How do you differentiate & integrate e to the x?

[pillar]

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

Homework Equations

ƒ b-a f(x)-g(x)

The Attempt at a Solution

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

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

 

[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.

 

[HallsofIvy]

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.

 

[pillar]

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.

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

 

[tiny-tim]

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

(I assume you mean e^2x differentiated would become 2e^2x?)

Yes, that's right.

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