersects the axes at points $$A (0, 1), B(1, 0), C(0, -1)$$ and $$D(-1, 0)$$.
It can be observed that the given curve is symmetrical about $$x$$-axis and $$y$$-axis.
$$\therefore Area\ ADCB = 4\times Area\ OBAO$$
$$=4\displaystyle \int_{0}^{1}(1-x)dx$$
$$=4\left (x-\dfrac {x^2}{2}\right )_{0}^{1}$$
$$=4\left [1-\dfrac {1}{2}\right ]$$
$$=4\left (\dfrac {1}{2}\right )$$
$$=2$$ sq. units

The graph of the function $$\sin^4x$$ is shown in figure

The graph of $$ cosx$$  is shown in image

Required area $$(A) =\displaystyle  \int_{0}^{a} y\ dx$$

Given curve is $$y = 2x - x^2$$
$$- y = x^2 - 2x$$
$$- y + 1 = x^2 - 2x + 1$$
$$-(y - 1) = (x - 1)^2$$
Which represents a downward parabola with vertex at $$(1, 1)$$

Equation of the parabola is
$$y^{2} = 16x$$ .... (i)
Required area $$= 2\displaystyle \int_{0}^{3} y\ dx$$
[by symmetry about $$x$$-axis]
$$= 2\displaystyle \int_{0}^{3}4\sqrt {x} dx$$
$$=8 \displaystyle \int_{0}^{3} x^{\frac {1}{2}} dx$$
$$= 8\left (\dfrac {x^{\frac {3}{2}}}{\dfrac {3}{2}} \right )_{0}^{3}$$
$$= \dfrac {16}{3}(3\sqrt {3})$$
$$= 16\sqrt {3}$$ square units.

Intersection point of the curves $$y^2 = 4x$$ and $$y=x$$

Area $$=\displaystyle \int _0^1|x-x^2|dx$$

Let the area bounded by the curves $$y=x^{2}$$ and $$y=4$$ be $$A$$

To find the intersection of parabola and the line, we substitute $$y = \cfrac{3x^2}{4}$$ in the line equation.

The required area 
$$\displaystyle =\int _{ 0 }^{ 1 }{ { x }^{ 2 }dx } +\int _{ 1 }^{ 4 }{ \sqrt { x } dx } ={ \left[ \frac { { x }^{ 3 } }{ 3 }  \right]  }_{ 0 }^{ 1 }+{ \left[ \frac { 2{ x }^{ 3/2 } }{ 3 }  \right]  }_{ 1 }^{ 4 }$$
$$\displaystyle =\frac { 1 }{ 3 } +\frac { 2 }{ 3 } \left( 8-1 \right) =5$$

$$x^2 + y^2 \leq 4$$ denotes the area inside the circle, whose center is $$(0,0)$$ and radius is $$2$$ units.

