How to Approach these problems..

[jetter2]

So we are doing integration, and I am somewhat confused on how to approach a problem that has a variable as an upper limit of integration. Also, I tried bringing sqrtX up from the bottom with a -1 exponet and it still didn't work out..I'm completely flustered

 

[galactus]

The first one involves the second fundamental theorem of calcarooney.

They are asking for the derivative in the first one, not the integration.

\(\displaystyle \frac{d}{dx}\int_{h(x)}^{g(x)}f(u)du=f(g(x))g'(x)-f(h(x))h'(x)\)

The second one:

\(\displaystyle \int_{1}^{9}\frac{x-1}{\sqrt{x}}dx\)

Rewrite:

\(\displaystyle \int_{1}^{9}x^{\frac{1}{2}}dx-\int_{1}^{9}x^{\frac{-1}{2}}dx\)

Now, it's easier, huh?.

 

[jetter2]

You move x^.5 from the denominator tot he numerator, then you end up with (x-1)(x^-.5) right?

So if you foil it out you get x^.5 - x^-.5

(2/3)x^1.5 - (2)x^.5 | @ 9 - @1...I'm not following the progression of how to solve it

 

[lookagain]

\(\displaystyle \int_{1}^{9}x^{\frac{1}{2}}dx\ - \ \int_{1}^{9}x^{\frac{-1}{2}}dx\)

The typo in this line is shown fixed in the above quote box.

 

[lookagain]

You move x^.5 from the denominator to the numerator, then you end up with (x-1)(x^-.5) right?

So if you [distribute] the x^(-.5), you get x^.5 - x^-.5

(2/3)x^1.5 - (2)x^.5 | @ 9 - @1 \(\displaystyle \ .\ . \ . \ \ I'm \ not \ following \ the \ progression \ of \ how \ to \ solve \ it.\)

\(\displaystyle \bigg (\frac{2}{3}\sqrt{x^3} - 2 \sqrt{x} \bigg ) \bigg|_1^9\)


For nonnegative \(\displaystyle x\)-values here, this is also equal to:


\(\displaystyle \bigg [\frac{2}{3} (\sqrt{x})^3 - 2 \sqrt{x} \bigg ] \bigg|_1^9\)


Plug \(\displaystyle \ x \ = \ 9\) into this and then subtract the result that you get when you plug in \(\displaystyle \ x \ = \ 1.\)

 

[galactus]

\(\displaystyle \int_{1}^{9}x^{\frac{1}{2}}dx\ - \ \int_{1}^{9}x^{\frac{-1}{2}}dx\)

The typo in this line is shown fixed in the above quote box.

Thanks for that, lookagain. That typo go by me last night.