# Integration: Areas Enclosed by the x- and y- axis

The definite integral gives the signed area under a curve. Areas above the x-axis give a positive definite integral. Areas below the x-axis gives a negative definite integral.

We normally think areas are positive, therefore it is important to separate positive and negative integration and use the magnitudes (instead of the signs). We use the absolute value of the definite integral.

Area = $\left| \int _{ a }^{ b }{ f\left( x \right) dx } \right|$

Remember to sketch the area!!!

### Examples

1. Find the area enclosed by the curve $y=2+x-{x}^{2}$ and the x-axis.

$Area=\int _{ -1 }^{ 2 }{ (2+x-{ x }^{ 2 })\quad dx } \\ \\ =\quad { \left[ 2x+\cfrac { { x }^{ 2 } }{ 2 } -\cfrac { { x }^{ 3 } }{ 3 } \right] }_{ -1 }^{ 2 }\\ \\ =\quad \left( 2(2)+\cfrac { { 2 }^{ 2 } }{ 2 } -\cfrac { { 2 }^{ 3 } }{ 3 } \right) -\left( 2(-1)+\cfrac { -1^{ 2 } }{ 2 } -\cfrac { -1^{ 3 } }{ 3 } \right) \\ \\ =\quad \left( 4+2-\cfrac { 8 }{ 3 } \right) -\left( -2+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } \right) \\ \\ = \quad 4\frac { 1 }{ 2 }$

So the area is 4.5 square units.

2. Find the area enclosed between the curve $y={x}^{3}$, the x-axis, and the lines x=-1 and x=2.

The sum of areas between x=-1 and 0 are calculated separately to x=0 and 3.

$\int _{ -1 }^{ 0 }{ { x }^{ 3 }\quad dx } \\ \\ =\quad { \left[ \cfrac { { x }^{ 4 } }{ 4 } \right] }_{ -1 }^{ 0 }\\ \\ =\quad \cfrac { { 0 }^{ 4 } }{ 4 } -\cfrac { { -1 }^{ 4 } }{ 4 } \\ \\ =\quad -\cfrac { 1 }{ 4 }$

So the area b = 0.25 square units

$\int _{ 0 }^{ 3 }{ { x }^{ 3 }\quad dx } \\ \\ =\quad { \left[ \cfrac { { x }^{ 4 } }{ 4 } \right] }_{ 0 }^{ 3 }\\ \\ =\quad \cfrac { 3^{ 4 } }{ 4 } -\cfrac { { 0 }^{ 4 } }{ 4 } \\ \\ =\quad \cfrac { 81 }{ 4 }$

So the area a = 20.25 square units.

Hence, the total area = 0.25 + 20.25 = 20.5 square units.

### Area Enclosed by the y-axis

To find the area between a curve and the y-axis, we change the subject of the equation to x. Therefore:

$x=f(y)$

So the integral is as follows:

$\int _{ a }^{ b }{ f(y)\quad dy }$

### Examples

1. Find the area enclosed by the curve $x={y}^{2}$, the y-axis and the lines y=1 and y=3.

Notice that the function x is now with variable y. Therefore, the curve will become a conic and in this case, a parabolic curve that is rotated 90 degrees to the right (positive).

Therefore, the area is given by:

$\int _{ 1 }^{ 3 }{ { y }^{ 2 }\quad dy } \\ \\ =\quad { \left[ \cfrac { { y }^{ 3 } }{ 3 } \right] }_{ 1 }^{ 3 }\\ \\ =\quad \cfrac { { 3 }^{ 3 } }{ 3 } -\cfrac { 1^{ 3 } }{ 3 } \\ \\ =\quad 8\frac { 2 }{ 3 } \quad square\quad units$

2. Find the area enclosed between the curve $y=\sqrt { x+1 }$, the y-axis, and the lines y=0 and y=3.

$y=\sqrt { x+1 } \\ { y }^{ 2 }=x+1\\ { y }^{ 2 }-1=x\\ \\ Area\quad =\quad \left| \int _{ 0 }^{ 1 }{ ({ y }^{ 2 }-1)\quad dy } \right| +\int _{ 1 }^{ 3 }{ ({ y }^{ 2 }-1)\quad dy } \\ \\ =\quad \left| { \left[ \cfrac { { y }^{ 3 } }{ 3 } -y \right] }_{ 0 }^{ 1 } \right| +{ \left[ \cfrac { { y }^{ 3 } }{ 3 } -y \right] }_{ 1 }^{ 3 }\\ \\ =\quad \left| \left( \cfrac { { 1 }^{ 3 } }{ 3 } -1 \right) -\left( \cfrac { { 0 }^{ 3 } }{ 3 } -0 \right) \right| +\left( \cfrac { { 3 }^{ 3 } }{ 3 } -3 \right) -\left( \cfrac { { 1 }^{ 3 } }{ 3 } -1 \right) \\ \\ =\quad \left| -\cfrac { 2 }{ 3 } \right| +6+\cfrac { 2 }{ 3 } \\ \\ =\quad \cfrac { 2 }{ 3 } +6\cfrac { 2 }{ 3 } \\ \\ =\quad 7\cfrac { 1 }{ 3 }$

## Sums and Differences of Areas

In calculus, you could also look for areas between 2 bounded regions. The general method is taking away the primitive values of the upper function (above) by the lower function (below). The resultant value will be the area of the desired region.

### Examples

1. Find the area enclosed between the curve $y={x}^{2}$, the y-axis, and the lines y=0 and y=4 in the first quadrant.

$A\quad =\quad Area\quad of\quad rectangle\quad -\quad \int _{ 0 }^{ 2 }{ { x }^{ 2 }\quad dx } \\ A\quad =\quad (4\times 2)-{ \left[ \cfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2 }\\ \\ A\quad =\quad 8-\left( \cfrac { 8 }{ 3 } -\cfrac { 0 }{ 3 } \right) \\ \\ A\quad =\quad 5\cfrac { 1 }{ 3 } \quad square\quad units$

Sometimes, you need to find the intersection of the two functions first in order to determine the boundaries.

2. Find the point of intersection of $y={x}^{2}$ and the line $y=x+2$. Hence, find the area enclosed between the two curves.

To find the boundaries we need to use simultaneous equations to solve, then find their primitive functions and insert the values in.

$y={ x }^{ 2 }\\ y=x+2\\ \\ So,\quad { x }^{ 2 }=x+2\\ \\ { x }^{ 2 }-x-2=0\\ (x+1)(x-2)=0\\ \\ x=-1\qquad or\qquad x=2\\ \\ Therefore:\\ \\ \int _{ -1 }^{ 2 }{ x+2-{ x }^{ 2 }\quad dx } \\ \\ =\quad { \left[ \cfrac { 1 }{ 2 } { x }^{ 2 }+2x-\cfrac { 1 }{ 3 } { x }^{ 3 } \right] }_{ -1 }^{ 2 }\\ \\ =\quad \left[ \cfrac { 1 }{ 2 } { (2) }^{ 2 }+2(2)-\cfrac { 1 }{ 3 } { (2) }^{ 3 } \right] -\left[ \cfrac { 1 }{ 2 } { (-1) }^{ 2 }+2(-1)-\cfrac { 1 }{ 3 } { (-1) }^{ 3 } \right] \\ \\ =\quad 4.5\quad sq.\quad units$