Question

Evaluating a definite integral without knowing the reverse chain rule?

Answer 1

use integral by parts ( http://en.wikipedia.org/wiki/Integrate_by_parts )

Answer 2

\[\int\limits_{3}^{4}\frac{ d }{ dt }\sqrt{4+4t^4}dt\]
I keep thinking that I can just evaluate sqrt{4+4t^4} at 4 and 3 and subtract, but this isn't yielding the right answer.

Answer 3

you have to use trig sub

Answer 4

I don't think we've learned that, yet.

Answer 5

\[\int\limits\limits_{3}^{4}\frac{ d }{ dt }\sqrt{4+4t^4}dt = 2\int\limits\limits_{3}^{4}\frac{ d }{ dt }\sqrt{1+t^4}dt\]

Answer 6

I still dont know where to go from there.

Answer 7

I mean it kinda looks like the reciprocal of the derivative of an arc trig function..

Answer 8

\[t^2 = \tan x\]
\[t = \sqrt {tan x}\]
\[dt = (1/2)tan^{-1/2} x dx\]
\[2\int\limits\limits\limits_{3}^{4}\frac{ d }{ dt }\sqrt{1+t^4}dt = 2\int\limits\limits\limits_{\sqrt {tan 3}}^{\sqrt {tan 4}}\frac{ d }{ dt }\sqrt{1+\tan^2 (x)} (1/2)tan^{-1/2} dx\]
\[=2\int\limits\limits\limits_{\sqrt {tan 3}}^{\sqrt {tan 4}}\frac{ d }{ dt }\sec(x) (1/2)tan^{-1/2} dx\]
\[=2\int\limits\limits\limits_{\sqrt {tan 3}}^{\sqrt {tan 4}}\frac{ d }{ dt }\sec(x) (1/2)tan^{-1/2} dx\] does this help at all?

Answer 9

That's the substitution method we haven't learned yet :(
I took calc last year in highschool and we did that, but we havent learned that in my college class yet, so Im not sure what Im supposed to do instead. Thanks anyways though...

Answer 10

limits are wrong...

Answer 11

Oh the boundaries? Yeah. But you can fiiix thaat. But I mean... we just havent covered this yet.

Answer 12

I thought there was a way around it.

Answer 13

But... no?

Answer 14

don't know...

Answer 15

I think they just want you to use the FTC Part 1, Fundamental Theorem of Calculus, on this problem.

It usually looks a little different, with the derivative operator on the OUTSIDE of the integral, and with a variable in the limits, like this,\[\huge \frac{d}{dx} \; \int\limits_0^x f(t)\; dt=f(x)\]
But using this idea, I think we can still make it work....
So we have...\[\huge \int\limits_3^4 \frac{d}{dt}F(t)\; dt\]Applying the differentiation first gives us,\[\huge \int\limits_3^4 f(t)dt\]Integrating will give us back what we started with,\[\huge F(t)|_3^4=F(4)-F(3)\]

Answer 16

Right. Which is what I had at the top, right? Where I said that I tried just evaluating the square root at the values and subtracting, but it wasn't right.

Answer 17

Hmm..

Answer 18

My thoughts exactly. :(

Answer 19

So you're entering it in online somewhere?
Does it instruct you to round to a decimal, or leave it exact?
I just wanna make sure you didn't make a silly mistake somewhere :o

Answer 20

Yeah I am. It says either to at least 4 decimals or exact.

Answer 21

I was putting it in as exact, and it has a preview thing where it shows what you typed in to make sure the syntax is correct.

Answer 22

\[\large \sqrt{4+4\cdot4^4}-\sqrt{4+4\cdot3^4}\]
\[\large \sqrt{1028}-\sqrt{328} \qquad = \qquad 13.9517\]Maybe try decimal? D: hmm seems like you're doing this correctly.. i dunnos :c

Answer 23

OKAY WELL OBVIOUSLY I MADE A SILLY MISTAKE. Decimal said that it was right. BAH. Thank you :)
I always make silly mistakes. I wonder what it was...

Answer 24

hah :3