Question

find the area bounded by the y axis and the curves y=e^2x and y =e^X+2

Answer 1

well we already know the region i guess

Answer 2

is it \[\int\limits_{1}^{3} \log2 .dy\] ?

Answer 3

oh no

Answer 4

you have to integrate from the y - axis which is x = 0 to the point of intersection of the curves

Answer 5

which was the previous exercise

Answer 6

you already found where these intersect. that is your upper limit of integration

Answer 7

lower limit is 0 since you are starting at the y - axis

Answer 8

oh i see

Answer 9

take bigger function minus smaller one on this region

Answer 10

\[\int_0^{\ln(2)} e^x+2-e^{2x}dx\] is the integral you want

Answer 11

you mean e^x -2-e^2x?

Answer 12

no you wrote
\[y=e^{2x}\] and
\[y=e^x+2\] yes? it is bigger one minus smaller one. on the interval
\[(0,\ln(2))\]
\[e^x+2\] is larger, so it is just what i wrote
\[\int e^x+2-e^{2x}dx\]

Answer 13

oh i understand, my apologies

Answer 14

now find the anti derivative, plug in ln(2) plug in 0 and subtract. done

Answer 15

what answer did you get cause i got 2 and according to the answers its wrong

Answer 16

oh wait
actuall nvm

Answer 17

disregard what i just said

Answer 18

antiderivative of
\[e^x\] is
\[e^x\]
antiderivative of
\[2\] is \[2x\]
antiderivative of
\[-e^{-2x}\] is \[\frac{1}{2}e^{-2x}\]

Answer 19

you have
\[e^x+2x+\frac{1}{2}e^{-2x}|_0^{\ln(2)}\]

Answer 20

isn't it 1/2e^2x

Answer 21

i mean sorry -e^2x

Answer 22

yes i don't know where i got the -x from

Answer 23

it is
\[-\frac{1}{2}e^{2x}\]

Answer 24

so it should be
\[e^x+2x-\frac{1}{2}e^{2x}|_0^{\ln(2)}\]

Answer 25

oh ok

Answer 26

i get
\[e^{\ln(2)}+2\ln(2)-\frac{1}{2}e^{\ln(2)}-(e^0-\frac{1}{2}e^0)\]

Answer 27

which cleans up to
\[2+2\ln(2)-1-1+\frac{1}{2}\]
\[=2\ln(2)+\frac{1}{2}\] if my arithmetic is correct

Answer 28

oh no

Answer 29

made some mistake somewhere

Answer 30

oh i see it. i should be
\[e^\ln(2)+2\ln(2)-\frac{1}{2}e^{2\times \ln(2)}-(e^0-\frac{1}{2}e^0)\]

Answer 31

now it is right. get
\[2+2\ln(2)-2-1+\frac{1}{2}\]
\[=2\ln(2)-\frac{1}{2}\]

Answer 32

YES YOUR ANSWER IS CORRECT ACCORDING TO THE SOLUTION PAPER !

Answer 33

Thank you!

Answer 34

that is the correct one. i was off by 1 because if forgot it was
\[e^{2x}\]

Answer 35

yw