## Bladerunner1122 Group Title 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=? one year ago one year ago

1. joemath314159 Group Title

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 Group Title

Thank you!!

3. Bladerunner1122 Group Title

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

4. joemath314159 Group Title

yes, thats correct :)

5. Bladerunner1122 Group Title

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