How do you calculate the left and right Riemann sum for the given function over the interval [0, ln2], using n=40 for #e^x#?

Answer 1

Left Riemann sum #= ln(2)/(40(2^(1/40)-1)) ~~ 0.99136068#

Right Riemann sum #= (2^(1/40)ln(2))/(40(2^(1/40)-1)) ~~ 1.00868936#

If #a_1, a_2, a_3,...# is a geometric series with initial term #a# and common ratio #r#, then the general term is:
#a_k = a r^(k-1)#

and the sum is given by:

#sum_(k=1)^n a_k = (a(r^n-1))/(r-1)#
#color(white)()# Given an exponential function #f(x)# and a sequence of equally spaced values of #x# at which we sample #f(x)#, the values at those points will be in geometric progression. So we can calculate the sum using the above formula.
#color(white)()# Dividing the interval #[0, ln(2)]# into #n# equal subintervals, the value of #f(x) = e^x# at the left hand end of each subinterval is:
#a_k = e^(1/n ((k-1) ln(2))) = 2^((k-1)/n)#
i.e. a geometric sequence with initial term #a=2^0 = 1# and common ratio #r = 2^(1/n)#

The left Riemann sum is:

#ln(2)/n sum_(k=1)^n 2^((k-1)/n) = ln(2)/n (2^(n/n) - 1)/(2^(1/n)-1) = ln(2)/(n(2^(1/n)-1))#
The right Rieman sum is given by multiplying this by #2^(1/n)# since every term is #2^(1/n)# times larger..
#(2^(1/n) ln(2))/(n(2^(1/n)-1)#
So with #n=40#, we find
Left Riemann sum #= ln(2)/(40(2^(1/40)-1)) ~~ 0.99136068#
Right Riemann sum #= (2^(1/40)ln(2))/(40(2^(1/40)-1)) ~~ 1.00868936#
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Answer 2

To calculate the left and right Riemann sums for the given function ( f(x) = e^x ) over the interval ([0, \ln 2]) using ( n = 40 ), follow these steps:

  1. Partition the Interval: Divide the interval ([0, \ln 2]) into ( n ) subintervals of equal width. Since ( n = 40 ), each subinterval will have a width of ( \frac{\ln 2}{40} ).

  2. Calculate Sample Points: For the left Riemann sum, choose the left endpoint of each subinterval as the sample point. For the right Riemann sum, choose the right endpoint of each subinterval.

  3. Evaluate Function Values: Evaluate the function ( f(x) = e^x ) at each of the chosen sample points.

  4. Sum Up the Areas: Multiply each function value by the width of the corresponding subinterval and sum up these products.

  5. Finalize: The sum obtained in step 4 for the left Riemann sum represents the approximation of the area under the curve using left endpoints, and for the right Riemann sum, it represents the approximation using right endpoints.

By following these steps, you can calculate the left and right Riemann sums for the given function over the specified interval using ( n = 40 ).

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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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