Calculus: (Anti)derivative

[ThisIsMyName]

Hey there.

I just have a few questions on the Calculus problems.... We're on a very difficult section right now and I just can't figure these out. It's taking the derivative/antiderivative :)

 

 

 

Derivative:

 

f(x) = x^2 – 3x

2x^2-1 – 3x^1-1

2x^1 – 3x^0

2x - 3

 

f(x) = 5x^3 (x^4-3x^2)

5 (3)x^3-1 ((4)x^4-1 – 3(2)x^2-1)

15x^2 (4x^3 – 6x^1)

 

 

Antiderivative:

 

f(x) = x^4 – 5x^3 + 2x – 6

x^4 – 5x^3 + 2x^1 – 6x^0

1 • 1/5x^5 + C1 – (5 • 1/4x^4 + C2) + 2 • 1/2x^2 + C3

1/5x^5 + 1 1/4x^4 + 1x^2 – 6 + C

 

Use limits to find the area between each curve and the x-axis for the given interval.

 

y = x^2 from x = 3 to x = 4

3∫4 x^2 dx = 0∫4 x^2 dx = 0∫3 x^2 dx

0∫4 x^2 dx

∆x = 4/n

xi = 4i/n

lim ∑ (4i/n)^2 • 4/n

n →∞ i = 1

256i^2/n^4

256 • 1^2/ n^4 + 256 • 2^2/ n^4 + ... + 256 • n^2/ n^4)

256/ n^4 • (1^2 + 2^2 + … + 256 • n^2/ n^4)

This is at far as I got

 

 

Define Integral:

 

2∫4 (x + 2)(2x + 3) dx

I need help starting out

 

Integral:

 

∫ (x^2 + 5x +2) dx

I need help starting out

 

Limit:

 

lim ( 1/t – 1/ t^2 + 1)

t → 0

(1/t + t^2/1 +1)

0 +1

1

I really think this is wrong

 

Can anyone help me?

[Pyrotex]

Hey there.

I just have a few questions on the Calculus problems....

Define Integral:

 

2∫4 (x + 2)(2x + 3) dx

I need help starting out

Okie-doke :) Integration is just the anti-derivative (more or less, definitions vary). So, we integrate, and then plug in x=4 to get the upper value. Then we plug in x=2 to get the lower value. Solution = upper value minus lower value. Like this:

 

2∫4 (x + 2)(2x + 3) dx

 

Multiply out to get a polynomial

 

= 2∫4 (2x^2 + 7x + 6) dx

 

Do the anti-derivative thing:

 

= 2/3 x^3 + 7/2 x^2 + 6x [2,4]

 

= { 2*64/3 + 7*16/2 + 24 } - { 2*8/3 + 7*4/2 + 12 }

 

Grouping like denominators while I multiply the numerators:

 

= 128/3 - 16/3 + 112/2 - 28/2 + 24 - 12

 

= 112/3 + 84/2 + 12 <eqn.1>

 

We can go two ways from here. One way is

Multiply terms by 2/2, 3/3, and 6/6 to get same denominators:

 

= 224/6 + 252/6 + 72/6

 

= 548/6

 

= 91 1/3

 

Or, starting with <eqn.1> again, we have

 

= 37 1/3 + 42 + 12

 

= 91 1/3

 

Did you follow that?? ;)

[sanctus]

Does this notation [imath]2\int 4[/imath] stand for [imath]\int_2^4[/imath]?

[Pyrotex]

I believe so. That would be the easiest interpretation of his original post.

[Symbology]

Wow that was alot of typing ThisIsMyName to get help. Lucky for you PyroTex did such a nice, clear job of explaining it. I am impressed with your risk taking and persistence, and my confidence factor for getting questions out here just went up

[ThisIsMyName]

Well, I just wanted to make sure everyone could get on the right page, you know? I don't require answers (even though I completely appreciate Pyrotex.... I sat there and stared at it for a while and then I solved it and actually figured it out and "got it" by myself.... so Thank you VERY MUCH, Pyrotex!)

 

I am impressed with your risk taking and persistence, and my confidence factor for getting questions out here just went up

 

I don't quite understand what you mean by this though, Symbology.

 

Thanks for the help again! And yes, sanctus... that's what I meant but I couldn't figure out how to get it to look like that :)

 

I've figured out the rest of those problems except for y = x^2 from x = 3 to x = 4.... If anyone could just start me off, I'd totally appreciate it... Unfortunately, my calc. teacher has been out lately, which is why I've come here for help instead of just asking him.

 

Thanks again, Pyrotex! :eek2:

[sanctus]

To know how i did it just quote my post and you see it :)

[Pyrotex]

....so Thank you VERY MUCH, Pyrotex!...I've figured out the rest of those problems except for y = x^2 from x = 3 to x = 4....

You are very welcome! :) :doh: :)

 

The area under the curve y = x^2 from x=3 to x=4.

 

It generally works that the area under any curve, y = f(x) is just the integration (or anti-derivative) of f(x).

 

In this case, the area is

[3,4] ∫ (x^2) dx

 

= 1/3 x^3 [3,4]

 

= {1/3 4^3} - {1/3 3^3}

 

= 1/3 {64 - 27}

 

= 1/3 * 37

 

= 12 1/3

 

now, since units, such as feet or meters were never specified, you are at liberty to assume your own default units. I suggest the "Fnortner" because it is so much fun to pronounce.

 

The area under the curve is 12 1/3 square Fnortners. :phones:

[alexander]

[3,4] ∫ (x^2) dx

 

= 1/3 x^3 [3,4]

 

= {1/3 4^3} - {1/3 3^3}

 

= 1/3 {64 - 27}

 

= 1/3 * 37

 

= 12 1/3

but you could prettify this by using some latex.... fairly simple

\begin{split}

\int^4_3{(x^2)dx}

&=\frac{x^3}{3} [3,4]\\

&=\frac{4^3}{3} - \frac{3^3}{3}\\

&=1/3(64 - 27)\\

&=1/3*37\\

&=12\frac{1}{3}

\end{split}

 

[math]

\begin{split}

\int^4_3{(x^2)dx}

&=\frac{x^3}{3} [3,4]\\

&=\frac{4^3}{3} - \frac{3^3}{3}\\

&=1/3(64 - 27)\\

&=1/3*37\\

&=12\frac{1}{3}

\end{split}

[/math]

 

prettyfying is cool :)

[math]f(x) \sim x^2, \quad x\to\infty[/math]