integral problem...

**Mathventure** · April 23rd 2009, 07:56 AM

Q: let f(x)=integral of e^(-t^2) limit from sin(x) to cos(x).Then find f'(pi/4)=

**Danny** · April 23rd 2009, 09:52 AM

Originally Posted by Mathventure

Q: let f(x)=integral of e^(-t^2) limit from sin(x) to cos(x).Then find f'(pi/4)=

$f(x) = \int_{\sin x}^{\cos x} e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt + \int_{\sin x}^a e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt - \int_a^{\sin x} e^{-t^2}\,dt$ so

$f'(x) = - e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x$

Then sub. in your value.

**Mathventure** · April 23rd 2009, 05:51 PM

Originally Posted by danny arrigo

$f(x) = \int_{\sin x}^{\cos x} e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt + \int_{\sin x}^a e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt - \int_a^{\sin x} e^{-t^2}\,dt$ so

$f'(x) = e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x$

Then sub. in your value.

after putting value i got zero but options available are
a)sqrt(1/e)
b)-sqrt(2/e)
c)sqrt(2/e)
d)-sqrt(1/e)

**NonCommAlg** · April 23rd 2009, 06:02 PM

Originally Posted by Mathventure

after putting value i got zero but options available are
a)sqrt(1/e)
b)-sqrt(2/e)
c)sqrt(2/e)
d)-sqrt(1/e)

that was just a typo and you should have noticed it. what danny meant was this: $f'(x) = -e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x.$

**Mathventure** · April 23rd 2009, 06:04 PM

Originally Posted by NonCommAlg

that was just a typo and you should have noticed it. what danny meant was this: $f'(x) = -e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x.$

but how he solved integral??

**skeeter** · April 23rd 2009, 06:09 PM

you don't need to solve the integral ... use the 2nd FTC.

$f(x) = \int_{\sin{x}}^{\cos{x}} e^{-t^2} \, dt$

$f'(x) = -e^{-\cos^2{x}} \sin{x} - e^{-\sin^2{x}} \cos{x}$

$f'\left(\frac{\pi}{4}\right) = -e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2} - e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2}<br />$

$f'\left(\frac{\pi}{4}\right) = -\sqrt{\frac{2}{e}}<br />$

**Mathventure** · April 23rd 2009, 06:12 PM

Originally Posted by skeeter

you don't need to solve the integral ... use the 2nd FTC.

$f(x) = \int_{\sin{x}}^{\cos{x}} e^{-t^2} \, dt$

$f'(x) = -e^{-\cos^2{x}} \sin{x} - e^{-\sin^2{x}} \cos{x}$

$f'\left(\frac{\pi}{4}\right) = -e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2} - e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2}<br />$

$f'\left(\frac{\pi}{4}\right) = -\sqrt{\frac{2}{e}}<br />$

i was really silly that i forgot FTC....

**NonCommAlg** · April 23rd 2009, 06:13 PM

Originally Posted by Mathventure

but how he solved integral??

let $\int_a^x f(t) \ dt = g(x),$ where $a$ is a constant. by the fundamental theorem of calculus we have $g'(x) = f(x).$ now if $u(x),v(x)$ are differentiable functions, then:

$h(x)=\int_{u(x)}^{v(x)} f(t) \ dt = g(v(x)) - g(u(x)).$ so by the chain rule: $h'(x)=v'(x)g'(v(x)) - u'(x)g'(u(x))=v'(x)f(v(x))-u'(x)f(u(x)).$

**derfleurer** · April 23rd 2009, 06:18 PM

Edit: It really takes too long typing latex sometimes. =p

Originally Posted by Mathventure

but how he solved integral??

What's to solve?

$f(x) = \int_{sinx}^{cosx} e^{-t^2}dt$

If that's f(x), that little function in the integral looks like f'(x)

$e^{-(cosx)^2}(-sinx) - e^{-(sinx)^2}(cosx)$

**Mathventure** · April 23rd 2009, 06:24 PM

Originally Posted by NonCommAlg

let $\int_a^x f(t) \ dt = g(x),$ where $a$ is a constant. by the fundamental theorem of calculus we have $g'(x) = f(x).$ now if $u(x),v(x)$ are differentiable functions, then:

$h(x)=\int_{u(x)}^{v(x)} f(t) \ dt = g(v(x)) - g(u(x)).$ so by the chain rule: $h'(x)=v'(x)g'(v(x)) - u'(x)g'(u(x))=v'(x)f(v(x))-u'(x)f(u(x)).$

is there any way to solve the value of integral directly??

**NonCommAlg** · April 23rd 2009, 07:02 PM

Originally Posted by Mathventure

is there any way to solve the value of integral directly??

NO!

**Danny** · April 24th 2009, 04:48 AM

Originally Posted by Mathventure

is there any way to solve the value of integral directly??

If you allow error functions then

$f(x) = \frac{\sqrt{\pi}}{2} \left( \text{erf}(\cos x) - \text{erf}(\sin x)\right)$

**Mathventure** · April 24th 2009, 06:24 AM

Originally Posted by NonCommAlg

NO!

then how can we solve double integral of x*e^(y^2) over region R, where R is region bounded by lines x=0,y=1 and the parabola y=x^2.

**mr fantastic** · April 24th 2009, 06:30 AM

Originally Posted by Mathventure

then how can we solve double integral of x*e^(y^2) over region R, where R is region bounded by lines x=0,y=1 and the parabola y=x^2.

By making sure you integrate in an order that makes the problem tractable:

$\int_{y = 0}^1 \int_{x = 0}^{x = + \sqrt{y}} x e^{y^2} \, dx \, dy$ .

There is a world of difference between double integrals and single integrals.

(I can't resist adding that you should have asked danny "How on erf do you do it?" .... Memo to baldeagle: I am the funniest

)

**Danny** · April 24th 2009, 07:42 AM

Originally Posted by mr fantastic

(I can't resist adding that you should have asked danny "How on erf do you do it?" .... Memo to baldeagle: I am the funniest

)

Now that's funny

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