Calc Question Help

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spartacus

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Hi guys, I really need help with the problem above. I don't understand what that bar thing is in front of the problem and why it has a square root x on top, and a 0 on the bottom? Also I'm not sure how to do the problem...is it an indefinite integral?

THANKS for the help in advance!
 
The bar thing? You mean the integral symbol \(\displaystyle \int\)? Do you know how to do integrals? Worries me that you have this problem in front of you to do but you don't know what this: \(\displaystyle \int\) is.
 
Okay so the bar thing means its an intergral, but does the square root of x on the top mean and the 0 on the bottom mean anything?
 
spartacus said:
Okay so the bar thing means its an intergral, but does the square root of x on the top mean and the 0 on the bottom mean anything?
Click to expand...
The question is stated incorrectly.
It should be \(\displaystyle F(x)=\int_0^{\sqrt x } {t\sin (t^2 )dt} \).
That is a proper integral function.
We never allow the same variable to be used in the way whoever wrote the question you posted has done. It is simply incorrect.
 
This is a definite integral, but usually the values on the integral symbol are numbers and not functions. Nevertheless, you treat the calculation the same. After you do the integral, call this F(x), you plug in the \(\displaystyle \sqrt{x}\) into F(x) then you subtract when you plug in 0 into F(x). So you do \(\displaystyle F(\sqrt{x})-F(0)\)
 
srmichael said:
This is a definite integral, but usually the values on the integral symbol are numbers and not functions. Nevertheless, you treat the calculation the same. After you do the integral, call this F(x), you plug in the \(\displaystyle \sqrt{x}\) into F(x) then you subtract when you plug in 0 into F(x). So you do \(\displaystyle F(\sqrt{x})-F(0)\)
Click to expand...
Read reply #4.
 
Hmm okay, I used the u-substitution, instead of the reverse chain rule and this is what I got..

I let u = cosx

and I let du/dx=x

So the integral is u^5 times du/dx times dx

And then the dx's cancel out

And your left with the integral of u^2 times du

And then you add one to the exponent and divide that number by whatever number is infront of u

So it would be 1/3u^3 + constant

Which is 1/3(cosx)^3 + constant ???
 
\(\displaystyle \displaystyle \int_{0}^{\sqrt{x}}tcos(t^{2})dt\)

Make the substitution \(\displaystyle u=t^{2}, \;\ du=2tdt, \;\ \frac{du}{2}=tdt\)

Now, make the subs and it becomes:

\(\displaystyle \displaystyle \frac{1}{2}\int_{0}^{x}cos(u)du\)

Now, integrate away.

Notice how the upper limit of integration, \(\displaystyle \sqrt{x}\), became an x.

It has to be changed accordingly when making the sub. Since \(\displaystyle u=t^{2}\), then

\(\displaystyle u=(\sqrt{x})^{2}=x\)
 
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