Integration Problem

**Em Yeu Anh** · November 16th 2009, 07:37 PM

There are two parts to this question. First, find $f(4)$ if $\int_0^{x^2}f(t)dt = x\cos{{\pi}x}$

The next part asks to Find $f({\pi}/2)$ if f satisfies the conditions (i) f is positive and continuous, and (ii) the area under the curve y = f(x) from x = 0 to x = a is:
$\frac{a^2}{2} + \frac{a}{2}sina+\frac{\pi}{2}cosa$

I don't think I'm doing this correctly. Same error for both parts, I tried applying the fundamental thereom but end up having to deal with f(0) which is unknown.
For part one I so far have:
$\int_0^{x^2}f(t)dt = x\cos{{\pi}x} = f(x^2) - f(0)$

$2cos2{\pi} = f(4) - f(0)$

$2 + f(0) = f(4)$ ??

Anyone mind guiding me in the correct direction? Thanks! Know it's a lot to read.

**mr fantastic** · November 16th 2009, 07:54 PM

Originally Posted by Em Yeu Anh

There are two parts to this question. First, find $f(4)$ if $\int_0^{x^2}f(t)dt = x\cos{{\pi}x}$

[snip]
For part one I so far have:
$\int_0^{x^2}f(t)dt = x\cos{{\pi}x} = f(x^2) - f(0)$

$2cos2{\pi} = f(4) - f(0)$

$2 + f(0) = f(4)$ ??

[snip]

I have time to do the first. You are on the wrong track. You need to differentiate both sides with respect to x (for the left hand side, use the chain rule and the Fundamental Theorem of Calculus). This will give you f(x).

**Em Yeu Anh** · November 16th 2009, 08:04 PM

Thank you,
alright I'll try that again:
$\frac{d}{dx}\int_0^{x^2}f(t)dt = -{\pi}xsin{\pi}x + cos{\pi}x$

$f(x^2)*2x = -{\pi}xsin{\pi}x + cos{\pi}x$

$f(x^2) = \frac{-{\pi}sin{\pi}x}{2} + \frac{cos{\pi}x}{2x}$

$f(4) = \frac{1}{4}$

Assuming the same method will be applicable to part 2,

$\frac{d}{dx}\int_0^af(x)dx = a + \frac{acosa}{2} - \frac{{\pi}sina}{2}$

$f({\pi}/2) = \frac{\pi}{2} + \frac{({\pi}/2)cos({\pi}/2)}{2} - \frac{{\pi}sin({\pi}/2)}{2}$

$f({\pi}/2) = 0$

**mr fantastic** · November 17th 2009, 01:47 AM

Originally Posted by Em Yeu Anh

Thank you,
alright I'll try that again:
$\frac{d}{dx}\int_0^{x^2}f(t)dt = -{\pi}xsin{\pi}x + cos{\pi}x$

$f(x^2)*2x = -{\pi}xsin{\pi}x + cos{\pi}x$

$f(x^2) = \frac{-{\pi}sin{\pi}x}{2} + \frac{cos{\pi}x}{2x}$

$f(4) = \frac{1}{4}$

Assuming the same method will be applicable to part 2,

$\frac{d}{d{\color{red}a}}\int_0^af(x)dx = a + \frac{acosa}{2} - \frac{{\pi}sina}{2}$ Mr F says: 1. Note the red a, and 2. Your derivative of the right hand side with respect to a is missing a term.

$f({\pi}/2) = \frac{\pi}{2} + \frac{({\pi}/2)cos({\pi}/2)}{2} - \frac{{\pi}sin({\pi}/2)}{2}$

$f({\pi}/2) = 0$

..

**Em Yeu Anh** · November 17th 2009, 05:07 PM

Not sure what I missed; Does $\frac{a}{2}sin(a)$ require a product rule?

**mr fantastic** · November 17th 2009, 05:36 PM

Originally Posted by Em Yeu Anh

Not sure what I missed; Does $\frac{a}{2}sin(a)$ require a product rule?

Well, a is the variable you're differentiating with respect to. And what you're differentiating is a product of two functions of a. So what do you think ....?

**Em Yeu Anh** · November 17th 2009, 06:03 PM

Of course, another problem from my error of thinking that it was differentiated with respect to x, I misinterpreted "a" as just being a constant.

Redo once again:
$<br /> \frac{d}{da}\int_0^af(x)dx = a + \frac{a}{2}cos(a) + \frac{sin(a)}{2} - \frac{\pi}{2}sin(a)$

$f(\frac{\pi}{2}) = \frac{1}{2}$

Hopefully no more mistakes this time.

**mr fantastic** · November 17th 2009, 06:07 PM

Originally Posted by Em Yeu Anh

Of course, another problem from my error of thinking that it was differentiated with respect to x, I misinterpreted "a" as just being a constant.

Redo once again:
$<br /> \frac{d}{da}\int_0^af(x)dx = a + \frac{a}{2}cos(a) + \frac{sin(a)}{2} - \frac{\pi}{2}sin(a)$

$f(\frac{\pi}{2}) = \frac{1}{2}$

Hopefully no more mistakes this time.

That looks OK.

**Em Yeu Anh** · November 17th 2009, 06:14 PM

Excellent! Thanks, I wouldn't have had a chance at getting this correct without your help.

Thread: Integration Problem

LinkBack

Thread Tools

Display