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Home >All Categories > Science & Mathematics >Mathematics >Open Question PC Member since:April 13, 2010Total points:107 (Level 1)I think it is self explanatory, I am just forgetting how to deal with the term that is inside the trigonometric function. obviously the integral of sin is -cos, but I forget the rule for the inside termAnswer Question

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Let u = x^2 + 2x^-1; then du = (2x - 2x^-2) dx, or (x - x^-2) dx = 1/2 du.

The integral becomes: (1/2)(integral) du/u = (1/2)ln(x^2 + 2x^-1) + C

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Let x^4 = u
4 x^3 dx = du
x^3 dx = (1/4) du

Integral x^3e^x^4 dx = (1/4) integral e^u du
= e^u / 4 + C
= e^x^4 / 4 + C

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Jump discontinuities are infinitesimally thin. They contribute NOTHING to the integral.

Plus, the limits of integration have no distinction between "=" and "<"...nor do they have any distinction between "=" and ">".

If you really want to be more comfortable, you can do the integration from 0 to 1 of f(x) = 3, and THEN add it to the integral of f(x) = 6 for integrating from 1+h to 2.

Then take the limit as h approaches zero. What do you get? You get exactly what you would get if you never even thought about what h is.

Integral of f(x)=3 from 0 to 1 = 3
Integral of f(x)=6 from 1+h to 2 = 6*(2 - (1 + h)) = 6*(1 - h)

Add them up:
3 + 6*(1 - h)

Now, take the limit as h approaches zero. What do you get? Why 3+6 of course, which equals 9. And that is exactly what you would get if you never thought about h.

To show you, suppose:
h=0.1...then 3+6*(1-h) = 8.4
h=0.01...then 3+6*(1-h) = 8.94
h=0.001...then 3+6*(1-h) = 8.994
h=0.0001...then 3+6*(1-h) = 8.9994

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I = ? [ v(3-5x) / (2x) ] dx ......................... (1)
_______________________

Let : x = (3/5) sin² F.

Then : dx = (3/5) 2 sin F cos F dF.

Also : v(3-5x) = v3. cos F.
________________________

From (1), then,

I = ? [ ( v3. cos F ) / ( 2(3/5) sin² F ) ] · (3/5) 2 sin F cos F dF

= v3. ? ( cos² F / sin F ) dF

= v3. ? [ ( 1 - sin² F ) / sin F ] dF

= v3. ? ( csc F - sin F ) dF

= v3. { ln | csc F - cot F | - ( - cos F ) } + C

= v3. { ln | csc F - cot F | + cos F } + C

= v3. ln | (v(5x/3) - v(3-5x)/v(5x) | + v(3-5x) + C .......... Ans.
________________________________

Happy To Help !
________________________________

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===Sister's Integral Formulas to Use===

?Substitution Method
?FTC: ?(x = a,b) f(x) dx = F(x) | x = a,b = F(b) - F(a)

* * * * * *

?(x = 0,1) (e^(x) + 1)/(e^(x) + x) dx

Let u = e^(x) + x. Then,

du/dx = e^(x) + 1
du = (e^(x) + 1) dx [Rearrange the terms to make (e^(x) + 1) dx the subject]

Note that you can substitute the limits for u function!

u(0) = e^(0) + 0
==> 1

u(1) = e^1 + 1
==> e

So:

?(u = 1,e) ((e^(x) + 1) dx)/u
==> ?(u = 1,e) du/u
==> ln|u| | u = 1,e

Finally, by FTC:

ln(e) - ln(1)
==> 1 - 0
==> 1

I hope this helps!

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x = (3y-y^2)^(1/2)

Then also by rotating the curve about the y-axis.

I got 18piy/12y-4y^2, for about the y-axis, but webassign keeps telling me I'm wrong. Any help would be very much appreciated!!!

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Please please show your steps. And I'm confused as to why this would be called an indefinite integral as opposed to just a regular (definite?) integral. I don't know if that's necessarily relevant to the solution, but I'd like to know either way.

THANK YOU!!

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Recall that 4^-2=1/4^2

So e^-x = 1/e^x

thus e^x^e^-x=e^x/e^x = 1

so you have integral of dx from 0 to 1 which has the anti-derivative of just x

so x evaluated at the limits is
1 - - 0 = 1

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