Question

graph y+4x-x^3. using integration, find the area bound by the curve and the x-axis. THANK YOU

Answer 1

A = integral ydx
= (4x-x^3)dx
= 2x^2 - x^4 /4

Answer 2

oh sorry it was y= not +

Answer 3

zero is the area

Answer 4

him can you show me how you did it please?

Answer 5

area or integral?

Answer 6

both would be very helpful

Answer 7

you need to find the bounds between f(x) and y=0 (the x-axis) and plug them into the integral to find the area

Answer 8

You first need to find the zeros (to find the bounds for you integral. Which, in this case, you have 0=x(4-x^2)=-x(x^2-4)=-x(x+2)(x-2). So, figure out which region you're talking about. I take it you're doing from -2 to 2? In that case, since the function has origin symmetry, the answer is zero.
This is the integral you want:
\[\int\limits_{-2}^{2} 4x-x^3 dx=2x^2-(1/4)x^4|_{-2}^{2}=2((2)^2-(-2)^2)-(1/4)((2^4)-(-2^4))=0\].

Answer 9

x^n dx = [x ^(n+1)]/(n+1)

Answer 10

Refresh the page and that latex will look better.

Answer 11

and malevolence is right about the area being zero...

Answer 12

http://www.wolframalpha.com/input/?i=plot+y%3D4x-x^3
You can see what I mean by origin symmetry. Then you can visually see why it would be zero. Because if the bounded region is under the x-axis, it is negative, if it is above the axis it is positive. So the INTEGRAL is zero. However, typically when you want to find area (since it is an ODD function) you do twice the area if you went from 0 to 2. But, as far as the integral is concerned, it is zero.

Answer 13

thanks

Answer 14

No problem :P

Answer 15

This may help a bit further:The integral of an odd function from −A to +A is zero (where A is finite, and the function has no vertical asymptotes between −A and A).
The integral of an even function from −A to +A is twice the integral from 0 to +A (where A is finite, and the function has no vertical asymptotes between −A and A).

Answer 16

So, to find the "area" you want to multiply the whole thing by 2 to avoid the "negative area" that is bounded under the axis.

Answer 17

ok. do have any idea on how to graph this on a graphing calculator?

Answer 18

Do you have a texas instruments?

Answer 19

nope i have a casio

Answer 20

Hmmm...let me see.

Answer 21

http://www.ehow.com/how_2246662_use-casio-graphing-calculator.html