Question

I really need help with this one problem:
g(x)=xe^(x^2)
find the slope of the tangent line to g at x=2 and find the area of the region bounded by the graph of , the x-axis and x=1

Answer 1

So I already started to answer it but I stopped because I am not sure if I was doing it right.

Answer 2

d/dx (x)e^(x^2)
=d/dx(e^(x^2))=(e^(x^2)(2x))
1(e^(x^2)+e^(x^2)(2xx)
=(e^(x^2)(2x^2)+1))
d/dx (xe^(x^2) = e^(x^2)((2x^2)+1)) using the product rule. I hope I got all of the parentheses right. :)
Next you have to plug in 2 into the equation to find the slope of the tangent line.
(e^(2^2)(2(2)^2)+1))= 9e^4

Answer 3

This is what I had so far...

Answer 4

the first asks for the derivative at 2

Answer 5

okay I think that I already did that??

Answer 6

the second is an integral, but there is something missing in the question

Answer 7

Find the area of the region bounded by the graph of g, the x-axis, and x = 1.

Answer 8

I am not sure what my integral is supposed to look like. How is it bounded?

Answer 9

\[\int_0^1xe^{x^2}dx\] the anti derivative is easily obtained by a u - sub\(u=x^2,du=2xdx\) etc