How to Use Riemann Sums and Integrals to Estimate and Evaluate Functions

In summary, the given table provides values of the function f for certain points on the closed interval 2 ≤ x ≤ 13. The problem involves finding the derivative and integral of f, using a left Riemann sum to approximate the integral, and using the properties of concavity to analyze the graph of f.
  • #1
gitty_678
7
0

Homework Statement


(x, f(x))
(2,1)
(3,4)
(5,-2)
(8,3)
(13,6)

A) Estimate f '(4). Show work.
B) Evaluate the Intergral from 2 to 13 of (3 - 5f '(x))dx. show work
C) Use left riemann sum with subintervals indicated bye the data in the table to apporoximate the intergral from 2 to 13 of (f '(x))dx. show work.
D) Suppose f '(5) = 3 and f ''(x) <0 for all x in the closed interval 5 <or= x <or= 8 to show graph of f at x=5 to show that f(7) <or= 4. use the secant lin for the graph of f on 5 <or= x <or= 8 to show that f(7) >or= 4/3.

The Attempt at a Solution


I have no idea where to even begin... if you could just give me a starting point or an equation or something to start with that would be amazing!
 
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  • #2
There's not really enough information to say for sure how to proceed. Are you given any information regarding [tex]f'(x)[/tex]? The answers to these questions will vary substantially depending on how you choose to interpolate between the given points.
 
  • #3
A. on the AP test, it's sufficient to just find the slope by m = f(b)-f(a) / b-a
B. integrating (3- 5f'(x))dx gives 3x - 5f(x) + C, and by fundamental theorem of calculus you can now plug in 13 and 2
C. just a left riemann sum 1*f(2) + 2*f(3) + 3*f(5) + 5*f(8)
D. on the actual exam it also mentioned being twice differentiable and says something about being concave down. from here, you can draw conclusions based on tangent line approximation, secant line approximation, and common sense with concavity.
 
  • #4
oh yeah i forgot to put this

Let f be a function that is twice differentiable for all real numbers. The table above gives values of f for selected ponts on the closed interval 2 <or= x <or= 13

and yes this is from the AP test... my teacher is making us do all of them from the test.
 

What is a Riemann sum?

A Riemann sum is a method for approximating the area under a curve by dividing the area into smaller rectangles and finding the sum of their areas. It is a fundamental concept in integral calculus.

What is the difference between left, right, and midpoint Riemann sums?

In a left Riemann sum, the height of each rectangle is determined by the left endpoint of the interval, while in a right Riemann sum, the height is determined by the right endpoint. In a midpoint Riemann sum, the height is determined by the midpoint of each interval. These different methods can lead to slightly different approximations of the area under a curve.

What is the connection between Riemann sums and definite integrals?

Riemann sums are used to approximate the value of a definite integral. As the number of rectangles used in the Riemann sum increases, the approximation becomes more accurate and approaches the exact value of the definite integral.

How can Riemann sums be used to find the area between a curve and the x-axis?

By using a Riemann sum with a specific number of rectangles, the area under a curve can be approximated. As the number of rectangles increases, the approximation becomes more accurate and can approach the exact area between the curve and the x-axis.

What is the relationship between Riemann sums and the midpoint rule?

The midpoint rule is a specific type of Riemann sum where the width of each rectangle is equal and the height is determined by the midpoint of each interval. It is a more accurate way of approximating the area under a curve compared to left and right Riemann sums, but it requires more calculations.

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