## anonymous 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=?

1. anonymous

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. anonymous

Thank you!!

3. anonymous

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

4. anonymous

yes, thats correct :)

5. anonymous

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