Question

How do you do these questions? With step by step instructions please

Answer 1

Answer:         n = 3                                              n = 4                

               Upper Sum ≈ 3.41                      Upper Sum ≈ 3.25

               Lower Sum ≈ 2.15                      Lower Sum ≈ 2.25

Step-by-step explanation:

You are trying to find the area under the curve.  Area = height x width.

Height is the y-value at the given coordinate --> f(x)

Width is the distance between the x-values --> dx

n = 3

First, let's figure out dx:  the distance from -1 to +1 is 2 units. We need to divide that into 3 sections because n = 3 --> dx = 2/3

So the points we will evaluate is when x = {-1, -1/3, 1/3, 1}

For the upper sum, we find the max y-value for each interval

For the lower sum, we find the min y-value for each interval

Next, let's find the height for each of the x-values:

f(x) = 1 + x²

f(-1) = 1 + (-1)²   = 2

f(-1/3) = 1 + (-1/3)² = 1 + 1/9 --> 10/9

f(1/3) =  1 + (1/3)² = 1 + 1/9 --> 10/9

f(1) = 1 + (1)²   = 2

Interval         Max                      Min

{-1, -1/3}          f(-1) = 2                   f(-1/3) = 10/9

{-1/3, 1/3}        f(-1/3) = 10/9           f(0) = 1   (vertex lies in this interval)

{1/3, 1}             f(1) = 2                   f(1/3) = 10/9

Now, let's find the Area: A = f(x) dx:

$\text{Upper Sum:}\quad A=\dfrac{2}{3}\bigg(2+\dfrac{10}{9}+2\bigg)\\\\.\qquad \qquad \qquad =\dfrac{2}{3}\bigg(\dfrac{46}{9}\bigg)\\\\.\qquad \qquad \qquad =\dfrac{92}{27}\\\\.\qquad \qquad \qquad =\large\boxed{3.41}$

$\text{Lower Sum:}\qquad A=\dfrac{2}{3}\bigg(\dfrac{10}{9}+1+\dfrac{10}{9}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{2}{3}\bigg(\dfrac{29}{9}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{58}{27}\\\\\\.\qquad \qquad \qquad \qquad =\large\boxed{2.15}$

*****************************************************************************************

n = 4

First, let's figure out dx:  the distance from -1 to +1 is 2 units. We need to divide that into 4 sections because n = 4 --> dx = 2/4 = 1/2 (simplified)

So the points we will evaluate is when x = {-1, -1/2, 0, 1/2, 1}

For the upper sum, we find the max y-value for each interval

For the lower sum, we find the min y-value for each interval

Next, let's find the height for each of the x-values:

f(x) = 1 + x²

f(-1) = 1 + (-1)²   = 2

f(-1/2) = 1 + (-1/2)² = 1 + 1/4 --> 5/4

f(0) = 1 + (0)² = 1

f(1/2) =  1 + (1/2)² = 1 + 1/4 --> 5/4

f(1) = 1 + (1)²   = 2

Interval         Max                      Min

{-1, -1/2}          f(-1) = 2                   f(-1/2) = 5/4

{-1/2, 0}          f(-1/2) = 5/4            f(0) = 1  

{0, 1/2}           f(1/2) = 5/4             f(0) = 1

{1/2, 1}            f(1) = 2                   f(1/3) = 5/4

Now, let's find the Area: A = f(x) dx:

$\text{Upper Sum:}\qquad A=\dfrac{1}{2}\bigg(2+\dfrac{5}{4}+\dfrac{5}{4}+2\bigg)\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{1}{2}\bigg(\dfrac{26}{4}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{13}{4}\\\\\\.\qquad \qquad \qquad \qquad =\large\boxed{3.25}$

$\text{Lower Sum:}\qquad A=\dfrac{1}{2}\bigg(\dfrac{5}{4}+1+1+\dfrac{5}{4}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{1}{2}\bigg(\dfrac{18}{4}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{9}{4}\\\\\\.\qquad \qquad \qquad \qquad =\large\boxed{2.25}$