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 2 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(x2)dx=?
 2 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(x2)dx=?

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joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1You need to do a u substitution. you have this right now:\[\int\limits_{2}^{5}h(x)dx = 11\]and the question is:\[\int\limits_{0}^{7}h(x2)dx=?\]So start with:\[\int\limits_{0}^{7}h(x2)dx\]Let u = x2, then du=dx, and your lower limit changes to:\[02=2\]while your upper limit becomes:\[72=5\]So with your substitution in place, you see that:\[\int\limits_{0}^{7}h(x2)dx=\int\limits_{2}^{5}h(u)du\]

Bladerunner1122
 2 years ago
Best ResponseYou've already chosen the best response.1Wow, ok so that means the answer IS 11.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1yes, thats correct :)

Bladerunner1122
 2 years ago
Best ResponseYou've already chosen the best response.1I'm in it for the "how to." This tells me how to solve the last fourth of my homework.
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