How do you use Riemann sums to evaluate the area under the curve of #f(x) = (e^x) − 5# on the closed interval [0,2], with n=4 rectangles using midpoints?

Answer 1

the answer
#S_p=-3.67701446661601#

The sketch of our function #f(x) = (e^x) − 5#

graph{e^x-5 [-16.02, 16.02, -8.01, 8.01]}

the width

#width=(2-0)/4=1/2#

The midpoints

#(0+1/2)/2=1/4#
#(1/2+1)/2=3/4#
#(1+3/2)/2=5/4#
#(3/2+2)/2=7/4#

now find the high
#f(1/4)=-3.71597458331226#
#f(3/4)=-2.88299998338733#
#f(5/4)=-1.50965704253816#
#f(7/4)=0.75460267600573#

The sketch of our function with midpoints

calculate Riemann sum

#S_p=width*high#

#S_p=(1/2)[-3.71597458331226-2.88299998338733-1.50965704253816+0.75460267600573]#

#S_p=-3.67701446661601#

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Answer 2

To use Riemann sums to evaluate the area under the curve of ( f(x) = e^x - 5 ) on the closed interval ([0,2]) with ( n=4 ) rectangles using midpoints, you would follow these steps:

  1. Determine the width of each rectangle: [ \Delta x = \frac{b-a}{n} = \frac{2-0}{4} = 0.5 ]

  2. Find the midpoint of each subinterval: [ x_i^* = a + \frac{(2i-1)\Delta x}{2} ] where ( i = 1, 2, 3, 4 ).

For ( i = 1 ), ( x_1^* = 0 + \frac{(2 \cdot 1 - 1) \cdot 0.5}{2} = 0.25 )
For ( i = 2 ), ( x_2^* = 0 + \frac{(2 \cdot 2 - 1) \cdot 0.5}{2} = 0.75 )
For ( i = 3 ), ( x_3^* = 0 + \frac{(2 \cdot 3 - 1) \cdot 0.5}{2} = 1.25 )
For ( i = 4 ), ( x_4^* = 0 + \frac{(2 \cdot 4 - 1) \cdot 0.5}{2} = 1.75 )

  1. Evaluate ( f(x) ) at each midpoint: [ f(x_i^) = e^{x_i^} - 5 ] [ f(0.25) = e^{0.25} - 5 \approx -3.942 ]
    [ f(0.75) = e^{0.75} - 5 \approx -2.192 ]
    [ f(1.25) = e^{1.25} - 5 \approx -0.745 ]
    [ f(1.75) = e^{1.75} - 5 \approx 0.189 ]

  2. Calculate the area of each rectangle: [ A_i = f(x_i^*) \cdot \Delta x ] [ A_1 = -3.942 \cdot 0.5 = -1.971 ]
    [ A_2 = -2.192 \cdot 0.5 = -1.096 ]
    [ A_3 = -0.745 \cdot 0.5 = -0.3725 ]
    [ A_4 = 0.189 \cdot 0.5 = 0.0945 ]

  3. Sum up the areas of all rectangles: [ \text{Area} = A_1 + A_2 + A_3 + A_4 ] [ \text{Area} = -1.971 - 1.096 - 0.3725 + 0.0945 ] [ \text{Area} \approx -3.345 ]

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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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