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Nina [5.8K]
3 years ago
15

Approximate the area under the curve over the specified interval by using the indicated number of subintervals (or rectangles) a

nd evaluating the function at the right-hand endpoints of the subintervals. (See Example 1.)
f(x) = 9 − x2 from x = 1 to x = 3; 4 subintervals
Mathematics
1 answer:
RideAnS [48]3 years ago
5 0

Split up [1, 3] into 4 subintervals:

[1, 3/2], [3/2, 2], [2, 5/2], [5/2, 3]

each with length (3 - 1)/2 = 1/2.

The right endpoints r_i are {3/2, 2, 5/2, 3}, which we can index by the sequence

r_i=1+\dfrac i2

with 1\le i\le4.

Evaluating the function at the right endpoints gives the sampling points f(r_i), {27/4, 5, 11/4, 0}.

Then the area is approximated by

\displaystyle\int_1^3f(x)\,\mathrm dx\approx\frac12\sum_{i=1}^4f(r_i)=\frac12\left(\frac{27}4+5+\frac{11}4+0\right)=\boxed{\frac{29}4}

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So Width of the rectangle = Area of the rectangle/ Length
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3 years ago
What is (12!) - (10!)?
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Hi my lil bunny!

❀ _____.______❀_______._____ ❀

(12!) - (10!)

= 479001600 - 10!

= 479001600 - 3628800

= 475372800

So the answer is = 475372800

❀ _____.______❀_______._____ ❀

Xoxo, , May

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