Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Bladerunner1122

  • 3 years ago

When the integral from -2 to 5 has h(x)dx=-11. How do you solve the integral from 0 to 7 has h(x-2)dx=?

  • This Question is Closed
  1. joemath314159
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You need to do a u substitution. you have this right now:\[\int\limits_{-2}^{5}h(x)dx = -11\]and the question is:\[\int\limits_{0}^{7}h(x-2)dx=?\]So start with:\[\int\limits_{0}^{7}h(x-2)dx\]Let u = x-2, then du=dx, and your lower limit changes to:\[0-2=-2\]while your upper limit becomes:\[7-2=5\]So with your substitution in place, you see that:\[\int\limits_{0}^{7}h(x-2)dx=\int\limits_{-2}^{5}h(u)du\]

  2. Bladerunner1122
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thank you!!

  3. Bladerunner1122
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Wow, ok so that means the answer IS -11.

  4. joemath314159
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, thats correct :)

  5. Bladerunner1122
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I'm in it for the "how to." This tells me how to solve the last fourth of my homework.

  6. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy