Question

Please help with a Calc problem!!

Answer 1

3? I haven't learned this yet.

Answer 2

@jim_thompson5910

Answer 3

do you have the graph set up?

Answer 4

I graphed it on my calculator

Answer 5

I'm going to use geogebra to graph this. One moment

Answer 6

Okay. And the function that I graphed on my calculator was sin(x^2). I am not sure if that is the correct function to graph.

Answer 7

yes it is

Answer 8

when you graph it, you should get something like this (see attached)

Answer 9

Yup! That's what I got

Answer 10

the area under the curve from 0 to 1 represents F(1) because x = 1 here

Answer 11

okay. So for part a of the question I have to just find the area under the curve between 0 and 1?

Answer 12

For part a), we have n = 4 and we're using the trapezoidal rule. Let's calculate delta x

\[\Large \Delta x = \frac{b-a}{n}\]

\[\Large \Delta x = \frac{1-0}{4}\]

\[\Large \Delta x = \frac{1}{4}\]

\[\Large \Delta x = 0.25\]

this means we'll have trapezoids set up like this (see attached)

Answer 13

from that last attachment, we see that the first two trapezoids are overapproximations and the next two are very close to the area under the curve (still are overapproximations though)

Answer 14

notice how each trapezoid has a width of 0.25, which is the value of \(\large \Delta x\)

Answer 15

yeah! And thats because we had to split the interval into four right?

Answer 16

correct

Answer 17

So after we have delta x, then we can use the area of a trapezoid to find the area under the curve?

Answer 18

yes you would find the area of each trapezoid and add them all up to get the total green area

Answer 19

to find the different sizes of the bases, for instance for the first trapezoid, would you just have to plug in 0 and 0.25 into the function sin(x^2)?

Answer 20

yes you need to find the function values to get the lengths of the bases

Answer 21

that first base at x = 0 is sin(x^2) = sin(0^2) = sin(0) = 0 units long

the second base at x = 0.25 is sin(x^2) = sin(0.25^2) = sin(0.0625) = 0.06245 units long (approximately)

I'm in radian mode

Answer 22

Okay. I am getting the same numbers, for the approximation of the area under the curve would it be 0.315?

Answer 23

I'm getting 0.31598 for the green area (approximate result)

Answer 24

so it's pretty close

Answer 25

so that is the approximate value of F(1)

Answer 26

\[\Large F(x) = \int_{0}^{x}\sin(t^2)dt\]

\[\Large F(1) = \int_{0}^{1}\sin(t^2)dt\]

\[\Large F(1) \approx 0.31598\]

Answer 27

Awesome!! Thanks so much! And for part b, how would I go about doing that?

Is the interval they want for when F is increasing on 0 to 3?

Answer 28

F is equal to the area under that blue curve

Answer 29

when the area (total area) increases as x increases, then F increases

Answer 30

so then the interval when F is increasing is from 0 to 1.75 and also from 2.5 to 3?

Answer 31

if you zoom in a bit, it's a little beyond 2.5, but you have the right idea

Answer 32

And then for part c, how do I find the average rate of change? I am confused about putting the answer in terms of k

Answer 33

I used geogebra to construct the antiderivative of sin(x^2) and it's the curve shown in purple

Answer 34

that purple curve confirms roughly what you wrote

Answer 35

for part c), you will use the formula

[ F(b) - F(a) ]/[ b - a ]

to compute the average rate of change on F(x)

Answer 36

in this case, a = 1 and b = 3

Answer 37

I wouldn't compute F(1) and F(3). I would leave them in integral form.

Answer 38

How would you write that integral? I thought that you have to solve F(1) by doing
(2*1)(cos(1^2))

Answer 39

and just plug 1 into the antiderivitive

Answer 40

the integral of sin(x^2) is not 2x*cos(x^2)

Answer 41

But the antiderivative of sin(x^2) is 2x*cos(x^2) right?

And what would the integral of sin(x^2) be?

Answer 42

the antiderivative of sin(x^2) is not 2x*cos(x^2)

you're thinking of the derivative

Answer 43

Ohhh!! Rightt! Hahaha So what would it be?

Answer 44

\[\Large F(x) = \int_{0}^{x}\sin(t^2)dt\]

\[\Large F(1) = \int_{0}^{1}\sin(t^2)dt\]

\[\Large F(3) = \int_{0}^{3}\sin(t^2)dt\]

-------------------------------------------------------

\[\Large A = \frac{F(b)-F(a)}{b-a}\]

\[\Large k = \frac{F(3)-F(1)}{3-1}\]

\[\Large k = \frac{[\int_{0}^{3}\sin(t^2)dt]-[\int_{0}^{1}\sin(t^2)dt]}{3-1}\]

\[\Large k = \frac{\int_{1}^{3}\sin(t^2)dt}{2}\]

Answer 45

I'm sure you see how to finish up

Answer 46

Yup! I do! Thanks soo much for helping me! I really appreciate your help! :)

Answer 47

no problem