Question

integral from 0 to 1 of e^2x *cos(e^2x)dx= integral from 1 to b of 0.5* cos(u)du is true if and only if b is?

Answer 1

u = sub for this one

Answer 2

\[\int\limits_{0}^{1} e^{2x} \cos(e^{2x})dx\]\[u = e^{2x}\]\[\frac {1}{2} du = e^{2x} dx\]\[\frac {1}{2} \int\limits\limits_{0}^{1} \cos udu\]

Answer 3

b=e^2 I think...

Answer 4

\[u=e^{2x}, du =2e^{2x}dx, u(0)=2, u(1)=e^2\]
\[\frac{1}{2}\int_0^{e^2}\cos(u)du\]

Answer 5

i meant
\[u(0)=1\] sorry

Answer 6

\[\frac {1}{2} \int\limits\limits\limits_{0}^{1} \cos udu = \left[ \frac {1}{2} \sin (e^{2x}) \right]_{0}^{1}\]

Answer 7

\[\frac{1}{2}\int_1^{e^2}\cos(u)du\] is probably better

Answer 8

You can transform the limits of integration like how satellite put it and just evaluate or substitute back alter and evaluate.

Answer 9

after*

Answer 10

i am confused about how to start...

Answer 11

Did you learn u-substitution yet?

Answer 12

yes

Answer 13

what is the value of b then? the two integral have different range..

Answer 14

\[\int_{0}^{1}e^{2x}\cos(e^{2x})dx=\int_{1}^{b}\frac12\cos udu\]u-sub\[u=e^{2x}\to du=2e^{2x}\]integrate\[\frac12\sin(e^{2x})|_{0}^{1}=\frac12\sin|_{1}^{b}\]evaluate\[\frac12\sin(e^2)-\frac12\sin1=\frac12\sin b-\frac12\sin1\]simplify\[\sin(e^2)=\sin b\]\[e^2=b\]