4.6. The Fundamental Theorem of Calculus http://www.ck12.org
We observe that the regions of interest are in the first and third quadrants fromx=−1 tox= 1 .We also observe the
symmetry of the graphs about the origin. From this we see that the total area enclosed is
2
∫ 1
0 (x−x(^3) )dx= 2
[∫ 1
0 xdx−∫ 1
0 x(^3) dx
]
= 2
x 22∣∣
∣∣
∣
10−x4
4∣∣
∣∣
∣
10=
= 2
[ 1
2 −
1
4
]
= 2
[ 1
4
]
=^12.
Example 2:
Find the area between the curves off(x) =|x− 1 |and thex−axis fromx=−1 tox= 3.
Solution:
We observe from the graph that we will have to divide the interval[− 1 , 3 ]into subintervals[− 1 , 1 ]and[ 1 , 3 ].
Hence the area is given by
∫ 1
− 1 (−x+^1 )dx+∫ 3
1 (x−^1 )dx=(
−x2
2 +x)∣∣
∣∣
+ 1
− 1+
(x 2
2 −x)∣∣
∣∣
+ 3
+ 1= 2 + 2 = 4.
Example 3:
Find the area enclosed by the curves off(x) =x^2 + 2 x+1 and