Given curves are
$$y=\sqrt { 5-{ x }^{ 2 } } \longrightarrow 1$$
$$y=\left| x-1 \right| $$
$$y=\sqrt { 5-{ x }^{ 2 } } $$ represents a semi circle of radius $$\sqrt { 5 }$$ and having center at $$(0,0)$$
(Image)
To find point of intersection
For point A
$${ y }^{ 2 }=5-{ x }^{ 2 }$$ (From $$1$$)
$$y=x-1$$
$${ x }^{ 2 }+{ (x-1) }^{ 2 }=5\rightarrow x=2,y=1$$
Point $$A(2,1)$$
For point B,
$${ y }^{ 2 }=5-{ x }^{ 2 }$$ (From $$1$$)
$$y=1-x$$
$${ x }^{ 2 }+{ (1-x) }^{ 2 }=5\rightarrow x=-1,y=2$$
Point $$B(-1,2)$$
$$A=\int _{ -1 }^{ 2 }{ \sqrt { 5-{ x }^{ 2 } } dx } -\int _{ -1 }^{ 1 }{ \left( 1-x \right) dx } -\int _{ 1 }^{ 2 }{ \left( x-1 \right) dx } $$
On integrating:
$$A={ \left[ \cfrac { x }{ 2 } \sqrt { 5-{ x }^{ 2 } } +\cfrac { 5 }{ 2 } \sin ^{ -1 }{ \left( \cfrac { x }{ \sqrt { 5 }  }  \right)  }  \right]  }_{ -1 }^{ 2 }-{ \left[ x-\cfrac { { x }^{ 2 } }{ 2 }  \right]  }_{ -1 }^{ 1 }-{ \left[ \cfrac { { x }^{ 2 } }{ 2 } -x \right]  }_{ 1 }^{ 2 }$$
$$\cfrac { 5 }{ 2 } \sin ^{ -1 }{ \left( \cfrac { 2 }{ \sqrt { 5 }  }  \right)  } +1-\cfrac { 5 }{ 2 } \sin ^{ -1 }{ \left( \cfrac { -1 }{ \sqrt { 5 }  }  \right)  } -1+\cfrac { 1 }{ 2 } -1-\cfrac { 1 }{ 2 } -2+2+\cfrac { 1 }{ 2 } -1$$
$$A=\cfrac { 5 }{ 2 } \left( \sin ^{ -1 }{ \cfrac { 2 }{ \sqrt { 5 }  }  } +\sin ^{ -1 }{ \left( \cfrac { 1 }{ \sqrt { 5 }  }  \right)  }  \right) -\cfrac { 1 }{ 2 } $$
$$\sin ^{ -1 }{ \cfrac { 1 }{ \sqrt { 5 }  }  } =\cos ^{ -1 }{ \cfrac { 2 }{ \sqrt { 5 }  }  } +\sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } =\cfrac { \pi  }{ 2 } $$
$$\therefore \cfrac { 5 }{ 2 } .\cfrac { \pi  }{ 2 } -\cfrac { 1 }{ 2 } =\left( \cfrac { 5x-2 }{ 2 }  \right) $$sq. units
$$=3.43$$ sq. units

Given $$x^2=4y$$ is a parabola with vertex (0,0) and it opens upwards.

Let the interior of the parabola be the required region $$R_1$$

The line x = 4y- 2 represents a straight line.

By putting x = 0 and y = 0 we obtain $$y= \frac{1}{2}$$ and $$x = -2$$ respectively. Hence the straight line meets the x-axis at (-2, 0) and at (0, 12) on the y-axis respectively.

Hence $$y=\dfrac{x+2}{4}$$

In order to find the points of intersection. Let us equate the equation of parabola and the equation of the straight line.

$$\dfrac{x+2}{4}=\dfrac{x^2}{4}\Rightarrow x^2-x-2=0$$

$$(x-2)(x+1)=0$$

Hence x = 2 and x = -1

If x = 2, y = -1 and if $$x = -1, y=\frac{1}{4}$$

Hence the points of intersection are (2, 1) and (-1, $$\frac{1}{4}$$).

Thus the area of the required region is bounded between the straight line and the parabola. This is shown in the figure.

Clearly the curve moves from the point -1 to 0 and from 0 to 2.

The required area is given by,

$$A=\int_{-1}^{0}(y_2-y_1)dx+\int_{0}^{2}(y_2-y_1)dx$$

$$=\int_{-1}^{0}\left ( \frac{x+2}{4} \right )dx-\int_{-1}^{0}\left ( \frac{x^2}{4} \right)dx+\int_{0}^{2}\left ( \frac{x+2}{4} \right)dx-\int_{0}^{2}\left ( \frac{x^2}{4} \right )dx$$

On integrating we get,

$$=\left [ \dfrac{1}{4}(\dfrac{x^2}{2}+2x)-\dfrac{1}{4}\dfrac{x^3}{3} \right]_{-1}^{0}+\left [ \dfrac{1}{4}(\dfrac{x^2}{2}+2x)-\dfrac{1}{4}\dfrac{x^3}{3} \right ]_{0}^{2}$$

On applying limits we get,

$$=\left [ 0-\dfrac{1}{4}(\dfrac{(-1)^2}{2}+2(-1)-(\dfrac{-1^3}{3})) +\dfrac{1}{4}(\dfrac{2^2}{2}+2(2))-\dfrac{2^3}{3}-0\right ]$$

$$=-\dfrac{1}{4}\left [ \dfrac{-3}{2}+\dfrac{1}{3} \right ]+\dfrac{1}{4}\left [ 6-\dfrac{8}{3} \right ]$$

$$=\dfrac{7}{24}+\dfrac{10}{12}=\dfrac{27}{24}=\dfrac{9}{8}$$ Sq.units

Therefore, the required area is $$\dfrac{9}{8}$$ sq.units.

