# Find the definite integral using substitution

• Sep 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
• Sep 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

$\int u^\frac{1}{2} * \frac{du}{4(u-1)^{3/4}}$
• Sep 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?
• Sep 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
• Sep 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 $u = t^4 + 1$

$u - 1 = t^4$

$(u - 1)^{\frac{1}{4}} = t$.
• Sep 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
• Sep 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.
• Sep 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
• Sep 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
• Sep 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.
• Sep 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

• Sep 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

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