Let f(x)=cos(x^2) . (a) Use a CAS to approximate the maximum value of |f^''(x)| on the interval

Let f(x)=cos(x^2) . (a) Use a CAS to approximate the maximum...

Let $f(x)=\cos \left(x^{2}\right) .$ (a) Use a CAS to approximate the maximum value of $\left|f^{\prime \prime}(x)\right|$ on the interval $[0,1] .$ (b) How large must $n$ be in the midpoint approximation of $\int_{0}^{1} f(x) d x$ to ensure that the absolute error is less than $5 \times 10^{-4}$ ? Compare your result with that obtained in Example $9 .$ Estimate the integral using the midpoint approximation with the value of $n$ obtained in part (b).

Let $f(x)=\cos \left(x^{2}\right) .$
(a) Use a CAS to approximate the maximum value of
$\left|f^{\prime \prime}(x)\right|$ on the interval $[0,1] .$
(b) How large must $n$ be in the midpoint approximation of
$\int_{0}^{1} f(x) d x$ to ensure that the absolute error is less than
$5 \times 10^{-4}$ ? Compare your result with that obtained in
Example $9 .$
Estimate the integral using the midpoint approximation with the value of $n$ obtained in part (b).

Key Concepts

Midpoint Rule

The midpoint rule is a method for approximating the definite integral of a function by dividing the interval into subintervals, computing the function's value at the midpoint of each subinterval, and then summing the contributions. It is a simple and effective technique in numerical integration, and its accuracy can be analyzed through error bounds that depend on the function's smoothness.

Error Estimation in Numerical Integration

Error estimation quantifies the difference between the numerical approximation of an integral and its true value. For the midpoint rule, the error bound typically involves the maximum absolute value of the second derivative of the function on the integration interval, highlighting the importance of the function's curvature in controlling the approximation error.

Use of Computer Algebra Systems (CAS) in Calculus

Computer Algebra Systems are valuable tools in calculus that enable symbolic differentiation, integration, and numerical approximation. In the context of numerical integration, a CAS can be used to compute derivatives, find maximum values of these derivatives on given intervals, and precisely evaluate or approximate integrals, thereby facilitating accurate error analysis and method selection.

Transcript

00:01 In this problem we have the function f of x equal cosine of x square.

00:08 In part a we want to approximate the maximum value of the absolute value of the second derivative of f on the interval 01 using a computer algebra system cas.

00:26 In part b we're going to find a natural number n in the midpoint approximation to the integral on the interval 0 1 of f, such that the absolute error is less than 5 times 10 to an 84, and we will compare the result we obtain here without obtain in example 9, and in part c we are going to approximate the integral between 0 and 1 of f, using the midpoint approximation with the value of n obtained in part b.

01:05 So in part a, we are going to use the 1.

01:11 Theorem 771, inequality 12, and so in part b, so in part a first we are going to use cas to find or approximate the maximum value of the absolute value of the second derivative of f.

01:42 So we have the the first derivative of first we write the function to remember that is cosine of x squared.

01:53 So the first derivative of the first derivative.

01:54 Is equal to negative 2x sine of x square and then the second derivative of f is equal to negative 2 times sine of x square plus plus 2x square cosine of x square.

02:39 So using this formula here we use math to draw or to plot the function on the interval 01 so we obtained this graph here and as we see here the maximum value of this function the interval 01 is around here that is somewhere between 0 .9 and 1.

03:16 So we have inspected numerically this interval making a very fine subdivision and taking the values on that subdivision.

03:30 So we found that the maximum value, the absolute value of the second derivative on the interval 01, that's what we call on theorem 771, which is used in inequality.

03:53 12 and 13 is about 3 .8 44880 and that's what we can see here we have a value that is close to 4 and is greater than the midpoint of this sub interval from 3 .5 and 4 so is and if indeed this value close to 3 .9...

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