Question

use the fundamental theorem of calculus to evaluate this equation

Answer 1

\[\int\limits_{0}^{2}(3-5x^4)dx\]

Answer 2

so what?!?
You actually get to use calculus to find the area or net area now?

Answer 3

lets talk about antiderivatives then

Answer 4

( what goes here)'=3

Answer 5

what can you take the derivative of that will give you 3?

Answer 6

hi again freckles, not exactly sure what to do here

Answer 7

well this is the short cut way to what you have been doing

Answer 8

thank god

Answer 9

we actually did this one day together on a problem

Answer 10

like we had
\[\int\limits_{0}^{1}(x^2+1)dx=(\frac{x^{3}}{3}+x)|_0^1=(\frac{1^3}{3}+1)-(\frac{0^3}{3}+0) \\ =\frac{1}{3}+1=\frac{1}{3}+\frac{3}{3}=\frac{4}{3}\]
remember this
I didn't do the last line because I was leaving the simplifying to you

Answer 11

like just like we found the antiderivative of (x^2+1) in that example

we need to find the antiderivative of 3-5x^4

Answer 12

Let c and d be some constant number.
The antiderivative of c is cx+d since (cx+d)'=(cx)'+(d)'=c(x)'+0=c(1)+0=c+0=c.

That is \[\int\limits_{}^{}c dx=cx+d\]
when we have definite integrals like:
\[\int\limits_{a}^{b}c dx=cx|_a^b=c(b-a)\]
we don't need to put the constant of integration (the one I called d above in; you can put it will just get canceled out)

Answer 13

so, 3x - x^5 is the antiderivative

Answer 14

wow yep that is right

Answer 15

so you have:
\[\int\limits_{0}^{2}(3-5x^4) dx =(3x-x^5)|_0^2\]

Answer 16

now I take the anti of F(1) minus the anti of F(0)

Answer 17

2, not 1

Answer 18

the next step is to evaluate that antiderivative expression at the upper limit 2
and then put a minus and then evaluate the antiderivative expression at the lower limit 0

Answer 19

and yes

Answer 20

well the anti if the F since the integrand was the f
where you know F'=f

Answer 21

but I think I got what you are saying

Answer 22

ok, here goes

Answer 23

well the ant is the F*

Answer 24

ok I'm ready when you are :p

Answer 25

came up with -58

Answer 26

hmm how did you get that exactly you did this:
\[\int\limits\limits_{0}^{2}(3-5x^4) dx =(3x-x^5)|_0^2 \\ =(3 \cdot 2-2^5)-(3 \cdot 0 -0^5)\]

Answer 27

oops, it's actually

Answer 28

-26

Answer 29

sounds beautiful
that is actually a net area

We know this because