The graph shows the velocity, in feet per second, of a car accelerating

The answer is 540 ft but I don't know how to get there.

>>220615
It says "*use the graph* to *estimate*", so just count the squares.

>>220624
you're retarded

>>220629
That's literally how integration works: area under the graph.

If OP was supposed to be calculating:

- It would say "calculate", not "estimate"
- he would already know what to do: homework is practice of stuff you've *already been taught*, to find out if it's sunk in or not.

>>220624
...and multiply that by the area of a square, which is 15 ft/s × 2 s.

>>220639
>>220642
Okay so i find the estimated area of each x = 2 increment. Add them up. Then what?

>>220648
Then nothing.

The area under a graph is the integral of the thing it's graphing.
The integral of velocity is distance.

You're done.

>>220652
well the right answer is 540 ft. idk how to get that from what you've said

File: 1479154395710.png (24KB, 232x218px) Image search: [Google]

24KB, 232x218px

>>220657
Just approximate the curve through lines and count the squares between 0 and 8 on the x-axis.

Here you get around 17.5 squares, and each square corresponds to 15 ft/s x 2 s = 30 ft.

17.5 x 30 ft = 525 ft.

If you want a more exact estimate you need to draw the lines more exactly and calculate the area of the triangles on the boundary instead of estimating the number of squares.

File: ops homework.png (917KB, 1399x1314px) Image search: [Google]

917KB, 1399x1314px

>>220615
so the basis for integration is to sum the area under a curve for a certain range
a way to approximate the integration would be to count the number of squares under the curves and approximate to whatever precision (probably to 1/4 at most), which is what you'd have to do without the equation and what the directions suggest ("estimate the distance")
if you can derive (note, not d/dx derivation) your own equation for this graph, you could then calculate it exactly, however, that does not appear to be the question