The prompt for this question is f(x) =sin(x^2)

A)A. Write the first four terms of the Maclaurin series for f(x)

B)Use the Maclaurin series found in Part A to approximate the integral from 0 to 1 of sin(x^2) dx

C)C. How many terms are needed to find the value of the integral given in Part B, correct to three decimal places? What is that value?

I do not understand anything in this problem and we have a long term substitute who is unable to explain it, please help!

Sure, I'd be happy to help you understand this problem! Let's break it down step by step.

A) In Part A, you are asked to write the first four terms of the Maclaurin series for the function f(x) = sin(x^2).

The Maclaurin series is a way to represent a function as an infinite sum of terms. It can be used to approximate the value of a function around a specific point, in this case, x = 0.

To find the Maclaurin series for f(x), we need to find the derivatives of f(x) at x = 0. Let's start by finding the first four derivatives:

f(x) = sin(x^2)
f'(x) = 2x * cos(x^2) (first derivative)
f''(x) = 2cos(x^2) - 4x^2sin(x^2) (second derivative)
f'''(x) = -8xsin(x^2) - 4x^3cos(x^2) (third derivative)
f''''(x) = -20x^2cos(x^2) + 16x^4sin(x^2) (fourth derivative)

Now, let's evaluate each of these derivatives at x = 0:

f(0) = sin(0^2) = 0
f'(0) = 2(0) * cos(0^2) = 0
f''(0) = 2cos(0^2) - 4(0)^2sin(0^2) = 2
f'''(0) = -8(0)sin(0^2) - 4(0)^3cos(0^2) = 0
f''''(0) = -20(0)^2cos(0^2) + 16(0)^4sin(0^2) = 0

Now, we can write the first four terms of the Maclaurin series for f(x):

f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + (f''''(0)/4!)x^4
= 0 + 0x + (2/2!)x^2 + (0/3!)x^3 + (0/4!)x^4
= x^2

Therefore, the first four terms of the Maclaurin series for f(x) = sin(x^2) are x^2.

B) In Part B, you are asked to use the Maclaurin series found in Part A to approximate the integral from 0 to 1 of sin(x^2) dx.

To do this, we'll use the fact that the integral of the Maclaurin series of a function is a good approximation of the integral of the original function.

The integral of x^2 from 0 to 1 is given by:

∫(0 to 1) x^2 dx = [x^3/3] (from 0 to 1)
= (1^3/3) - (0^3/3)
= 1/3

So, the integral from 0 to 1 of sin(x^2) dx is approximately equal to 1/3.

C) In Part C, you are asked to determine how many terms are needed in the Maclaurin series to find the value of the integral given in Part B, correct to three decimal places.

To approximate the integral with a desired accuracy, we need to add more terms to the Maclaurin series until the difference between the approximation and the actual value of the integral is within the desired range. In this case, we want the approximation to be correct to three decimal places.

We can start by using the Maclaurin series with 1 term:

Approximation = 1/3 (using x^2 as the Maclaurin series for sin(x^2))

The difference between this approximation and the actual value of the integral (1/3) is 0.

To check if it's accurate to three decimal places, we can calculate the error using the formula:

Error = |Actual Value - Approximation|
= |1/3 - 1/3|
= 0

Since the error is zero, we have already achieved the desired accuracy of three decimal places and we don't need to add more terms.

Therefore, to find the value of the integral given in Part B, correct to three decimal places, we need only the first term of the Maclaurin series, which is x^2. The value of the integral is 1/3.