Question

∫_0^4▒(4e^5+ 3x)dx evaluate

Answer 1

\[∫_0^4(4e^5+ 3x)dx \]

Answer 2

again!!

Answer 3

yes didnt understand it sorry!

Answer 4

do you know how to find the anti derivative or are you supposed to use geometry?

Answer 5

anti deri

Answer 6

ok then this is straight forward application of power rule backwards
the anti derivative of
\[x^n\] is
\[\frac{x^{n+1}}{n+1}\]

Answer 7

i think combination of both though anti + geometry

Answer 8

is that possible?

Answer 9

so the anti derivative of
\[3x\] is
\[\frac{3x^2}{2}\] and the anti derivative of
\[4e^5\] is
\[4e^5x\] is this ok?

Answer 10

yes

Answer 11

ok so anti derivative of
\[4e^5+3x\] is
\[4e^5x+\frac{3x^2}{2}\]

Answer 12

so we are almost done. last job is to replace x by 4

Answer 13

\[∫_0^4(4e^5+ 3x)dx \]

Answer 14

you get
\[4e^5\times 4+\frac{3\times 4^2}{2}\]
\[16e^5+24\]

Answer 15

then replace x by 0. you get 0

Answer 16

so the "final answer" is
\[16e^5+24\] done

Answer 17

plug in 0 to original form?

Answer 18

oh no, plug 4 into anti- derivative, plug 0 into anti derivative and subtract

Answer 19

wish you can write out everything man in one piece i am confused

Answer 20

you are using this"
\[\int_a^b f(x)dx =F(b)-F(a)\] where F is an anti derivative of f

Answer 21

i can teach myself then

Answer 22

k that makes sense

Answer 23

ok here goes.

Answer 24

\[f(x)=4e^5+3x\]
\[F(x)=4e^5x +\frac{3x^2}{2}\]
\[\int_0^4 (4e^5+3x) dx= F(4)-F(0)\]
\[=4e^5\times 4+\frac{3\times 4^2}{2}-(4e^5\times 0+\frac{3\times 0^2}{2})\]
\[=16e^5+24-0=16e^5+24\]

Answer 25

all details there

Answer 26

thank you

Answer 27

yw