Question

Find f. f ''(t) = 5et + 3 sin t, f(0) = 0, f(π) = 0

Answer 1

\[f''(t)=5e^t+3sin(t)\]?

Answer 2

if so integrate these two seperately and it'll be easier

\[f'(t)=\int{}{}{5e^t}dt+\int{}{}{3sin(t)dt}\]

Answer 3

yepp. my bad !

it's ′′(t)=5e^t+3sin(t)

but could youplz use Antiderivatives to solve it....??

cuz I havent studied about integrals.... ==''

Answer 4

integrals = anti derivatives

Answer 5

integrals are basically finding the area of something or finding volume and such but it uses the anti derivative to do this... basically mean't take the anti derivative of each

Answer 6

do you understand how to use the anti derivative?

Answer 7

i mean , for example,

f′′(t)=5e^t+3sin(t)

then F'(t)=5e^t-3cos(t)+C

then F(x)=5e^t-3sin(t)+Cx+D

Answer 8

that's correct

Answer 9

so I just put 0 into F(x) then solve the C and D , right?

Answer 10

now you just use your initial values to figure out what the constants are

Answer 11

yep =]

Answer 12

thanksssss :DDDDD

Answer 13

0 will get you

0=5+0+0+d

d=-5

and pi gets you

0=5e^{pi}+0+Cpi-5

Answer 14

\[5e^{\pi}+C\pi\]=0

\[-5e^{\pi}=C\pi\]

\[\frac{-5e^{\pi}}{\pi}=c\]

Answer 15

you can try cleaning it up more if you'd like

Answer 16

i'd just leave it

Answer 17

hold on .

I think my anti derivative of F(x) is not correct, dont I

cuz if I put 0 and pi into the F(x)

what I get are

F(pi)=5e^pi-3sin(pi)+Cpi+D=0

F(0)=5e^0-3sin(0)+C*0+D=0------ 5-0+0+D=0---D=-5