sugarxsweet said:Homework Statement
I am having issues figuring out the error involved in the trapezoidal and midpoint methods for ∫cos(x^2) from 0 to 1 with n=8
Homework Equations
|Et|<= k(b-a)^3/(12n^2)
|Em|<=k(b-a)^3/(24n^2)
The Attempt at a Solution
f(x)=cos(x^2)
f'(x)=-2xsin(x^2)
f''(x)=-4x^2cos(x^2)
f''(1) = -3.844
Is it right given that f''(1) is negative? Just wanted to make sure! Thanks
The formula for the approximate integral of cos(x^2) is: ∫cos(x^2) ≈ ∑cos(x^2)Δx, where Δx is the width of each rectangle used in the approximation, and the summation is taken over all the rectangles.
The approximate integral of cos(x^2) is useful in scientific research because it allows us to estimate the area under the curve of a function without having to find the exact value of the integral. This can save time and resources, and is especially helpful when dealing with complex functions.
The accuracy of the approximation for the integral of cos(x^2) depends on the width of the rectangles used in the approximation. The smaller the width, the more accurate the approximation will be. However, using too many rectangles can also lead to computational errors.
Yes, the approximate integral of cos(x^2) can be used for any range of values. However, it is important to choose an appropriate number of rectangles and width for the approximation in order to ensure accuracy.
One limitation of the approximate integral of cos(x^2) is that it can only be used for continuous functions. It also may not be as accurate as finding the exact value of the integral using other methods, such as integration by parts or substitution. Additionally, it is important to choose an appropriate number of rectangles and width for the approximation in order to minimize errors.