Approximate the value of integral of exp (-x^2) over interval of (0,3)

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We are asked to approximate the definite integral `int_(0)^(3) e^(-x^2)dx`

One method is to use upper and lower sums. We divide the region into n rectangles bounded by the x-axis and the curve such that the widths of the rectangles is `Delta x` and the height of the rectangles is determined by the value of the function within the interval defined by the width of the rectangle.

For the upper sum we choose the largest value in the interval and for the lower sum we choose the smallest value. The given function is decreasing on [0,3] so the largest value is the left endpoint of the interval, and the smallest value is the right endpoint. The area of each rectangle is given by`f(c_i)Delta x_i` where `c_i` is the point chosen in the `i^(th)` subinterval and `Delta x_i` is the width of the `i^(th)` subinterval.

Suppose we choose to make each rectangle of equal width. Let n be the number of rectangles.

(a) If n=6 the width of each rectangle is 3/6=1/2.

For the upper sum we evaluate the function at the leftmost endpoint of each subinterval. So the upper approximation for the area with 6 rectangles is: `1/2(e^(-0)+e^(-1/2)+e^(-1)+e^(-3/2)+e^(-2)+e^(-5/2))~~1.136`

Note that the lower endpoints are 0,1/2,1,3/2,2,5/2.

For the lower sum we choose the right endpoints in each subinterval (1/2,1,3/2,2,5/2,3) to get:

`/2(e^(-1/2)+e^(-1)+e^(-3/2)+e^(-2)+e^(-5/2)+e^(-3))~~0.636`

(b) For a better approximation we can use 12 rectangles. Now the width of each interval is 1/4. The upper sum is `U~~1.074` while the lower sum is `L~~0.761`

(c) For n=24 the width of each interval is 1/8. `~~0.932, L~~0.735`

We can continue, using more rectangles as needed to get better approximations. My calculator gives `A~~0.88620735`

Instead of rectangles you can use trapezoids or Simpson's method (using parabolic sections) to get better approximations with fewer intervals.

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