antiderivative

candy101

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Oct 4, 2009
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43
i need help with this problem
∫√x + 1/ 2√x dx

this is the answer but i don't understand the steps,


∫√x dx+∫ 1/ 2√x dx ==> i understand this part and i know that the derivative of √x is 1/2 , but everything else i don't understand

∫x^1/2 +1/2 ∫ x-1/2 dx

=2/3x^3/2 +1/2 * 2x1/2 +C

= 2/3x^3/2 +x1/2+c


can you please explain the steps


thank you.
 
i need help with this problem
∫√x + 1/ 2√x dx
Your formatting is ambigous. Is the integral either of the following?

. . . . .\(\displaystyle \mbox{a) }\, \displaystyle{\int\, \left[\, \sqrt{x}\, +\, \frac{1}{2\sqrt{x}}\,\right]\, dx}\)

. . . . .\(\displaystyle \mbox{b) }\,\displaystyle{\int\, \left[\,\sqrt{x}\, +\, \frac{1}{2}\sqrt{x}\,\right]\, dx}\)

i know that the derivative of √x is 1/2...
No. The derivative of (1/2)x is 1/2. The derivative of the square root of x is different.

∫x^1/2 +1/2 ∫ x-1/2 dx
How did the square root of x turn into the difference of x and 1/2? :shock:
 
When writing down equations or math symbols, it is important to make sure what you write is clear. I know, I've been accused of taking 'short cuts' myself. Anyway what it looks like to me is the problem is, what is the value of
∫( √x + 1/(2√x) ) dx
and your answer of
(2/3)x3/2 +x1/2+c
or, as you wrote it (with a slight correction),
(2/3)x^(3/2) +x^(1/2)+c
is correct.

Possibly you had a different question and I misunderstood.
 
Code:
Your formatting is ambigous. Is the integral either of the following?

. . . . .\(\displaystyle \mbox{a) }\, \displaystyle{\int\, \left[\, \sqrt{x}\, +\, \frac{1}{2\sqrt{x}}\,\right]\, dx}\)

. . . . .\(\displaystyle \mbox{b) }\,\displaystyle{\int\, \left[\,\sqrt{x}\, +\, \frac{1}{2}\sqrt{x}\,\right]\, dx}\)


No. The derivative of (1/2)x is 1/2. The derivative of the square root of x is different.


How did the square root of x turn into the difference of x and 1/2? :shock:

I am sorry I don't know how to get that formula that you got. It is the first one A


Thank you
 
Your formatting is ambigous. Is the integral either of the following?

. . . . .\(\displaystyle \mbox{a) }\, \displaystyle{\int\, \left[\, \sqrt{x}\, +\, \frac{1}{2\sqrt{x}}\,\right]\, dx}\)
I am sorry I don't know how to get that formula that you got. It is the first one A
Um... I don't know what you mean by "that formula that got". Did you maybe mean the "formatting" that I "did"? If so, then I understand your reply to mean that the integral, in your assignment, is formatted as in my option (a). Okay....

So your integral is this:

. . . . .\(\displaystyle \mbox{a) }\, \displaystyle{\int\, \left[\, \sqrt{x}\, +\, \frac{1}{2\sqrt{x}}\,\right]\, dx}\)

It is often helpful, when integrating with radicals, to convert radical form to exponential form. In this case, the square roots become one-half powers:

. . . . .\(\displaystyle \mbox{a) }\, \displaystyle{\int\, \left[\, x^{1/2}\, +\, \frac{1}{2 x^{1/2}}\,\right]\, dx}\)

Then, of course, apply what you learned back in algebra to get that "x" in the second fraction up on top, with a negative exponent. Then apply the Power Rule for integrating, and you're pretty much done! ;)
 
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