The required area is twice the area bounded by the curve $$y = 2\sqrt a \sqrt x, x = 0, x = a$$ (latus rectum) and $$x$$-axis. [Since it is symmetrical about $$x$$ - axis].
Required area $$= 2\displaystyle  \int_0^a y dx$$
$$= 2\displaystyle \int_0^a 2 \sqrt a. \sqrt x dx $$

The radius of circle is $$a$$ , The perpendicular distance between center of circle and the given line is $$\dfrac{a}{\sqrt 2}$$

Equation of the latus rectum are $$x = \pm ae$$
$$a^{2} = 9, b^{2} = 5$$
Therefore, $$e = \sqrt {1 - \dfrac {b^{2}}{a^{2}}} = \sqrt {1 - \dfrac {5}{9}} = \dfrac {2}{3}$$
$$\Rightarrow ae = 3. \dfrac {2}{3} = 2$$
Thus the equation of the L.R. are $$x = \pm 2$$.
The required area is bounded by the ellipse, $$x = -2$$ and $$x = 2$$.
Since the curve is symmetrical about both axes the required area is $$4$$ times the area in the first quadrant. i.e., the area bounded by the curve $$\dfrac {x^{2}}{9} + \dfrac {y^{2}}{5} = 1$$ (or) $$y = \dfrac {\sqrt {5}}{3}\sqrt {9 - x^{2}}, x = 0, x = 2$$ and $$x$$-axis.
Required area $$= 4\displaystyle \int_{0}^{2} y dx = 4\displaystyle \int_{0}^{2} \dfrac {\sqrt {5}}{3} . \sqrt {9 - x^{2}} dx$$
$$= \dfrac {4\sqrt {5}}{3}\left [\dfrac {x}{2}\sqrt {9 - x^{2}} + \dfrac {9}{2}\sin^{-1} \left (\dfrac {x}{3}\right )\right ]_{0}^{2}$$
$$= \dfrac {4\sqrt {5}}{3} \left [\sqrt {5} + \dfrac {9}{2}\sin^{-1} \left (\dfrac {2}{3}\right )\right ]$$

Centre of circle is $$(0,0)$$
Radius of circle $$=4$$
$${x}^{2}+{y}^{2}=16$$ .....(i)
$$y=x$$ ......(ii)
By equation (i)$ and (ii), we get
$${x}^{2}+{x}^{2}=16\Rightarrow$$ $${x}^{2}=8$$
$$\therefore$$ $$x=2\sqrt{2},y=2\sqrt{2}$$
Shaded area $$=$$ Area of $$OMPO+$$ Area of $$PMQP$$
$$=\displaystyle \int _{ 0 }^{ 2\sqrt { 2 }  }{ ydx } +\int _{ 2\sqrt { 2 }  }^{ 4 }{ ydx } $$
by line by circle
$$\quad =\displaystyle \int _{ 0 }^{ 2\sqrt { 2 }  }{ xdx } +\int _{ 2\sqrt { 2 }  }^{ 4 }{ \sqrt { 16-{ x }^{ 2 } } dx } $$
$$={ \left[ \cfrac { { x }^{ 2 } }{ 2 }  \right]  }_{ 0 }^{ 2\sqrt { 2 }  }+{ \left[ \cfrac { x }{ 2 } \sqrt { 16-{ x }^{ 2 } } +\cfrac { 16 }{ 2 } \sin ^{ -1 }{ \cfrac { x }{ 4 }  }  \right]  }_{ 2\sqrt { 2 }  }$$
$$=\left( \cfrac { 8 }{ 2 } -0 \right) +\left[ 0+8\sin ^{ -1 }{ 1 }  \right] -\left[ \cfrac { 2\sqrt { 2 }  }{ 2 } \sqrt { 16-8 } +8\sin ^{ -1 }{ \cfrac { 1 }{ \sqrt { 2 }  }  }  \right] $$
$$=4+\left[ 8\sin ^{ -1 }{ 1 } -\sqrt { 2 } \times 2\sqrt { 2 } -8\sin ^{ -1 }{ \cfrac { 1 }{ \sqrt { 2 }  }  }  \right] $$
$$=4+\left( 8\times \cfrac { \pi  }{ 2 } -4-8\times \cfrac { \pi  }{ 4 }  \right) \quad =4+4\pi -4-2\pi =2\pi $$ square unit

