How to find area under the cureve of x^2= y and y=x+1
Answers
Answer:
We must find the crossing points of the two curves; in other words, we find
the values of x and y that satisfy both equations simultaneously.
x = y2 and y = x − 2
so:
y − 2 2 = y
y2 − y − 2 = 0
(y − 2)(y + 1) = 0
We conclude that:
y = 2 or y = −1.
We can plug these values of y back in to either equation to find the associated
x values:
y = x − 2
2 = x − 2
4 = x.
If we perform a similar equation with y = −1 we’ll find that the two points of
intersection are (1, −1) and (4, 2).
The equation of the upper half of the sideways parabola is y = √x and that
of the lower half is y = −√x. The equation of the lower right hand boundary
of the region is just y = x − 2.
We find the area A between the two curves by integrating the difference
between the top curve and the bottom curve in each region:
top bottom-l top bottom-r
� 1 ���� � �� � � 4 ���� � �� �
A = (
√x − (−√x) )) dx + (
√x − (x − 2) ) dx
� 0 �� � � 1 �� �
left right
The rest of this calculation is easy; just evaluate the integrals.
� 1 � 4
A = 2 √x dx + (−x + √x + 2) dx
0 1
� �1 � �4
= 2
2
3
x3/2 + −1
2
x2 +
2
3
x3/2 + 2x
� �
0
� �
1
� � 2 42 2 1 2 = 2
3 − 0 + − 2 + 3 · 43/2 + 8 − −2 + 3 + 2
4 16 1 2 = 3 − 8 + 3 + 8 + 2 − 3 − 2
9 A = .
2
2
Easy Way: Slice it horizontally
There’s a much quicker way to complete this area calculation; you should look
for an easier way as soon as you notice the need to split the region into parts.
The quicker way is similar in principle but reverses the roles of x and y; in this
method we slice the area in question into horizontal rectangles.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
−3
−2
−1
0
1
2
3
X
Y
x=y
2
y=x−2
(1,−1)
(4,2)
dy
Figure 3: The area between x = y2 and y = x − 2 and one horizontal rectangle.
The height of these rectangles is dy; we get their width by subtracting the
x-coordinate of the edge on the left curve from the x-coordinate of the edge on
the right curve. (If you get mixed up and subtract the right from the left you’ll
get a negative answer.) The left curve is the sideways parabola x = y2. The
right curve is the straight line y = x − 2 or x = y + 2.
The limits of integration come from the points of intersection we’ve already
calculated. In this case we’ll be adding the areas of rectangles going from the
bottom to the top (rather than left to right), so from y = −1 to y = 2.
� y=2
� � A = (y + 2) − y2 dy y=−1
� 3 2 �2
= −y + 2y + y
3 2
� � �−1
� 4 8 1 1 = 2 + 4 − 3 − 2 − 2 + 3
9 A = 2
You’ll notice that if you plug the limits of integration into th
Step-by-step explanation: