Find the definite integral using substitution

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September 23rd 2009, 11:45 PM
The Power

Find the definite integral using substitution

Suppose I have this function where a = 0 and b = .5 not sure on the latex tags for that

$I(x) = \int \sqrt{t^4+1}$

and I let

$u = t^4+1$

and

$du = 4t^3dx$

this is the part where I get confused...

Do I

$\frac{1}{4}du = t^3dx$

or solve for dx completely

If I do it at the step above and continue on to

$\int u^\frac{1}{2} * du$

and plug in the corresponding values my answers do not come out correctly
September 24th 2009, 01:46 AM
tanujkush

Quote:

Originally Posted by The Power

Suppose I have this function where a = 0 and b = .5 not sure on the latex tags for that

$I(x) = \int \sqrt{t^4+1}$

and I let

$u = t^4+1$

and

$du = 4t^3dx$

this is the part where I get confused...

Do I

$\frac{1}{4}du = t^3dx$

or solve for dx completely

If I do it at the step above and continue on to

$\int u^\frac{1}{2} * du$

and plug in the corresponding values my answers do not come out correctly

When you substitute $u = t^4+1$
you need to get dt completely in terms of du, so you do

$du = 4t^3dt$
Further,
$t = (u-1)^{1/4}$
so
$t^3 = (u-1)^{3/4}$

and $dt = \frac{du}{4(u-1)^{3/4}}$

and your integral now looks like

$\int u^\frac{1}{2} * \frac{du}{4(u-1)^{3/4}}$
September 24th 2009, 02:18 AM
mr fantastic

Quote:

Originally Posted by The Power

Suppose I have this function where a = 0 and b = .5 not sure on the latex tags for that

$I(x) = \int \sqrt{t^4+1}$

and I let

$u = t^4+1$

and

$du = 4t^3dx$

this is the part where I get confused...

Do I

$\frac{1}{4}du = t^3dx$

or solve for dx completely

If I do it at the step above and continue on to

$\int u^\frac{1}{2} * du$

and plug in the corresponding values my answers do not come out correctly

I doubt an answer exists in a closed form using elementary functions. Where has this integral come from?
September 24th 2009, 05:41 AM
The Power

From a text book we were told to make a table for the values of x = 0, 5, 1, 1.5

if the integral is from b = x and a = 0

And to the previous fella you lost me at t = (u-1)^1/4 where did that come from
September 24th 2009, 05:45 AM
Prove It

Quote:

Originally Posted by The Power

From a text book we were told to make a table for the values of x = 0, 5, 1, 1.5

if the integral is from b = x and a = 0

And to the previous fella you lost me at t = (u-1)^1/4 where did that come from

If $u = t^4 + 1$

$u - 1 = t^4$

$(u - 1)^{\frac{1}{4}} = t$ .
September 24th 2009, 09:50 AM
The Power

What other ways can I get the anti-derivative of this function to make a table of values
September 24th 2009, 02:09 PM
mr fantastic

Quote:

Originally Posted by The Power

From a text book we were told to make a table for the values of x = 0, 5, 1, 1.5

if the integral is from b = x and a = 0

And to the previous fella you lost me at t = (u-1)^1/4 where did that come from

Please type the question exactly as it appears in your textbook. Because it looks like you've omitted important imformation.

Memo to all members: It helps if you type all of the problem rather than just the bits of it you think are important. (The fact that important information is either overlooked or considered unimportant is often a major reason why people can't do the question they post). This saves time for everyone.
September 24th 2009, 02:15 PM
artvandalay11

You aren't going to find an elementary antiderivative for that function... so unless you use estimation techniques or just put in into a calculator for the values of b=x that you want, I don't think you're going to get an answer

You could use taylor series to get as good an estimation as you'd like
September 24th 2009, 04:41 PM
The Power

I later stated the whole problem find I(x) which is the integral when x is = 0, .5 , 1, 1.5 and make a table of values it is in the section pertaining to the Second FTC
September 24th 2009, 04:49 PM
mr fantastic

Quote:

Originally Posted by The Power

I later stated the whole problem find I(x) which is the integral when x is = 0, .5 , 1, 1.5 and make a table of values it is in the section pertaining to the Second FTC

As has been said - twice - the integral cannot be evaluated in closed form using elementary functions. So either important information from the problem statement is missing or you're expected to use technology to get approximate values for the integral (in which case the question will have said so). The fact that you might be expected to apply the Second FTC only lends weight to the fact that something is missing from the question as posted.
September 24th 2009, 05:22 PM
The Power

Here Ill post the question in it's entirety instead of bits and pieces in this thread

For x = 0, .5, 1, 1.5, 2 make a table of values for

$I(x)=\int\sqrt{t^4+1}$

and x is the upper limit and the lower limit is 0

Thanks for your patience Mr.F
September 24th 2009, 09:32 PM
mr fantastic

Quote:

Originally Posted by The Power and edited by Mr F (including an important omission in red)

Here Ill post the question in it's entirety instead of bits and pieces in this thread

For x = 0, .5, 1, 1.5, 2 make a table of values for

$I(x)=\int_{0}^{x} \sqrt{t^4+1} \, {\color{red}dt}$

Thanks for your patience Mr.F

Perhaps the following link will put in context the many attempts that have been made to tell you that the integral cannot be done in closed form using elementary functions: Wolfram|Alpha

You will have to use technology to evaluate I(x) for x = 0.5, 1, 1.5, 2. Was that made clear in the question?

Except for when x = 0 in which case obviously I(0) = 0.
September 25th 2009, 12:25 AM
tanujkush

Lets see...