Example 3
Example:
Find the area enclosed by the function f(x)=4x3 between x=−2 and x=2\\
Answer:
If we integrate this as usual, we will get answer that doesn't make sense. ∫2−24x3dx=[x4]2−2=24−(−2)4=0
Clearly , as you can see in the graph below, the area enclosed by the function isn't zero. The integration has given us an answer of zero because it counts area below the x-axis as negative area. To find the absolute value of the area, we must calculate the area enclosed by the function from
x=−2 to x=0 and
x=0 to x=−2 separately.
∫204x3dx=[x4]20=24−(0)4=16
∫0−24x3dx=[x4]0−2=0−(−2)4=−16
Now, to find the area, we sum the absolute value of these two integrands. Area=16+∣−16∣=32