Question

Area under the curve

Answer 1

I got 8/3 but it is not listed. I wonder if I am doing something wrong

Answer 2

(2/3)+(-1)+0+3

Answer 3

Where did the negative 1 come from?

Answer 4

slope 3 to 4

Answer 5

you may use \(\large \int_a^b f'(x) dx = f(x) \Bigg|_a^b\)

Answer 6

slope value from x=3 to x=4

Answer 7

isn't it just
\(f(7)-f(0)\)
oh..ganeshe got it

Answer 8

Oh I didn't notice the \(f'(x)\) in the integral.

Answer 9

4-0=0 correct

Answer 10

lol
4-0=4

Answer 11

I did another similar to this one but I had to go from 3 to 7 f'(x)dx

Answer 12

f(7)-f(3)

Answer 13

I added up the slopes on those and got the correct solution

Answer 14

why did it not work on this one from 0 to 7

Answer 15

both methods should work

Answer 16

are my slopes incorrect?

Answer 17

or maybe co-incidence
as f(7) -f (3) = 4-2 =2
and also
-1+3 =2

Answer 18

true, I am still confused what I did incorrect?

Answer 19

btw you are correct according to the key it is 4

Answer 20

you forgot to multiply the width of rectangle in your first method

Answer 21

try below :

Area under f' = (2/3)(3)+(-1)(1)+(0)(2)+3(1)

Answer 22

graph given is f(x)

Answer 23

yes since f(x) is piecewise linear function, slopes give you the constant functions \(f'\)

Answer 24

it is because of that "dx", that we need to multiply with width of the boxes, right ?

Answer 25

I don't know anymore

Answer 26

\[\int \limits_0^7 f'(x) dx = \int \limits_0^3 f'(x) dx + \int \limits_3^4 f'(x) dx + \int \limits_4^6 f'(x) dx + \int \limits_6^7 f'(x) dx \]

Answer 27

evaluate each integral and add up

Answer 28

thats exactly what you're trying to do in your first method ^

Answer 29

ok thanks, I guess I bypass it with the second one especially since the dx from 3 to 7 was 1, 2, 1 and the 2 was multiplied by 0. Thanks

Answer 30

yeah for this problem, you may treat \(dx\) as \(\Delta x\) (width of the rectangle) and work it... just acknowledge the fact that we have been implicitly using the definition : \(\text{slope of f(x)} = f'(x) \)

Answer 31

Ok I got to get better at reading this notation. Thank you, I redid the other problem. I understand now.

Answer 32

it may make more sense if we overlap the graph of f'(x) over f(x) and take a look

Answer 33

no, my brain can not handle that extension. Thanks anyways. It is very obvious that you clearly have a great understanding of Calculus. I feel like a novice at times. But I know that I just need to keep solving problems. Eventually the lightbulb gets turned on in my head. Once again, thanks a lot.

Answer 34

see if that graph helps ^ our interest is in the area below(above if function is negatve) the curve in red color

Answer 35

Thanks, I think I will stick to the original. I still have a bunch of problems to solve. Thanks once again.