What is the integral of h''(u) between 1 and 2?

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Click For Summary
The integral of h''(u) between 1 and 2 can be evaluated using the Fundamental Theorem of Calculus. Given that h''(u) is continuous, the integral from 1 to 2 is equal to h'(2) - h'(1). Substituting the provided values, h'(2) is 5 and h'(1) is 2, resulting in an integral value of 3. The correct answer reflects the relationship between the function's derivatives at the specified points. This demonstrates the application of calculus principles to solve the problem effectively.
Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
Thanks to those who participated in last week's POTW! Here's this week's problem.

-----

Problem: Suppose that $h$ is a function such that $h(1) = -2$, $h^{\prime}(1) = 2$, $h^{\prime\prime}(1) = 3$, $h(2) = 6$, $h^{\prime}(2) = 5$, $h^{\prime\prime}(2) = 13$, and $h^{\prime\prime}$ is continuous everywhere.

Evaluate $\displaystyle\int_1^2 h^{\prime\prime}(u)\,du$.

-----

 
Physics news on Phys.org
This question was correctly answered by thesurfmaster, BAhdi, Reckoner, and Sudharaka. Here's the solution.

By the fundamental theorem of calculus, we have $\displaystyle\int_1^2 h^{\prime\prime}(u)\,du = h^{\prime}(2) - h^{\prime}(1)=5-2=3$.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K