Question

Would someone take a look at my integration homework? TIA

Answer 1

what have you done so far?

Answer 2

it's on the second page

Answer 3

hard to read your work

can you explain your process for #1

Answer 4

#1 and #2 are the same process so if you can do one you can do the other.

Answer 5

find the antiderivative, subtract the upper interval from the lower, that's the answer

Answer 6

good, and for the record, what was the antiderivative? of 1 and 2 ... and 3 for that matter

Answer 7

1x,2x,3x?

Answer 8

2x,3x,4x, sorry

Answer 9

not what im asking ... the first 3 questions, what did you find as antiderivatives ...

Answer 10

each is listed in the first line of the solution

Answer 11

and as i stated to start with, those are hard to read ....

Answer 12

3x + x^5

Answer 13

good, next

Answer 14

(3/2)t^(4/3) + (5/9)t^(9/5)

Answer 15

2 3/4 t^(4/3) + 5/9 t^(9/5)

3/2 t^(4/3) + 5/9 t^(9/5) agreed

and #3?

Answer 16

tanx

Answer 17

tan(y) but yeah ok. then as long as you input the values right you did fine.

Answer 18

now explain how you work #4

Answer 19

using 'like' an inverse chain rule, I received help from someone here, but would love a replete explanation of what makes it so

Answer 20

ok

lets start with the fun thrm of calc

\[F(x)=\int_{a}^{b}f(t)~dt=F(b)-F(a)\]
now what if a and b are functions?
\[F(x)=\int_{a(x)}^{b}f(x)(t)~dt=F(b(x))-F(a(x))\]
do we agree so far?

Answer 21

forgot to do b(x) as the top limit .....

Answer 22

yes, agreed

Answer 23

thats got too many typos, let me redo it

Answer 24

the (x) got attached to f instead of b in the coding

\[F(x)=\int_{a(x)}^{b(x)}f(t)~dt=F(b(x))-F(a(x))\]

thats better

Answer 25

now the idea is to take the derivative of it all

\[\frac d{dx}F(x)=\frac d{dx}\int_{a(x)}^{b(x)}f(t)~dt=\frac d{dx}F(b(x))-\frac d{dx}F(a(x))\]
\[\frac d{dx}\int_{a(x)}^{b(x)}f(t)~dt=f(b(x))\frac {d}{dx}b(x)-f(a(x))\frac d{dx}a(x)\]

Answer 26

to clean it up ..

\[g'(x)=f(b)b'-f(a)a'\]

Answer 27

tell me if this loses you

Answer 28

i think I understand

Answer 29

since f is already defined in the functions integration, and a and b are the limits

g'(x) = (x^2)' x^2 sin(x^2) - 1' 1 sin(1)

g'(x) = 2x x^2 sin(x^2) - 0 1 sin(1)

g'(x) = 2x^3 sin(x^2)

Answer 30

if we see f(t) as the derivative it is ....f = F'

\[\int_{a}^{b}F'(t)dt=F(b)-F(a)\]
\[\frac d{dx}\int_{a}^{b}F'(t)dt=\frac d{dx}F(b)-\frac d{dx}F(a)\]
\[\frac d{dx}\int_{a}^{b}F'(t)dt=\underbrace{F'(b)b'}_{chain~rule}-\underbrace{F'(a)a'}_{chain rule}\]

notice that F' is already defined in the integral setup. its the function being integrated.