How do you use the trapezoidal rule with n=9 to approximate the area between the curve #y=x^2 -2x +2# from 0 to 3?

Answer 1

We use the formula #1/6[f(x_0)+2f(x_1)+2f(x_2)+ cdots+ 2f(x_8) + f(x_9)]# where #f(x) = x^2 -2x +2# and #x_0 cdots x_9# are chosen to evenly divide the interval #[0,3]# into 9 pieces.

The trapezoidal rule works by dividing the area under the curve into #n# trapezoids, then calculating the area of each trapezoid, and summing them up. The formula is:
#int_a^b f(x)dx = (b-a)/(2n)[f(x_0)+2f(x_1)+2f(x_2)+cdots2f(x_{n-1})+f(x_n)]#
In our case, #n=9#, #a = 0#, and #b=3#, so we have:
#int_0^3 f(x)dx = (3-0)/(2*9)[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+2f(x_4)+2f(x_5)+2f(x_6) + 2f(x_7) + 2f(x_8) + f(x_9)]#
We choose the #x_0 cdots x_9# so that they divide up the interval from 0 to 3 evenly. So #x_0 = 0, x_1 = 1/3, x_2 = 2/3, x_3 = 1, x_4 = 4/3, x_5 = 5/3, x_6 = 2, x_7 = 7/3, x_8 = 8/3, x_9 = 9#.
From there it's just a matter of plugging these numbers into #f(x)# and solving the formula.
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Answer 2

To use the trapezoidal rule with ( n = 9 ) to approximate the area between the curve ( y = x^2 - 2x + 2 ) from ( x = 0 ) to ( x = 3 ), follow these steps:

  1. Divide the interval ( [0, 3] ) into ( n = 9 ) subintervals of equal width. Since ( n = 9 ), each subinterval width is ( \Delta x = \frac{3 - 0}{9} = \frac{1}{3} ).
  2. Compute the function values at the endpoints of each subinterval. That is, calculate ( y_0 = f(0), y_1 = f(\frac{1}{3}), y_2 = f(\frac{2}{3}), \ldots, y_9 = f(3) ), where ( f(x) = x^2 - 2x + 2 ).
  3. Use the trapezoidal rule formula to find the approximate area: [ A \approx \frac{\Delta x}{2} \left[ y_0 + 2(y_1 + y_2 + \ldots + y_8) + y_9 \right] ]
  4. Plug in the values obtained in steps 1 and 2 into the formula and compute the result.
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Answer 3

To use the trapezoidal rule with ( n = 9 ) to approximate the area between the curve ( y = x^2 - 2x + 2 ) from ( x = 0 ) to ( x = 3 ), follow these steps:

  1. Determine the interval width, ( \Delta x ), which is given by: [ \Delta x = \frac{b - a}{n} ] where ( a = 0 ) (lower limit), ( b = 3 ) (upper limit), and ( n = 9 ) (number of subintervals).

  2. Calculate ( \Delta x ): [ \Delta x = \frac{3 - 0}{9} = \frac{1}{3} ]

  3. Create the x-values for the endpoints of each subinterval. Since ( n = 9 ), there will be 10 points including the endpoints: [ x_0 = 0, , x_1 = \frac{1}{3}, , x_2 = \frac{2}{3}, , x_3 = 1, , x_4 = \frac{4}{3}, , x_5 = \frac{5}{3}, , x_6 = 2, , x_7 = \frac{7}{3}, , x_8 = \frac{8}{3}, , x_9 = 3 ]

  4. Evaluate the function ( y = x^2 - 2x + 2 ) at each of these x-values to get the corresponding y-values.

  5. Use the trapezoidal rule formula to calculate the area of each trapezoid and sum them up: [ A \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n) \right] ]

  6. Substitute the values into the formula and compute the sum.

  7. The result will be the approximation of the area between the curve ( y = x^2 - 2x + 2 ) and the x-axis from ( x = 0 ) to ( x = 3 ) using the trapezoidal rule with ( n = 9 ).

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Answer from HIX Tutor

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.

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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