Question

Let g(x) = integration from 0 to x, f(t)dt, where f is the function whose graph is shown. Evaluate g(x) for x = 0, 5, 10, 15, 20, 25, 30.

Answer 1

I don't know how to solve for g(25) and g(30).

Answer 2

you do the same thing you've been doing with the other values

if it helps, draw in 25 between 20 and 30 like this (see attached)

Answer 3

I'm not sure what formula to use, would I made the integration from 20 to 25 by adding a rectangle + triangle?

Answer 4

how did you compute something like g(5) or g(15) ?

Answer 5

I use the triangle formula (1/2)bh, up till now, they were all triangles

Answer 6

\[\int\limits_{0}^{5}f(t)dt = \frac{ 5*5 }{ 2 }=\frac{ 25 }{ 2 }\]

Answer 7

yeah you can think of it as

g(25) = g(20) + T

where T is the area of the trapezoid shown in the attached image

Answer 8

so you can build off of what you have already

Answer 9

Ohhhhhhh, I didn't see it as a trapezoid. I got it now. :)

Answer 10

you could do a rectangle+triangle, but that seems like more work than needed

besides, you'll learn about the trapezoid rule later on anyway

Answer 11

Last question, if you may help me on this, how would I estimate g(35) using the midpoint for the best estimate?

Answer 12

g(35) = g(30) + Q

where
Q = area of blue triangle (see attached)

Answer 13

this will be an underestimate and probably the closest we can get

Answer 14

so I would use (1/2)bh = (1/2)(5)(15) to get the area estimate for the blue triangle?

Answer 15

correct

Answer 16

Okay thanks, appreciate it :)

Answer 17

you're welcome

Answer 18

btw if you must use the midpoint approximation, to estimate g(35), then it would probably look something like this. See attached

Answer 19

The question only wanted me to use the midpoint approximation to estimate g(35), for the rest I just add areas using the respective geometric formulas. But, noted your picture. :)

Answer 20

ok great