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\(\int_{0}^{5} h(x)dx = 0\) 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!!! angry

 Jun 26, 2017
 #1
avatar+520 
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First, the integral of 5 with respect to x is 5x

Next, you evaluate this 5x at x=5 and at x=0

 

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 Jun 26, 2017
edited by Badinage  Jun 26, 2017
 #2
avatar+99322 
+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.   wink

 Jun 26, 2017

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