Estimate the area under the graph of f(x)=7/x from x=1 to x=7 by using 6 approximating rectangular
Estimate with right endpoints=_____________________________
Estimate with left endpoints=___________________________
To estimate the area under the graph of \(f(x) = \frac{7}{x}\) from \(x=1\) to \(x=7\) using right endpoints and left endpoints, we can divide the interval \([1,7]\) into six equal subintervals.
Step 1: Find the width of each subinterval.
The width of each subinterval can be calculated by dividing the total width of the interval \([1,7]\) by the number of subintervals, which is 6 in this case. So, the width of each subinterval is:
\[\text{Width} = \frac{{\text{Total Width}}}{{\text{Number of Subintervals}}} = \frac{{7-1}}{6} = \frac{6}{6} = 1\]
Therefore, each subinterval has a width of 1.
Step 2: Determine the right and left endpoints of each subinterval.
To estimate using the right endpoints, we choose the rightmost point of each subinterval as the height of the rectangle. For the left endpoints, we choose the leftmost point of each subinterval as the height of the rectangle.
For the given interval \([1,7]\), the right and left endpoints of each subinterval are as follows:
Subinterval 1: Right endpoint = 2, Left endpoint = 1
Subinterval 2: Right endpoint = 3, Left endpoint = 2
Subinterval 3: Right endpoint = 4, Left endpoint = 3
Subinterval 4: Right endpoint = 5, Left endpoint = 4
Subinterval 5: Right endpoint = 6, Left endpoint = 5
Subinterval 6: Right endpoint = 7, Left endpoint = 6
Step 3: Calculate the area of each rectangle.
The area of each rectangle is given by the product of its width and height. In this case, the width is 1 for all rectangles, so we only need to find the height for each rectangle using the given function \(f(x)=\frac{7}{x}\).
To estimate the area using the right endpoints, we substitute the right endpoints into the function:
Area with right endpoints = \(\frac{7}{2} \times 1 + \frac{7}{3} \times 1 + \frac{7}{4} \times 1 + \frac{7}{5} \times 1 + \frac{7}{6} \times 1 + \frac{7}{7} \times 1\).
To estimate the area using the left endpoints, we substitute the left endpoints into the function:
Area with left endpoints = \(\frac{7}{1} \times 1 + \frac{7}{2} \times 1 + \frac{7}{3} \times 1 + \frac{7}{4} \times 1 + \frac{7}{5} \times 1 + \frac{7}{6} \times 1\).
Now, simply calculate the sums:
Area with right endpoints = \(\frac{7}{2} + \frac{7}{3} + \frac{7}{4} + \frac{7}{5} + \frac{7}{6} + 7\).
Area with left endpoints = \(7 + \frac{7}{2} + \frac{7}{3} + \frac{7}{4} + \frac{7}{5} + \frac{7}{6}\).
Calculating the sums will give you the estimated areas under the graph of \(f(x) = \frac{7}{x}\) from \(x=1\) to \(x=7\) using right endpoints and left endpoints, respectively.
is this the simplified answer because when I simplify this I get 9f, which isn't the right answer
the intervals are all of width 1, so
right: 1(f(2)+f(3)+... + f(7))
left: 1(f(1)+f(2)+...+f(6))
so plug and chug
oh dear !
f is not a number
f(x) = 7/x
so,
f(1) = 7/1
f(2) = 7/2
and so on
If you are taking calculus, I find it incredible that you would make such a blunder.