As per point of intersection equation of circle is taken as
$$x^2+y^2=4$$
Equations of the circle and the line are
$$x^2+y^2=4$$ and $$x+y=2$$.
The curves intersect at $$(2, 0)$$ and $$(0, 2)$$.
Shaded area is the smaller area enclosed by the two curves and is
$$\displaystyle\int  ^{2}_{0}\sqrt{4-x^2} dx-\int  ^{2}_{0}(2-x)dx$$      ...$$(i)$$

Equation of the latus rectams are $$x = \pm ae$$
Given, $$a^2 =9; b^2 = 5$$
$$\Rightarrow e = \sqrt{\dfrac{a^2 - b^2}{a^2}} = \sqrt{\dfrac{4}{9}} = \dfrac{2}{3}$$
$$\Rightarrow ae = 2$$
Thus the equation of L.R. are
$$x = \pm 2$$
The required area is bounded by the curve
$$\dfrac{x^2}{9} + \dfrac{y^2}{5} = 1$$ (or) $$y = \dfrac{\sqrt 5}{3} \sqrt{9 - x^2}, x = 0, x = 2$$ and $$x$$ - axis.
Required area $$= 4 \displaystyle \int_0^2 y  dx$$
$$= 4 \displaystyle \int_0^2 \dfrac{\sqrt 5}{3} \sqrt{9^2 - x^2 dx}$$
$$= \dfrac{4 \sqrt 5}{3} \left [ \sqrt 5 + \dfrac{9}{2} \sin^{-1} \left ( \dfrac{2}{3} \right) \right ]$$

The given curve $$y={ x }^{ 2 }-2x-3$$
The curve meet $$x$$ axis
Put $$y=0$$
$$\Rightarrow { x }^{ 2 }-2x-3=0$$
$$\Rightarrow \left( x-3 \right) \left( x+1 \right) =0$$
$$\Rightarrow x=3,-1$$
The curve lies above the $$x$$-axis between $$x=-3$$ and $$x=-1$$ and $$x=3$$ and $$x=5$$.
The curve lies below $$x$$-axis between $$x=-1$$ and $$x=3$$.
The required area $$=\displaystyle\int _{ a }^{ b }{ ydx } +\displaystyle\int _{ c }^{ d }{ -ydx } +\displaystyle\int _{ e }^{ f }{ ydx }$$
$$=\displaystyle\int _{ -3 }^{ -1 }{ \left( { x }^{ 2 }-2x-3 \right) dx } -\displaystyle\int _{ -1 }^{ 3 }{ \left( { x }^{ 2 }-2x-3 \right) dx } +\displaystyle\int _{ 3 }^{ 5 }{ \left( { x }^{ 2 }-2x-3 \right) dx }$$
$$={ \left[ \dfrac { { x }^{ 3 } }{ 3 } -{ x }^{ 2 }-3x \right]  }_{ -3 }^{ -1 }-{ \left[ \dfrac { { x }^{ 3 } }{ 3 } -{ x }^{ 2 }-3x \right]  }_{ -1 }^{ 3 }+{ \left[ \dfrac { { x }^{ 3 } }{ 3 } -{ x }^{ 2 }-3x \right]  }_{ 3 }^{ 5 }$$
$$=\dfrac { -1 }{ 3 } -1+3-\left( \dfrac { -27 }{ 3 } -9+9 \right) -\left( \dfrac { +27 }{ 3 } -9-9 \right) +\left( \dfrac { -1 }{ 3 } -1+3 \right) +\dfrac { 125 }{ 3 } -25-15-\left( \dfrac { -27 }{ 3 } -9-9 \right)$$
$$=\left( \dfrac { -1 }{ 3 } +2 \right) -\left( -9 \right) -\left( -9 \right) +\left( \dfrac { -1 }{ 3 } +2 \right) +\left( \dfrac { 125 }{ 3 } -40 \right) -\left( -a \right)$$
$$ =-\dfrac { 1 }{ 3 } +2+9+9-\dfrac { 1 }{ 3 } +2+\dfrac { 125 }{ 3 } -40+9$$
$$=\left( -\dfrac { 1 }{ 3 } -\dfrac { 1 }{ 3 } +\dfrac { 125 }{ 3 }  \right) +\left( -9 \right) $$
$$=\dfrac { -1-1+125 }{ 3 } -9=\dfrac { 123 }{ 3 } -9$$
$$=41-9=32$$ square units.