Example:

Find the area enclosed by the function \(f(x) = 4x^3\) between \(x=-2 \text{ and } x=2\)\\

Answer:

If we integrate this as usual, we will get answer that doesn't make sense. \[ \int ^2_{-2} 4x^3dx=\Big[x^4 \Big]^2_{-2}=2^4-(-2)^4=0\]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 \(x=-2 \text{ to } x=0\) and \(x=0 \text{ to } x=-2\) separately.

\[\int ^2_{0} 4x^3dx=\Big[x^4 \Big]^2_{0}=2^4-(0)^4=16\]

\[\int ^0_{-2} 4x^3dx=\Big[x^4 \Big]^0_{-2}=0-(-2)^4=-16\]

Now, to find the area, we sum the absolute value of these two integrands. \[Area = 16+\lvert -16\rvert =32\]

Definite Integrals