\(\int_{0}^{5} h(x)dx = 0\) That is what I got after trying to figure out the area of a square using an integral. h(x) = 5 creating a square... the area is 25 but why this integral no work? A line is a curve too!!!

Guest Jun 26, 2017

#2**+2 **

\(\int_{0}^{5} h(x)dx = 0\)

That is what I got after trying to figure out the area of a square using an integral. h(x) = 5 creating a square... the area is 25 but why this integral no work? A line is a curve too!!!

h(x)=5

\(A=\int_{0}^{5} h(x)dx \\ =\int_{0}^{5} \;5\:dx \\ =\left[5x\right]_0^5\\ =25-0\\ =25 units^2 \)

My integral works just fine.

Melody
Jun 26, 2017