1. ## integral problem...

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

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

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

3. Originally Posted by danny arrigo
$\displaystyle f(x) = \int_{\sin x}^{\cos x} e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt + \int_{\sin x}^a e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt - \int_a^{\sin x} e^{-t^2}\,dt$ so

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

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)

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

5. Originally Posted by NonCommAlg
that was just a typo and you should have noticed it. what danny meant was this: $\displaystyle f'(x) = -e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x.$
but how he solved integral??

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

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

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

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

$\displaystyle f'\left(\frac{\pi}{4}\right) = -\sqrt{\frac{2}{e}}$

7. Originally Posted by skeeter
you don't need to solve the integral ... use the 2nd FTC.

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

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

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

$\displaystyle f'\left(\frac{\pi}{4}\right) = -\sqrt{\frac{2}{e}}$
i was really silly that i forgot FTC....

8. Originally Posted by Mathventure

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

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

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

Originally Posted by Mathventure
but how he solved integral??
What's to solve?

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

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

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

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

$\displaystyle h(x)=\int_{u(x)}^{v(x)} f(t) \ dt = g(v(x)) - g(u(x)).$ so by the chain rule: $\displaystyle h'(x)=v'(x)g'(v(x)) - u'(x)g'(u(x))=v'(x)f(v(x))-u'(x)f(u(x)).$
is there any way to solve the value of integral directly??

11. Originally Posted by Mathventure
is there any way to solve the value of integral directly??
NO!

12. Originally Posted by Mathventure
is there any way to solve the value of integral directly??
If you allow error functions then

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

13. ## integral problem...

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.

14. 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:

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

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 )

15. 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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