Calculating Midpoint Riemann Sum with Two Intervals for Function f(x) = x^2 * e^(-x)

Let f be the function given by f(x)=x2e−x. It is known that ∫10f(x)ⅆx=0.160603. If a midpoint Riemann sum with two intervals of equal length is used to approximate ∫10f(x)ⅆx, what is the absolute difference between the approximation and ∫10f(x)ⅆx ?

To compute the midpoint Riemann sum with two intervals, we need to divide the interval into two equal subintervals and evaluate the function at the midpoint of each subinterval

To compute the midpoint Riemann sum with two intervals, we need to divide the interval into two equal subintervals and evaluate the function at the midpoint of each subinterval.

First, let’s determine the interval size. The given interval is [1, 0], so the interval size is 10 – 1 = 1.

Since we have two intervals, each interval size will be 1/2.

Now, we can calculate the midpoint values and evaluate the function at those points.

For the first subinterval:
Midpoint = (1/2) + (1/2)(1/2) = 3/4
f(3/4) = (3/4)^2 * e^(-3/4)

For the second subinterval:
Midpoint = (1/2)(1/2) = 1/4
f(1/4) = (1/4)^2 * e^(-1/4)

Next, we need to calculate the areas of each subinterval. The area of a single subinterval is given by the function value multiplied by the interval size.

For the first subinterval:
Area = (1/2) * [(3/4)^2 * e^(-3/4)]

For the second subinterval:
Area = (1/2) * [(1/4)^2 * e^(-1/4)]

Now, we can add the two areas together to get the overall approximation of the integral:

Approximation = Area of first subinterval + Area of second subinterval

The absolute difference between the approximation and the actual integral value can be calculated by taking the absolute value of the difference:

Absolute difference = |Actual value – Approximation|

Given that the actual value is 0.160603, we can substitute this value along with the calculated approximation values to find the absolute difference.

It’s important to note that the given function is f(x) = x^2 * e^(-x).

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