## lilisa27 Group Title Integrate e^(2x) on the interval [-1, 1] one year ago one year ago

1. ByteMe Group Title

let u=2x so du=2dx or $$\large \frac{1}{2}du=dx$$ so, $$\large \int_{-1}^{1}e^{2x}dx=\int_{-2}^{2}e^udu=e^u|_{-2}^{2}=$$ ???

2. ByteMe Group Title

oops... there should be a factor of 1/2 outside of that second integral.....

3. lilisa27 Group Title

So the limits do change..... was curious about that.... So it should be.... 27.2899172?

4. ByteMe Group Title

this is what it should look like: $$\large \int_{-1}^{1}e^{2x}dx=\frac{1}{2}\int_{-2}^{2}e^udu=\frac{1}{2}e^u|_{-2}^{2}=\frac{1}{2}(e^2-e^{-2})=\frac{1}{2}(\frac{e^4-1}{e^2})=\frac{e^4-1}{2e^2}$$ that's the exact form.

5. lilisa27 Group Title

..... isn't u standing in for "2x".... therefore it would be e^4 - e^(-4)?

6. ByteMe Group Title

yes it is but we made a u-substitution that enables us to evaluate the integral in terms of u so we use -2 and 2 as the lower/upper limits of integration

7. ByteMe Group Title

if you wanted to back substitute so you could use x , then the limits would not change.

8. lilisa27 Group Title

....Pretty sure in class, even when making u substitutions, we always would convert back to x form after integrating, keeping the new limits of the integral, and substituting those numbers in for x....

9. lgbasallote Group Title

Fun fact: did you know you can actually perform definite integrals in scientific calculators?

10. swissgirl Group Title

Yes someone like a half a yr ago informed me. Just cant remember who

11. lilisa27 Group Title

Uh, yeah.... but when I do it by hand I'm not getting the right answer. And I have to show my work as though I did it by hand.

12. ByteMe Group Title

Since this is a definite integral, the answer is numeric. so i don't see why you would back substitute back to x if we already made the transition from x to u. But either way, the numeric answer should be the same if you did a back subsitution to x or used the new limits involving u.