# Thread: Inaccurate Area Under a Curve Calculations... What's wrong?!

1. ## Inaccurate Area Under a Curve Calculations... What's wrong?!

This question has been driving me nuts, and multiple colleagues have not been able to help. I expect its something simple, but the solution has not yet presented itself. I came across this issue trying to write a spreadsheet that will help quickly identify limits to a physical issue… so far, no luck.

Key issue is: Calculating the area under a curve with an integral is inaccurate.

If you look at the graph below, using the equation y = 50 cos2 (x) you can infer that the area under the curve would roughly be (and even exceed) a rectangle of 25x45 plus a triangle of 25h x 45b… therefore the approx Area = 25x45 + (1/2)(25x45) = 1687.5

If you use the integral of the equation: ½ (50)(x + sin(x)cos(x))

With the limit of 45 and 0, you get
Area = ½ (50)(45 + sin(45)cos(45))
= (25)(45 + (0.7071)(0.7071))
= (25)(45.5)
= 1137.5

To find exact area… Using a polynomial curve fit to the equation, and then integrating that between 0 and 45, I find an area of 1840.95

What is going on?... why is the method using an integral finding 1137.5 when the actual area under the curve is closer to 1840.95 ???

just to clarify, the base equation is y = 50 x [cosine squared of x] ... NOT the [cosine of 2x]

2. $\displaystyle A = \frac{180}{\pi} \int_0^{\frac{\pi}{4}} 50\cos^2{x} \, dx \approx 1841$

3. Ahhhh, convert after integrating... Thanks Skeeter, big help

4. The derivative formulas, (sin x)'= cos(x) and (cos x)'= -sin x, and the integral formulas that are derived from them, are based on the limit formula $\lim_{x\to 0}\frac{sin x}{x}= 1$ and $\lim_{x\to 0}\frac{1- cos x}{x}= 0$ are based on x being in radians.