Question

sketch the area enclosed by the curves and find its area.

y=e^x , y=e^2x , x=0, x= ln2

Answer 1

\[\int\limits_{0}^{\ln(2)}e^{2x} - \int\limits_{0}^{\ln(2)} e^x\]

3/2 - 1 = 1/2

Answer 2

\[ \int\limits_{0}^{\ln2}(e^{2x}-e^{x})dx = [(1/2)e^{2x} -e^{x}] =\] (from 0 to ln2)

Answer 3

can you guys show me each step? i need to be refreshed in calculus

Answer 4

Area bounded by two functions y=e^x , y=e^2x and two vertical lines x=0, x= ln2 , therefore the area = integal ( f(x) -g(x)) , such that f(x) > g(x) [from x=0 to x =ln2]

Answer 5

here is an important point, that how to decide which function is f(x) and which one is g(x) ===> actually we have to check which one is greater than the other.

Answer 6

and that was give in the graph been posted.

Answer 7

hope that helps.

Answer 8

the answer in the back of the book is 1/2, so was this problem solved algebraically or graphically?

Answer 9

it's been solved algebraically.

Answer 10

did we calculate it ( algebraically) or use the picture to read the answer (graphically)?

Answer 11

in fact the graphs only as a helping tool in deciding which function is higher in y

Answer 12

i want to see the each little step algebraically... i don't how you got 3/2-1= 1/2

Answer 13

but everything here is solved algebraically

Answer 14

[(1/2)e^2x−e^x] , here 1st subtitute by x=ln2 , then subtract the result at x=0

Answer 15

oh!!!!! i get it now!!!!!!

Answer 16

duh!!!! lol thanks guys, sorry i'm a little slow

Answer 17

no worries :)

Answer 18

:)

Answer 19

no problem , any time

Answer 20

For a complete answer:
\[\int\limits_{0}^{\ln2}e^{2x}dx = \left[ e^{2x} \over 2 \right]_{0}^{\ln2} = \left[ e^{2 \times \ln 2} - e^0 \over 2 \right] = \left[ 4 - 1 \over 2 \right] = { 3 \over 2 }\]

Answer 21

that' right

Answer 22

3/2-"1"=1/2, how did you get "1"? i keep getting 0