How would you find the area of the shaded space above the axis but below the curve?

You mean dividing by dx?

with a double integral of course
[eqn]\int_{-2}^{2}\int_{0}^{-x^2 + 4} 1\ \text{d}y\ \text{d}x[/eqn]

This works. I used tiny triangles and counted eleven minus one third square units.

Its boundless there is infinite area

It is boundless there is infinite area

Pleb.

[math] \int 1_{0\leq y\leq -x^2+4} d\lambda [/math] where [math] \lambda [/math] is Lebesgue measure on [math]\mathbb{R}^2[/math]

here ya go op
desmos.com/calculator/xu9l0gnibu

You're all overcomplicating this. What you do is make a bitmap out of it in paint. You can then plug it into Mathematica, and make a histogram of white versus gray pixels. If you know the overall dimensions of the image, BOOM, there's your answer.

would you include the curve itself as apart of the area?

That's approximation, not solution.

>would you include the curve itself as apart of the area?
course not! integral is muh area under the curve.

2*integral from 0 to 2 of (-x^2 + 4) dx

>above the axis but below the curve
You mean "above the axis *AND* below the curve"
Lrn2and fgt pls

How would you find the rate of change at a point along the curve?

...

underrated

engineer here.

first make about 30 copies of the curve on a copy machine using high quality paper. Measure the area and weight of each piece of paper after printing, being careful to tabulate your results in correct order.

Carefully cut out the shaded area and weigh again. Take the weight ratios and multiply by the corresponding area measurement to get the desired metric.

Apply standard statistical methods to determine the precision and accuracy of your measurements.

find zeros
find convex
area of the "hull" - /int^{x_0}_{x_1}f(x)dx

If I would have wanted to find area of a traingle using MC method my result would be of by 8%

there's no way you're an engineer
there was no mention of dicks in your post

here you go OP
just throw a bunch of random points in there, and compare the total number of random points to the number of random points in the shaded area

They rediscovered Riemann Sums

Differentiate for that point

This is going above and beyond at this point...
Talk about all your problems looking like nails.

my uni didn't require us to show work - it's how i solved all my calc problems. i lost a point on most stuff because i got one of the decimal numbers wrong

Monte Carlo is more like a multitool

>call others pleb
>cant tex
hahaha well memed my friend

I am fucking retarded, studying calc III and spent 5 minutes trying to remember what the double integral represented, thinking it was a prank answer...

Use a protracter.

6 < area < 16

Monte Carlo simulation :^)

how the FUCK did you find that

Multiplied by 2, dumbass

>not a triple integral

you just say its infinite

substitute your x for e in this equation (see Fermat's Rule for why) and your constant for one of the integrals (Because the dx of x is any constant)

Whoops saged without meaning to

>That's approximation, not solution.

True, but Mathematica shitpost == priceless.

Best homework thread evar.

>Acknowledgments—I would like to dedicate Tai's Model to my late parents Mr. and Mrs. T. C. Tai.

awww :(
Bet they'd be proud!

32/3?

did that in my head

took 7 minutes though

12

integrate between the roots
\int_{-2}^{2}x^2+4\text{ d}x

integrate between the roots
[tex]\int_{-2}^{2}x^2+4\text{ d}x[/text]

You're a failure, give up on life

fuck im never browsing sci again

use ["math"] without the quotes

thx b

A square plus b square equals see square

2 square plus 3 square equals see square

4plus9equals13...square?
26 26 26 26

Integrate[-x^2 + 4, {x, -2, 2}]

...

>no one mentioned the method of exhaustion
Has Veeky Forums finally gone full brainlet?

50%, it either happens or it doesn't

draw a rectangle around the curve and fill the space around it with infinite triangles, add up the area of those rectangles

I still can't believe this is real

Find the area of the entire box

Then subtract from it the area outside of the shaded regionf

that worked really well, probabilistic methods are neat

delet this

"But" is LITERALLY the exact same fucking thing as "And" logically speaking.

[eqn]\int_{-2}^{2} (-x^2 + 4)\, dx = \left[ \frac{-x^3}{3} + 4x \right]_{-2}^{2} = \frac{-8}{3} + 8 - \left( \frac{8}{3} - 8 \right) = 16 - \frac{16}{3} = \frac{32}{3}.[/eqn]

Now fucking KYS and never come back

This only works for glucose tolerance and other metabolic curves, you retard :^)

>Mary
fugging wymmyn amirite

math.uconn.edu/~kconrad/math1132s14/handouts/taicomments.pdf

She responded to some comments in this

The voice of reason has spoken

>using jewish integration
top kek, JIDF pls go