# A Complicated Integral

• Jan 17th 2011, 04:59 PM
PPSSAGW
A Complicated Integral
$\displaystyle d/dx \int_3^{x^2} (2t^2+5)^2 dx$

Can someone please see if I did it right or not?

$\displaystyle x^2=2x$

Substitute $\displaystyle t$ with $\displaystyle x^2$

$\displaystyle (2(x^2)^2+5)^2$

$\displaystyle 2x(2x^4+5)^2$

Do I have to add +C?

Thanks a lot
• Jan 17th 2011, 05:03 PM
dwsmith
Quote:

Originally Posted by PPSSAGW
$\displaystyle d/dx \int_3^x^2 (2t^2+5)^2 dx$

Do you mean

$\displaystyle \displaystyle\frac{d}{dx}\int_3^{x^2}(2*t^2+5)^2dx$
• Jan 17th 2011, 05:04 PM
PPSSAGW
Yeah I later changed it.
• Jan 17th 2011, 05:04 PM
dwsmith
Quote:

Originally Posted by PPSSAGW
Yeah I later changed it.

Your x^2 needs to be entered in as ^{x^2} because you have a floating 2.
• Jan 17th 2011, 05:05 PM
PPSSAGW
Thanks. I'm new to LaTex.
• Jan 17th 2011, 05:07 PM
PPSSAGW
Quote:

Originally Posted by dwsmith
Your x^2 needs to be entered in as ^{x^2} because you have a floating 2.

Did I solve it correctly though? Do I have to add +C?
• Jan 17th 2011, 05:10 PM
dwsmith
Quote:

Originally Posted by PPSSAGW
Did I solve it correctly though? Do I have to add +C?

$\displaystyle \displaystyle\int_3^{x^2}(2*t^2+5)^2dx\Rightarrow (2*t^2+5)^2\int_3^{x^2} dx=(2*t^2+5)^2\left[x\right]_3^{x^2}=(2*t^2+5)^2[x^2-3]$

No plus C and now differeniate.
• Jan 18th 2011, 07:11 AM
HallsofIvy
There is no need to actually do the integral- this is an exercise in using the "Fundamental Theorem of Calculus":
$\displaystyle \frac{d}{dx}\int_a^x f(t)dt= f(x)$
Here, because the upper limit is $\displaystyle x^2$ rather than x, you need to use the substitution $\displaystyle u= \sqrt{t}= t^{1/2}$. That way the upper limit is $\displaystyle \sqrt{x^2}= x$ (assuming x is positive). Now $\displaystyle du= (1/2)t^{1/2}dt$ and $\displaystyle t= u^2$ so that $\displaystyle dt= \frac{2du}{t^{1/2}}= 2\frac{du}{u}$ and the integral becomes
$\displaystyle 2\int_{\sqrt{3}}^x (2u+ 5)^2\left(\frac{du}{u}\right)$

and, by the Fundamental Theorem of Mathematics, the derivative of that is
$\displaystyle 2\frac{(2x+ 5)^2}{x}$
just by replacing u in the integrand by x.
• Jan 18th 2011, 10:58 AM
dwsmith
Quote:

Originally Posted by HallsofIvy
There is no need to actually do the integral- this is an exercise in using the "Fundamental Theorem of Calculus":
$\displaystyle \frac{d}{dx}\int_a^x f(t)dt= f(x)$
Here, because the upper limit is $\displaystyle x^2$ rather than x, you need to use the substitution $\displaystyle u= \sqrt{t}= t^{1/2}$. That way the upper limit is $\displaystyle \sqrt{x^2}= x$ (assuming x is positive). Now $\displaystyle du= (1/2)t^{1/2}dt$ and $\displaystyle t= u^2$ so that $\displaystyle dt= \frac{2du}{t^{1/2}}= 2\frac{du}{u}$ and the integral becomes
$\displaystyle 2\int_{\sqrt{3}}^x (2u+ 5)^2\left(\frac{du}{u}\right)$

and, by the Fundamental Theorem of Mathematics, the derivative of that is
$\displaystyle 2\frac{(2x+ 5)^2}{x}$
just by replacing u in the integrand by x.

I thought that too but the integral isn't with respect to t.

Is it ok to change dx to dt?
• Jan 18th 2011, 12:45 PM
HallsofIvy
Good point but since, unless the intent was
$\displaystyle \frac{d}{dx}\int_{3}{x^2} (2t^2+ 5)^2 dt$
It really is very close to a trivial question!
• Jan 18th 2011, 12:46 PM
dwsmith
Quote:

Originally Posted by HallsofIvy
Good point but since, unless the intent was
$\displaystyle \frac{d}{dx}\int_{3}{x^2} (2t^2+ 5)^2 dt$
It really is very close to a trivial question!

I wanted to do that first but decided to just go what was there. It probably is supposed to be dt.