# How do you approximate the given integral with the specified value of "n" for the integral from 0 to 1/2 of #sin (x^2) dx# (n=4)?

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To approximate the given integral with ( n = 4 ) for the integral from ( 0 ) to ( \frac{1}{2} ) of ( \sin(x^2) ) dx, you can use the midpoint rule formula:

[ \Delta x = \frac{b - a}{n} ]

[ x_i = a + \frac{(2i - 1) \Delta x}{2} ]

[ \text{Approximation} = \Delta x \sum_{i=1}^{n} f(x_i) ]

Where:

- ( a = 0 ) (lower limit of integration)
- ( b = \frac{1}{2} ) (upper limit of integration)
- ( n = 4 ) (number of subdivisions)
- ( \Delta x ) is the width of each subdivision
- ( x_i ) are the midpoints of each subdivision
- ( f(x_i) = \sin(x_i^2) )

Calculate ( \Delta x ), then find ( x_i ) for each ( i ), evaluate ( f(x_i) ), and finally sum up the results to get the approximation.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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