# How do you differentiate & integrate e to the x?

1. Sep 6, 2009

### pillar

1.a Evaluate the following. Show all work.
http://img90.imageshack.us/img90/289/problemu.png [Broken]
1.b Differentiate.
http://img293.imageshack.us/img293/1440/problem2.png [Broken]

2. Relevant equations
ƒ b-a f(x)-g(x)

3. The attempt at a solution

1.a.http://img511.imageshack.us/img511/1205/problemews.png [Broken]

1.b.http://img512.imageshack.us/img512/4250/problem2ans.png [Broken]

Last edited by a moderator: May 4, 2017
2. Sep 6, 2009

### rootX

e^2x does not integrate to 2e^2x because differentiation of 2e^2x is 4e^2x which is not equal to the e^2x.

3. Sep 6, 2009

### HallsofIvy

Staff Emeritus
Presumably, you know that $(e^x)'= e^x$ and $\int e^x dx= e^x+ C$.

To differentiate $e^{2x}$, use the chain rule, df/dx= (df/du)(du/dx), with u= 2x. Then $f(u)= e^u$ so $df/du= e^u$ and u= 2x so du/dx= 2.

To integrate $e^{2x}$, use the "inverse" of the chain rule- substitution. Let u= 2x so du= 2dx or dx= (1/2)du.

The one multiplies by 2, the other divides by 2.

4. Sep 6, 2009

### pillar

So $e^2x$ differentiated would become $2e^2x$?

5. Sep 6, 2009

### tiny-tim

(I assume you mean e2x differentiated would become 2e2x?)

Yes, that's right.

(To see why, apply the chain rule with f(x) = ex and g(x) = 2x.)