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Bladerunner1122

  • one year 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=?

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  1. joemath314159
    • one year ago
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    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
    • one year ago
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    Thank you!!

  3. Bladerunner1122
    • one year ago
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    Wow, ok so that means the answer IS -11.

  4. joemath314159
    • one year ago
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    yes, thats correct :)

  5. Bladerunner1122
    • one year ago
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    I'm in it for the "how to." This tells me how to solve the last fourth of my homework.

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